\displaystyle \textbf{Question 1: } \text{Form the differential equation representing the family of curves } y=A\cos(x+B),\text{ where }A\text{ and }B\text{ are parameters.}\hspace{4.0cm} \text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is}
\displaystyle y=A\cos(x+B)\ \ \ ...(i)
\displaystyle \text{This equation contains two arbitrary constants. So, let us differentiate it two times to} \\ \text{obtain a differential equation of second order.}
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle \frac{dy}{dx}=-A\sin(x+B)\ \ \ ...(ii)
\displaystyle \text{Differentiating (ii) with respect to }x,\text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=-A\cos(x+B)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=-y\ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}+y=0,\text{ which is the required differential equation of the given family of curves.}

\displaystyle \textbf{Question 2: } \text{Form the differential equation corresponding to } y^{2}=a(b-x)(b+x)\text{ by eliminating parameters }a\text{ and }b.\hspace{4.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is}
\displaystyle y^{2}=a\left(b^{2}-x^{2}\right)\ \ \ ...(i)
\displaystyle \text{Clearly, there are two arbitrary constants in this equation. So, we shall differentiate it} \\ \text{two times to get a differential equation of second order.}
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle 2y\frac{dy}{dx}=-2ax\Rightarrow y\frac{dy}{dx}=-ax\ \ \ ...(ii)
\displaystyle \text{Differentiating (ii) with respect to }x,\text{ we get}
\displaystyle y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=-a \Rightarrow a=-\left\{y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}\right\}\ \ \ ...(iii)
\displaystyle \text{Substituting the value of }a\text{ obtained from (iii) in (ii), we get}
\displaystyle x\left\{y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}\right\} =y\frac{dy}{dx},\text{ which is the required differential equation.}

\displaystyle\textbf{Question 3: } \text{Find the differential equation of all the circles in the first quadrant which} \\ \text{touch the coordinate axes.}\hspace{4.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of circles in the first quadrant which touch the coordinate axes is}
\displaystyle (x-a)^{2}+(y-a)^{2}=a^{2}\ \ \ ...(i)
\displaystyle \text{where }a\text{ is a parameter.}
\displaystyle \text{This equation contains one arbitrary constant, so we shall differentiate it once only to} \\ \text{get a differential equation of first order.}
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle 2(x-a)+2(y-a)\frac{dy}{dx}=0
\displaystyle \Rightarrow x-a+(y-a)\frac{dy}{dx}=0
\displaystyle \Rightarrow a=\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}
\displaystyle \Rightarrow a=\frac{x+py}{1+p},\text{ where }p=\frac{dy}{dx}
\displaystyle \text{Substituting the value of }a\text{ in (i), we get}
\displaystyle \left(x-\frac{x+py}{1+p}\right)^{2} +\left(y-\frac{x+py}{1+p}\right)^{2} =\left(\frac{x+py}{1+p}\right)^{2}
\displaystyle \Rightarrow (xp-py)^{2}+(y-x)^{2}=(x+py)^{2}
\displaystyle \Rightarrow (x-y)^{2}p^{2}+(x-y)^{2}=(x+py)^{2}
\displaystyle \Rightarrow (x-y)^{2}(p^{2}+1)=(x+py)^{2}
\displaystyle \Rightarrow (x-y)^{2}\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\} =\left(x+y\frac{dy}{dx}\right)^{2},\text{ which is the required differential equation.}

\displaystyle \textbf{Question 4: } \text{Form the differential equation of family of parabolas having vertex at the} \\ \text{origin and axis along positive }y\text{-axis.}\hspace{4.0cm} \text{[CBSE 2010, 11]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of parabolas having vertex at the origin and axis along positive }y\text{-axis is}
\displaystyle x^{2}=4ay,\text{ where }a\text{ is a parameter.}\ \ \ ...(i)
\displaystyle \text{This is a one parameter family of curves. So, we differentiate it once only.}
\displaystyle \text{Differentiating with respect to }x,\text{ we get}
\displaystyle 2x=4a\frac{dy}{dx}\Rightarrow a=\frac{x}{2\frac{dy}{dx}}
\displaystyle \text{Substituting the value of }a\text{ in (i), we get}
\displaystyle x^{2}=4\times\frac{x}{2\left(\frac{dy}{dx}\right)}\times y \Rightarrow x\frac{dy}{dx}=2y,\text{ which is the required differential equation.}

\displaystyle \textbf{Question 5: } \text{Represent the following family of curves by forming the corresponding} \\ \text{differential equations }(a,b\text{ are parameters}):
\displaystyle (i)\ \frac{x}{a}+\frac{y}{b}=1\ \ \ \ \ (ii)\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\ \ \ \ \ (iii)\ (y-b)^{2}=4(x-a)\hspace{2.0cm} \text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of curves is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1\ \ \ ...(i)
\displaystyle \text{where }a,b\text{ are parameters.}
\displaystyle \text{It is a two parameter family of curves. So, we will differentiate it twice with respect to }x.
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle \frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\ \ \ ...(ii)
\displaystyle \text{Differentiating (ii) with respect to }x,\text{ we get}
\displaystyle \frac{1}{b}\frac{d^{2}y}{dx^{2}}=0\Rightarrow \frac{d^{2}y}{dx^{2}}=0,\text{ which is the required differential equation.}
\displaystyle (ii)\ \text{The equation of the family of curves is}
\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\ \ \ ...(i)
\displaystyle \text{It is a two parameter family of curves. So, we will differentiate it twice to obtain the} \\ \text{differential equation.}
\displaystyle \text{Differentiating with respect to }x,\text{ we get}
\displaystyle \frac{x}{a^{2}}+\frac{y}{b^{2}}\frac{dy}{dx}=0\ \ \ ...(ii)
\displaystyle \text{Differentiating (ii) with respect to }x,\text{ we get}
\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}\left(\frac{dy}{dx}\right)^{2} +\frac{y}{b^{2}}\frac{d^{2}y}{dx^{2}}=0
\displaystyle \text{Multiplying both sides by }xy,\text{ we get}
\displaystyle \frac{x}{a^{2}}+\frac{x}{b^{2}}\left(\frac{dy}{dx}\right)^{2} +\frac{xy}{b^{2}}\frac{d^{2}y}{dx^{2}}=0\ \ \ ...(iv)
\displaystyle \text{Subtracting (ii) from (iv), we get}
\displaystyle \frac{1}{b^{2}}\left\{x\left(\frac{dy}{dx}\right)^{2} +xy\frac{d^{2}y}{dx^{2}}-y\frac{dy}{dx}\right\}=0
\displaystyle \Rightarrow x\left(\frac{dy}{dx}\right)^{2} +xy\frac{d^{2}y}{dx^{2}}-y\frac{dy}{dx}=0,\text{ which is the required differential equation.}
\displaystyle (iii)\ \text{The equation of the family of curves is}
\displaystyle (y-b)^{2}=4(x-a)\ \ \ ...(i)
\displaystyle \text{It is a two parameter family of curves. So, we will differentiate it twice with respect to }x.
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle 2(y-b)\frac{dy}{dx}=4\Rightarrow (y-b)\frac{dy}{dx}=2\ \ \ ...(ii)
\displaystyle \text{Differentiating with respect to }x,\text{ we get}
\displaystyle (y-b)\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0\ \ \ ...(iii)
\displaystyle \text{From (ii), we get }y-b=\frac{2}{\frac{dy}{dx}}.
\displaystyle \text{Substituting this value of }y-b\text{ in (iii), we get}
\displaystyle \frac{2}{\frac{dy}{dx}}\times\frac{d^{2}y}{dx^{2}} +\left(\frac{dy}{dx}\right)^{2}=0 \Rightarrow 2\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{3}=0,\text{ which is the required differential equation.}

\displaystyle \textbf{Question 6: } \text{Obtain the differential equation of all circles of radius }r. \text{[CBSE 2010, 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{ The equation of the family of circles of radius }r\text{ is}
\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}\ \ \ ...(i)
\displaystyle \text{where }a\text{ and }b\text{ are parameters.}
\displaystyle \text{Clearly equation (i) contains two arbitrary constants. So, let us differentiate it two times} \\ \text{ with respect to }x.
\displaystyle \text{Differentiating (i) with respect to }x,\text{ we get}
\displaystyle 2(x-a)+2(y-b)\frac{dy}{dx}=0
\displaystyle \Rightarrow (x-a)+(y-b)\frac{dy}{dx}=0\ \ \ ...(ii)
\displaystyle \text{Differentiating (ii) with respect to }x,\text{ we get}
\displaystyle 1+(y-b)\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0\ \ \ ...(iii)
\displaystyle \Rightarrow y-b=-\frac{1+\left(\frac{dy}{dx}\right)^{2}}{\frac{d^{2}y}{dx^{2}}}\ \ \ ...(iv)
\displaystyle \text{Putting this value of }(y-b)\text{ in (ii), we obtain}
\displaystyle x-a=\frac{\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}\frac{dy}{dx}}{\frac{d^{2}y}{dx^{2}}}\ \ \ ...(v)
\displaystyle \text{Substituting the values of }(x-a)\text{ and }(y-b)\text{ in (i), we get}
\displaystyle \frac{\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{2}\left(\frac{dy}{dx}\right)^{2}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{2}} +\frac{\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{2}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{2}} =r^{2}
\displaystyle \Rightarrow \left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{3} =r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}
\displaystyle \text{This is the required differential equation.}

\displaystyle \textbf{Question 7: } \text{Solve:}
\displaystyle (i)\ \sec^{2}x\tan y\,dx+\sec^{2}y\tan x\,dy=0\hspace{4.0cm} [\text{CBSE 2007}]
\displaystyle (ii)\ e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0\hspace{4.0cm} [\text{CBSE 2012, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{ (i) We have,}
\displaystyle \sec^{2}x\tan y\,dx+\sec^{2}y\tan x\,dy=0
\displaystyle \Rightarrow \sec^{2}x\tan y\,dx=-\sec^{2}y\tan x\,dy
\displaystyle \Rightarrow \frac{\sec^{2}x}{\tan x}\,dx=-\frac{\sec^{2}y}{\tan y}\,dy
\displaystyle \Rightarrow \int\frac{\sec^{2}x}{\tan x}\,dx=-\int\frac{\sec^{2}y}{\tan y}\,dy\ \ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log\left|\tan x\right|=-\log\left|\tan y\right|+\log C
\displaystyle \Rightarrow \log\left|(\tan x)(\tan y)\right|=\log C
\displaystyle \Rightarrow |\tan x\tan y|=C,\text{ which is the solution of the given differential equation.}
\displaystyle \text{(ii) We are given that}
\displaystyle e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0
\displaystyle \Rightarrow e^{x}\sqrt{1-y^{2}}\,dx=-\frac{y}{x}\,dy
\displaystyle \Rightarrow xe^{x}\,dx=-\frac{y}{\sqrt{1-y^{2}}}\,dy
\displaystyle \Rightarrow \int xe^{x}\,dx=-\int\frac{y}{\sqrt{1-y^{2}}}\,dy\ \ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow xe^{x}-\int e^{x}\,dx=\frac{1}{2}\int\frac{dt}{\sqrt{t}},\text{ where }t=1-y^{2}
\displaystyle \Rightarrow xe^{x}-e^{x}=\frac{1}{2}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C
\displaystyle \Rightarrow xe^{x}-e^{x}=\sqrt{t}+C
\displaystyle \Rightarrow xe^{x}-e^{x}=\sqrt{1-y^{2}}+C\text{ is the required solution.}

\displaystyle \textbf{Question 8: } \text{Solve the differential equation:} \\ (1+e^{2x})\,dy+(1+y^{2})e^{x}\,dx=0\text{ given that when }x=0,\ y=1.\hspace{1.0cm} \text{[CBSE 2004, 05]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle (1+e^{2x})\,dy+(1+y^{2})e^{x}\,dx=0
\displaystyle \Rightarrow (1+e^{2x})\,dy=-(1+y^{2})e^{x}\,dx
\displaystyle \Rightarrow \frac{1}{1+y^{2}}\,dy=-\frac{e^{x}}{1+e^{2x}}\,dx
\displaystyle \Rightarrow \int\frac{1}{1+y^{2}}\,dy=-\int\frac{e^{x}}{1+e^{2x}}\,dx\ \ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \int\frac{1}{1+y^{2}}\,dy=-\int\frac{dt}{1+t^{2}},\text{ where }t=e^{x}
\displaystyle \Rightarrow \tan^{-1}y=-\tan^{-1}(t)+C
\displaystyle \Rightarrow \tan^{-1}y=-\tan^{-1}(e^{x})+C\ \ \ ...(i)
\displaystyle \text{It is given that }y=1,\text{ when }x=0.\text{ So, putting }x=0,y=1\text{ in (i), we get}
\displaystyle \tan^{-1}1=-\tan^{-1}(e^{0})+C\Rightarrow \frac{\pi}{4}=-\frac{\pi}{4}+C\Rightarrow C=\frac{\pi}{2}
\displaystyle \text{Putting }C=\frac{\pi}{2}\text{ in (i), we obtain}
\displaystyle \tan^{-1}y=-\tan^{-1}(e^{x})+\frac{\pi}{2}
\displaystyle \Rightarrow \tan^{-1}y+\tan^{-1}(e^{x})=\frac{\pi}{2}
\displaystyle \Rightarrow \tan^{-1}y=\frac{\pi}{2}-\tan^{-1}(e^{x})
\displaystyle \Rightarrow \tan^{-1}y=\cot^{-1}(e^{x})
\displaystyle \Rightarrow \tan^{-1}y=\tan^{-1}\left(\frac{1}{e^{x}}\right)\Rightarrow y=\frac{1}{e^{x}},\text{ which is the required solution.}

\displaystyle \textbf{Question 9: } \text{Solve the differential equation:} \\ (1+y^{2})(1+\log x)\,dx+x\,dy=0\text{ given that when }x=1,y=1.\hspace{1.0cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle (1+y^{2})(1+\log x)\,dx+x\,dy=0
\displaystyle \Rightarrow (1+\log x)(1+y^{2})\,dx=-x\,dy
\displaystyle \Rightarrow \frac{1+\log x}{x}\,dx=-\frac{1}{1+y^{2}}\,dy
\displaystyle \Rightarrow \int\frac{1+\log x}{x}\,dx=-\int\frac{1}{1+y^{2}}\,dy\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \int t\,dt=-\int\frac{1}{1+y^{2}}\,dy,\text{ where }1+\log x=t
\displaystyle \Rightarrow \frac{t^{2}}{2}=-\tan^{-1}y+C
\displaystyle \Rightarrow \frac{1}{2}(1+\log x)^{2}=-\tan^{-1}y+C\ \ \ ...(i)
\displaystyle \text{It is given that when }x=1,y=1.\text{ So, putting }x=1,y=1\text{ in (i), we obtain}
\displaystyle \frac{1}{2}(1+\log 1)^{2}=-\tan^{-1}1+C\Rightarrow \frac{1}{2}=-\frac{\pi}{4}+C\Rightarrow C=\frac{1}{2}+\frac{\pi}{4}
\displaystyle \text{Putting }C=\frac{1}{2}+\frac{\pi}{4}\text{ in (i), we obtain}
\displaystyle \frac{1}{2}(1+\log x)^{2}=-\tan^{-1}y+\frac{1}{2}+\frac{\pi}{4}
\displaystyle \Rightarrow \tan^{-1}y=\frac{\pi}{4}+\frac{1}{2}-\frac{1}{2}(1+\log x)^{2}
\displaystyle \Rightarrow y=\tan\left\{\frac{\pi}{4}+\frac{1}{2}-\frac{1}{2}(1+\log x)^{2}\right\},\text{ which is the solution of the given differential equation.}

\displaystyle \textbf{Question 10: } \text{Solve the differential equation:} \\ x(1+y^{2})\,dx-y(1+x^{2})\,dy=0,\text{ given that }y=0,\text{ when }x=1.\hspace{1.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle x(1+y^{2})\,dx-y(1+x^{2})\,dy=0
\displaystyle \Rightarrow x(1+y^{2})\,dx=y(1+x^{2})\,dy
\displaystyle \Rightarrow \frac{x}{1+x^{2}}\,dx=\frac{y}{1+y^{2}}\,dy
\displaystyle \Rightarrow \frac{2x}{1+x^{2}}\,dx=\frac{2y}{1+y^{2}}\,dy
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{2x}{1+x^{2}}\,dx=\int\frac{2y}{1+y^{2}}\,dy
\displaystyle \Rightarrow \log|1+x^{2}|=\log|1+y^{2}|+\log C
\displaystyle \Rightarrow \log\left|\frac{1+x^{2}}{1+y^{2}}\right|=\log C
\displaystyle \Rightarrow \frac{1+x^{2}}{1+y^{2}}=C\ \ \ ...(i)
\displaystyle \text{It is given that when }x=1,y=0.\text{ So, putting }x=1,y=0\text{ in (i), we get}
\displaystyle (1+1)=(1+0)C\Rightarrow C=2
\displaystyle \text{Putting }C=2\text{ in (i), we get}
\displaystyle (1+x^{2})=2(1+y^{2}),\text{ which is the required solution.}

\displaystyle \textbf{Question 11: } \text{Solve the following differential equations:}
\displaystyle (i)\ \frac{dy}{dx}=1+x+y+xy\hspace{4.0cm} [\text{CBSE 2014}]
\displaystyle (ii)\ y-x\frac{dy}{dx}=a\left(y^{2}+\frac{dy}{dx}\right)\hspace{4.0cm} [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) We are given that}
\displaystyle \frac{dy}{dx}=1+x+y+xy
\displaystyle \Rightarrow \frac{dy}{dx}=(1+x)+y(1+x)
\displaystyle \Rightarrow \frac{dy}{dx}=(1+x)(1+y)
\displaystyle \Rightarrow \frac{1}{1+y}\,dy=(1+x)\,dx
\displaystyle \Rightarrow \int\frac{1}{1+y}\,dy=\int(1+x)\,dx\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log|1+y|=x+\frac{x^{2}}{2}+C,\text{ which is the general solution of the given differential equation.}
\displaystyle \text{(ii) The given differential equation is}
\displaystyle y-x\frac{dy}{dx}=a\left(y^{2}+\frac{dy}{dx}\right)
\displaystyle \Rightarrow y-ay^{2}=\frac{dy}{dx}(a+x)
\displaystyle \Rightarrow (y-ay^{2})\,dx=(a+x)\,dy
\displaystyle \Rightarrow \frac{dx}{a+x}=\frac{dy}{y-ay^{2}}
\displaystyle \Rightarrow \int\frac{1}{a+x}\,dx=\int\frac{1}{y-ay^{2}}\,dy\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \int\frac{1}{a+x}\,dx=\int\left(\frac{1}{y}+\frac{a}{1-ay}\right)\,dy\ \ \ [\text{By using partial fractions on RHS}]
\displaystyle \Rightarrow \log|x+a|=\log|y|-\log|1-ay|+\log C
\displaystyle \Rightarrow \log\left|\frac{(x+a)(1-ay)}{y}\right|=\log C
\displaystyle \Rightarrow \frac{(x+a)(1-ay)}{y}=C
\displaystyle \Rightarrow (x+a)(1-ay)=Cy,\text{ which is the general solution of the given differential equation.}

\displaystyle \textbf{Question 12: } \text{Solve:}
\displaystyle (i)\ (x^{2}-yx^{2})\,dy+(y^{2}+x^{2}y^{2})\,dx=0\hspace{4.0cm} [\text{CBSE 2012, 2014}]
\displaystyle (ii)\ 3e^{x}\tan y\,dx+(1-e^{x})\sec^{2}y\,dy=0\hspace{4.0cm} [\text{CBSE 2011, 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{ (i) The given differential equation is}
\displaystyle x^{2}(1-y)\,dy+y^{2}(1+x^{2})\,dx=0
\displaystyle \Rightarrow x^{2}(1-y)\,dy=-y^{2}(1+x^{2})\,dx
\displaystyle \Rightarrow \frac{1-y}{y^{2}}\,dy=-\left(\frac{1+x^{2}}{x^{2}}\right)\,dx,\ \text{if }x,y\neq 0
\displaystyle \Rightarrow \left(\frac{1}{y}-\frac{1}{y^{2}}\right)\,dy=\left(\frac{1}{x^{2}}+1\right)\,dx
\displaystyle \Rightarrow \int\left(\frac{1}{y}-\frac{1}{y^{2}}\right)\,dy=\int\left(\frac{1}{x^{2}}+1\right)\,dx\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log|y|+\frac{1}{y}=-\frac{1}{x}+x+C,\text{ which is the general solution of the differential equation.}
\displaystyle \text{(ii) We are given that}
\displaystyle 3e^{x}\tan y\,dx+(1-e^{x})\sec^{2}y\,dy=0
\displaystyle \Rightarrow 3e^{x}\tan y\,dx=-(1-e^{x})\sec^{2}y\,dy
\displaystyle \Rightarrow 3\frac{e^{x}}{1-e^{x}}\,dx=-\frac{\sec^{2}y}{\tan y}\,dy
\displaystyle \Rightarrow 3\int\frac{e^{x}}{1-e^{x}}\,dx=-\int\frac{\sec^{2}y}{\tan y}\,dy\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow 3\int\frac{e^{x}}{e^{x}-1}\,dx=\int\frac{\sec^{2}y}{\tan y}\,dy
\displaystyle \Rightarrow 3\log|e^{x}-1|=\log|\tan y|+\log C
\displaystyle \Rightarrow \log\left|\frac{(e^{x}-1)^{3}}{\tan y}\right|=\log C
\displaystyle \Rightarrow \frac{(e^{x}-1)^{3}}{\tan y}=C
\displaystyle \Rightarrow (e^{x}-1)^{3}=C\tan y,\text{ which the general solution of the given differential equation.}

\displaystyle \textbf{Question 13: } \text{Solve: }\frac{dy}{dx}=y\sin 2x,\text{ it being given that }y(0)=1.\hspace{1.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=y\sin 2x
\displaystyle \Rightarrow \frac{1}{y}\,dy=\sin 2x\,dx
\displaystyle \Rightarrow \int\frac{1}{y}\,dy=\int\sin 2x\,dx\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log|y|=-\frac{1}{2}\cos 2x+C\ \ \ ...(i)
\displaystyle \text{It is given that }y(0)=1,\text{ i.e. }y=1\text{ when }x=0.\text{ Putting }x=0,y=1\text{ in (i), we get}
\displaystyle 0=-\frac{1}{2}+C\Rightarrow C=\frac{1}{2}
\displaystyle \text{Putting }C=\frac{1}{2}\text{ in (i), we get}
\displaystyle \log|y|=-\frac{1}{2}\cos 2x+\frac{1}{2}
\displaystyle \Rightarrow \log|y|=\frac{1}{2}(1-\cos 2x)
\displaystyle \Rightarrow \log|y|=\sin^{2}x
\displaystyle \Rightarrow |y|=e^{\sin^{2}x}
\displaystyle \Rightarrow y=\pm e^{\sin^{2}x}
\displaystyle \Rightarrow y=e^{\sin^{2}x}\ \text{or }y=-e^{\sin^{2}x}
\displaystyle \text{But }y=-e^{\sin^{2}x}\text{ is not satisfied by }y(0)=1.\text{ Hence, }y=e^{\sin^{2}x}\text{ is the required solution.}

\displaystyle \textbf{Question 14: } \text{Solve the following initial value problems:}
\displaystyle (i)\ (x+1)\frac{dy}{dx}=2e^{-y}-1,\ y(0)=0\hspace{4.0cm} [\text{CBSE 2012}]
\displaystyle (ii)\ y-x\frac{dy}{dx}=2\left(1+x^{2}\frac{dy}{dx}\right),\ y(1)=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle (x+1)\frac{dy}{dx}=2e^{-y}-1
\displaystyle \Rightarrow (x+1)\,dy=(2e^{-y}-1)\,dx
\displaystyle \Rightarrow \frac{1}{x+1}\,dx=\frac{1}{2e^{-y}-1}\,dy\ \ [\text{On separating the variables}]
\displaystyle \Rightarrow \int\frac{1}{x+1}\,dx=\int\frac{1}{2e^{-y}-1}\,dy
\displaystyle \Rightarrow \int\frac{1}{x+1}\,dx=\int\frac{e^{y}}{2-e^{y}}\,dy
\displaystyle \Rightarrow \int\frac{1}{x+1}\,dx=-\int\frac{e^{y}}{e^{y}-2}\,dy
\displaystyle \Rightarrow \log|x+1|=-\log|e^{y}-2|+\log C
\displaystyle \Rightarrow \log|x+1|+\log|e^{y}-2|=\log C
\displaystyle \Rightarrow \log|(x+1)(e^{y}-2)|=\log C
\displaystyle \Rightarrow |(x+1)(e^{y}-2)|=C\ \ \ ...(i)
\displaystyle \text{It is given that }y(0)=0\text{ i.e. }y=0\text{ when }x=0.\text{ Putting }x=0,y=0\text{ in (i), we get}
\displaystyle |(0+1)(1-2)|=C\Rightarrow C=1
\displaystyle \text{Putting }C=1\text{ in (i), we get}
\displaystyle |(x+1)(e^{y}-2)|=1
\displaystyle \Rightarrow (x+1)(e^{y}-2)=\pm 1
\displaystyle \Rightarrow e^{y}-2=-\frac{1}{x+1}
\displaystyle \Rightarrow e^{y}=\left(2-\frac{1}{x+1}\right)
\displaystyle \Rightarrow y=\log\left(2-\frac{1}{x+1}\right),\text{ which is the required solution.}
\displaystyle \text{(ii)\ \ }y-x\frac{dy}{dx}=2\left(1+x^{2}\frac{dy}{dx}\right)
\displaystyle \Rightarrow y-2=2x^{2}\frac{dy}{dx}+x\frac{dy}{dx}
\displaystyle \Rightarrow y-2=x(2x+1)\frac{dy}{dx}
\displaystyle \Rightarrow (y-2)\,dx=x(2x+1)\,dy
\displaystyle \Rightarrow \frac{1}{x(2x+1)}\,dx=\frac{1}{y-2}\,dy
\displaystyle \Rightarrow \int\frac{1}{x(2x+1)}\,dx=\int\frac{1}{y-2}\,dy
\displaystyle \Rightarrow \int\left(\frac{1}{x}-\frac{2}{2x+1}\right)\,dx=\int\frac{1}{y-2}\,dy
\displaystyle \Rightarrow \log|x|-\log|2x+1|=\log|y-2|+\log C
\displaystyle \Rightarrow \log\left|\frac{x}{2x+1}\right|-\log|y-2|=\log C
\displaystyle \Rightarrow \log\left|\frac{x}{(2x+1)(y-2)}\right|=\log C
\displaystyle \Rightarrow \frac{x}{(2x+1)(y-2)}=C\ \ \ ...(i)
\displaystyle \text{It is given that }y(1)=1\text{ i.e. }y=1\text{ when }x=1.\text{ Putting }x=1,y=1\text{ in (i), we get}
\displaystyle \left|\frac{1}{3(-1)}\right|=C\Rightarrow C=\frac{1}{3}
\displaystyle \text{Putting }C=\frac{1}{3}\text{ in (i), we get}
\displaystyle \left|\frac{x}{(2x+1)(y-2)}\right|=\frac{1}{3}
\displaystyle \Rightarrow \frac{x}{(2x+1)(y-2)}=\pm\frac{1}{3}
\displaystyle \Rightarrow y-2=\pm\frac{3x}{2x+1}\Rightarrow y=2\pm\frac{3x}{2x+1}
\displaystyle \text{But, }y=2+\frac{3x}{2x+1}\text{ is not satisfied by }y(1)=1.
\displaystyle \text{Hence, }y=2-\frac{3x}{2x+1},\text{ where }x\neq -\frac{1}{2}\text{ is the required solution.}

\displaystyle \textbf{Question 15: } \text{Find the particular solution of the differential equation:} \\ \log\left(\frac{dy}{dx}\right)=3x+4y\text{ given that }y=0\text{ when }x=0.\hspace{4.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \log\left(\frac{dy}{dx}\right)=3x+4y
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x+4y}
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x}e^{4y}
\displaystyle \Rightarrow e^{-4y}\,dy=e^{3x}\,dx
\displaystyle \text{On integrating, we get}
\displaystyle -\frac{1}{4}e^{-4y}=\frac{1}{3}e^{3x}+C
\displaystyle \Rightarrow 4e^{3x}+3e^{-4y}+12C=0\ \ \ ...(i)
\displaystyle \text{It is given that }y=0\text{ when }x=0.\text{ Substituting }x=0\text{ and }y=0\text{ in (i), we get}
\displaystyle 4+3+12C=0\Rightarrow C=-\frac{7}{12}
\displaystyle \text{Substituting the value of }C\text{ in (i), we get } 4e^{3x}+3e^{-4y}-7=0\text{ as a particular solution} \\ \text{of the given differential equation.}

\displaystyle \textbf{Question 16: } \text{Solve the differential equation }(x^{2}-y^{2})\,dx+2xy\,dy=0;\text{ given that } \\ y=1\text{ when }x=1.\hspace{4.0cm} \text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle (x^{2}-y^{2})\,dx+2xy\,dy=0
\displaystyle \Rightarrow (x^{2}-y^{2})\,dx=-2xy\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x^{2}-y^{2}}{2xy}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}\ \ \ ...(i)
\displaystyle \text{Since each of the functions }y^{2}-x^{2}\text{ and }2xy\text{ is a homogeneous function of degree }2,\text{ the given} \\ \text{differential equation is therefore homogeneous.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-x^{2}}{2x\cdot vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v^{2}-1}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-1}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-1-2v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\left(\frac{v^{2}+1}{2v}\right)
\displaystyle \Rightarrow \frac{2v}{v^{2}+1}\,dv=-\frac{dx}{x}\ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \int\frac{2v}{v^{2}+1}\,dv=-\int\frac{dx}{x}\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log(v^{2}+1)=-\log|x|+C
\displaystyle \Rightarrow \log(v^{2}+1)+\log|x|=\log C
\displaystyle \Rightarrow (v^{2}+1)|x|=C
\displaystyle \Rightarrow \left\{\left(\frac{y^{2}}{x^{2}}\right)+1\right\}|x|=C\ \ \ [\because v=\frac{y}{x}]
\displaystyle \Rightarrow (x^{2}+y^{2})=C|x|\ \ \ ...(ii)
\displaystyle \text{It is given that }y=1\text{ when }x=1.\text{ So, putting }x=1,y=1\text{ in (ii), we get: }C=2.
\displaystyle \text{Substituting }C=2\text{ in (ii), we obtain }x^{2}+y^{2}=2|x|\Rightarrow x^{2}+y^{2}=\pm 2x
\displaystyle \text{But, }x=1,y=1\text{ do not satisfy }x^{2}+y^{2}=-2x.\text{ Hence, }x^{2}+y^{2}=2x\text{ is the required solution.}

\displaystyle \textbf{Question 17: } \text{Solve: }x^{2}y\,dx-(x^{3}+y^{3})\,dy=0\hspace{4.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle x^{2}y\,dx-(x^{3}+y^{3})\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}y}{x^{3}+y^{3}}\ \ \ ...(i)
\displaystyle \text{Since each of the functions }x^{2}y\text{ and }x^{3}+y^{3}\text{ is a homogeneous function of degree }3,\text{ so the given} \\ \text{differential equation is homogeneous.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx^{3}}{x^{3}+v^{3}x^{3}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v}{1+v^{3}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v}{1+v^{3}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v-v-v^{4}}{1+v^{3}}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{4}}{1+v^{3}}
\displaystyle \Rightarrow x(1+v^{3})\,dv=-v^{4}\,dx
\displaystyle \Rightarrow \frac{1+v^{3}}{v^{4}}\,dv=-\frac{dx}{x}\ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \left(\frac{1}{v^{4}}+\frac{1}{v}\right)\,dv=-\frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{3v^{3}}+\log|v|=-\log|x|+C\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow -\frac{1}{3v^{3}}+\log|v|+\log|x|=C
\displaystyle \Rightarrow -\frac{1}{3}\frac{x^{3}}{y^{3}}+\log\left|\frac{y}{x}x\right|=C\ \ \ [\because v=\frac{y}{x}]
\displaystyle \Rightarrow -\frac{x^{3}}{3y^{3}}+\log|y|=C,\text{ which is the required solution.}

\displaystyle \textbf{Question 18: } \text{Find the particular solution of the differential equation:} \\ (x^{2}+xy)\,dy=(x^{2}+y^{2})\,dx\text{ given that }y=0\text{ when }x=1.\hspace{2.0cm} \text{[CBSE 2005, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle (x^{2}+xy)\,dy=(x^{2}+y^{2})\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}\ \ \ ...(i)
\displaystyle \text{Since each of the functions }x^{2}+y^{2}\text{ and }x^{2}+xy\text{ is a homogeneous function of degree }2,\text{ so the} \\ \text{given differential equation is homogeneous.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in (i), we get}
\displaystyle v+x\frac{dv}{dx} =\frac{x^{2}+v^{2}x^{2}}{x^{2}+vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx} =\frac{1+v^{2}}{1+v}
\displaystyle \Rightarrow x\frac{dv}{dx} =\frac{1+v^{2}-v-v^{2}}{1+v}
\displaystyle \Rightarrow x\frac{dv}{dx} =\frac{1-v}{1+v}
\displaystyle \Rightarrow \frac{1+v}{1-v}\,dv=\frac{dx}{x} \ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \left(\frac{2}{1-v}-1\right)dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\left(\frac{2}{1-v}-1\right)dv=\int\frac{dx}{x}
\displaystyle \Rightarrow -2\log|1-v|-v=\log|x|+\log C
\displaystyle \Rightarrow \log\{C|x|(1-v)^{2}\}=-v
\displaystyle \Rightarrow C|x|(1-v)^{2}=e^{-v}
\displaystyle \Rightarrow C|x|\!\left(1-\frac{y}{x}\right)^{2}=e^{-y/x}
\displaystyle \Rightarrow C(x-y)^{2}=|x|e^{-y/x}\ \ \ ...(ii)
\displaystyle \text{It is given that }y=0\text{ when }x=1.\text{ Putting these values in (ii), we get }
\displaystyle C(1-0)^{2}=e^{0}\Rightarrow C=1.
\displaystyle \text{Putting }C=1\text{ in (ii), we obtain }(x-y)^{2}=|x|e^{-y/x},\text{ which is the particular} \\ \text{solution of the given differential equation.}

\displaystyle \textbf{Question 19: } \text{Find the particular solution of the differential equation} \\ (3xy+y^{2})\,dx+(x^{2}+xy)\,dy=0;\text{ for }x=1,y=1.\hspace{2.0cm} \text{[CBSE 2008, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle (3xy+y^{2})\,dx+(x^{2}+xy)\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx} =-\left(\frac{3xy+y^{2}}{x^{2}+xy}\right)\ \ \ \text{...(i)}
\displaystyle \text{Since each of the functions }(3xy+y^{2})\text{ and }(x^{2}+xy) \text{ is a homogeneous function of} \\ \text{degree }2,\text{ the given equation is homogeneous.}
\displaystyle \text{Putting }y=vx\text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}\textbf{ in (i), we get}
\displaystyle v+x\frac{dv}{dx} =-\left(\frac{3vx^{2}+v^{2}x^{2}}{x^{2}+vx^{2}}\right)
\displaystyle \Rightarrow v+x\frac{dv}{dx} =-\left(\frac{3v+v^{2}}{1+v}\right)
\displaystyle \Rightarrow x\frac{dv}{dx} =-\left(\frac{3v+v^{2}}{1+v}+v\right)
\displaystyle \Rightarrow x\frac{dv}{dx} =-\left(\frac{2v^{2}+4v}{v+1}\right)
\displaystyle \Rightarrow (v+1)x\,dv=-(2v^{2}+4v)\,dx
\displaystyle \Rightarrow \frac{v+1}{2v^{2}+4v}\,dv=-\frac{dx}{x} \ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \frac{2v+2}{v^{2}+2v}\,dv=-4\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{2v+2}{v^{2}+2v}\,dv =-4\int\frac{dx}{x} \ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log|v^{2}+2v| =-4\log|x|+\log C
\displaystyle \Rightarrow \log|v^{2}+2v| =\log\!\left(\frac{C}{x^{4}}\right)
\displaystyle \Rightarrow |v^{2}+2v| =\frac{C}{x^{4}}
\displaystyle \Rightarrow \left|\frac{y^{2}}{x^{2}}+\frac{2y}{x}\right| =\frac{C}{x^{4}}\ \ \ [\because v=\frac{y}{x}]
\displaystyle \Rightarrow |y^{2}+2xy| =\frac{C}{x^{2}}\ \ \ \textbf{...(ii)}
\displaystyle \text{It is given that }y=1\text{ when }x=1. \text{ Putting these values in (ii), we get }C=3.
\displaystyle \text{Putting }C=3\text{ in (ii), we obtain }
\displaystyle y^{2}+2xy=\frac{3}{x^{2}}, \text{ which is the required solution.}

\displaystyle \textbf{Question 20: } \text{Solve: }x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx\hspace{4.0cm} \text{[CBSE 2005, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation can be written as}
\displaystyle \frac{dy}{dx}=\frac{\sqrt{x^{2}+y^{2}}+y}{x},\ x\neq0
\displaystyle \text{Clearly, it is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in it, we get}
\displaystyle v+x\frac{dv}{dx}=\frac{\sqrt{x^{2}+v^{2}x^{2}}+vx}{x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\sqrt{1+v^{2}}+v
\displaystyle \Rightarrow x\frac{dv}{dx}=\sqrt{1+v^{2}}
\displaystyle \Rightarrow \frac{dv}{\sqrt{1+v^{2}}}=\frac{dx}{x}\ \ \ [\text{By separating the variables}]
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{dv}{\sqrt{1+v^{2}}}=\int\frac{dx}{x}
\displaystyle \Rightarrow \log\!\left|v+\sqrt{1+v^{2}}\right|=\log|x|+\log C
\displaystyle \Rightarrow \left|v+\sqrt{1+v^{2}}\right|=|Cx|
\displaystyle \Rightarrow \left|\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}\right|=|Cx|\ \ \ [\because v=\frac{y}{x}]
\displaystyle \Rightarrow \left|y+\sqrt{x^{2}+y^{2}}\right|^{2}=C^{2}x^{4},\text{ which gives the required solution.}

\displaystyle \textbf{Question 21:} \text{Solve: } \\ y\!\left\{x\cos\!\left(\frac{y}{x}\right)+y\sin\!\left(\frac{y}{x}\right)\right\}dx -x\!\left\{y\sin\!\left(\frac{y}{x}\right)-x\cos\!\left(\frac{y}{x}\right)\right\}dy=0\hspace{1.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{ The given differential equation can be written as}
\displaystyle \frac{dy}{dx}= \frac{y\{x\cos(\tfrac{y}{x})+y\sin(\tfrac{y}{x})\}} {x\{y\sin(\tfrac{y}{x})-x\cos(\tfrac{y}{x})\}}\ \ \ ...(i)
\displaystyle \text{It can be checked that RHS does not change when }x\text{ is replaced by }\lambda x\text{ and }y\text{ by }\lambda y. \\ \text{ So, the equation is homogeneous.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in (i), we get}
\displaystyle v+x\frac{dv}{dx} =\frac{v\{\cos v+v\sin v\}}{v\sin v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx} =\frac{v\cos v+v^{2}\sin v-v^{2}\sin v+v\cos v} {v\sin v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx} =\frac{2v\cos v}{v\sin v-\cos v}
\displaystyle \Rightarrow \frac{v\sin v-\cos v}{v\cos v}\,dv =2\frac{dx}{x}\ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \int\frac{v\sin v-\cos v}{v\cos v}\,dv =2\int\frac{dx}{x}\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow -\log|v\cos v|=2\log|x|+\log C
\displaystyle \Rightarrow \log\!\left(\frac{1}{|v\cos v|}\right) =\log|Cx^{2}|
\displaystyle \Rightarrow \frac{1}{|v\cos v|}=|Cx^{2}|
\displaystyle \Rightarrow \frac{x}{|y|\cos(\tfrac{y}{x})}=|Cx^{2}|\ \ \ [\because v=\tfrac{y}{x}]
\displaystyle \Rightarrow |xy\cos(\tfrac{y}{x})|=\frac{1}{|C|}
\displaystyle \Rightarrow xy\cos\!\left(\frac{y}{x}\right)=k,\ \text{where }k=\frac{1}{|C|}
\displaystyle \text{Hence, }xy\cos\!\left(\frac{y}{x}\right)=k,\ x\neq0,\ k>0\text{ is the required solution.}

\displaystyle \textbf{Question 22: } \text{Solve: }x\frac{dy}{dx}=y-x\tan\!\left(\frac{y}{x}\right)\hspace{4.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle x\frac{dy}{dx}=y-x\tan\!\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\tan\!\left(\frac{y}{x}\right)\ \ \ ...(i)
\displaystyle \text{Clearly, the given differential equation is homogeneous. Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in (i), we get}
\displaystyle v+x\frac{dv}{dx}=v-\tan v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\tan v
\displaystyle \Rightarrow \cot v\,dv=-\frac{dx}{x}\ \ \ [\text{By separating the variables}]
\displaystyle \Rightarrow \int\cot v\,dv=-\int\frac{dx}{x}\ \ \ [\text{Integrating both sides}]
\displaystyle \Rightarrow \log|\sin v|=-\log|x|+\log C
\displaystyle \Rightarrow |\sin v|=\left|\frac{C}{x}\right|
\displaystyle \Rightarrow \left|\sin\!\left(\frac{y}{x}\right)\right|=\left|\frac{C}{x}\right|
\displaystyle \text{Hence, }\left|\sin\!\left(\frac{y}{x}\right)\right|=\frac{C}{x}\text{ gives the required solution.}

\displaystyle \textbf{Question 23: } \text{Solve: }2y e^{x/y}\,dx+(y-2x e^{x/y})\,dy=0\hspace{2.0cm} \text{[CBSE 2012, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle 2y e^{x/y}\,dx+(y-2x e^{x/y})\,dy=0
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{2x e^{x/y}-y}{2y e^{x/y}}\ \ \ ...(i)
\displaystyle \text{Clearly, the given differential equation is a homogeneous differential equation. As the RHS of (i) is expressible} \\ \text{as a function of }\frac{x}{y},\text{ we put }x=vy\text{ and }\frac{dx}{dy}=v+y\frac{dv}{dy}\text{ to get}
\displaystyle v+y\frac{dv}{dy}=\frac{2v e^{v}-1}{2e^{v}}
\displaystyle \Rightarrow y\frac{dv}{dy}=\frac{2v e^{v}-1}{2e^{v}}-v
\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{1}{2e^{v}}
\displaystyle \Rightarrow 2y e^{v}\,dv=-dy
\displaystyle \Rightarrow 2e^{v}\,dv=-\frac{1}{y}\,dy\ \ \ [\text{On integrating}]
\displaystyle \Rightarrow 2e^{v}=-\log|y|+\log C
\displaystyle \Rightarrow 2e^{v}=\log\!\left|\frac{C}{y}\right|
\displaystyle \Rightarrow 2e^{x/y}=\log\!\left|\frac{C}{y}\right|
\displaystyle \text{Hence, }2e^{x/y}=\log\!\left|\frac{C}{y}\right|\text{ gives the general solution of the given differential equation.}

\displaystyle \textbf{Question 24:} \text{Solve the following initial value problems:}
\displaystyle \text{(i)}\ x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)+x-y\sin\!\left(\frac{y}{x}\right)=0,\ y(1)=\frac{\pi}{2}
\displaystyle \text{(ii)}\ xe^{y/x}-y\sin\!\left(\frac{y}{x}\right)+x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)=0,\ y(1)=0\hspace{4.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i) We have,}
\displaystyle x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)+x-y\sin\!\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x-y\sin(\tfrac{y}{x})}{x\sin(\tfrac{y}{x})}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{1-v\sin v}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v\sin v}{\sin v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1}{\sin v}
\displaystyle \Rightarrow \sin v\,dv=-\frac{dx}{x},\ x\neq0
\displaystyle \Rightarrow \int\sin v\,dv=-\int\frac{dx}{x}\ \ [\text{On integrating}]
\displaystyle \Rightarrow -\cos v=-\log|x|+C
\displaystyle \Rightarrow -\cos\!\left(\frac{y}{x}\right)+\log|x|=C\ \ ...(i)
\displaystyle \text{It is given that }y(1)=\frac{\pi}{2}.\text{ Putting }x=1,y=\frac{\pi}{2}\text{ in (i), we get }C=0.
\displaystyle \text{Putting }C=0\text{ in (i), we get }
\displaystyle \log|x|=\cos\!\left(\frac{y}{x}\right).
\displaystyle \text{Hence, }\log|x|=\cos\!\left(\frac{y}{x}\right)\text{ is the required solution.}
\displaystyle \text{(ii) We have,}
\displaystyle xe^{y/x}-y\sin\!\left(\frac{y}{x}\right)+x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y\sin(\tfrac{y}{x})-xe^{y/x}}{x\sin(\tfrac{y}{x})}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{v\sin v-e^{v}}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\sin v-e^{v}}{\sin v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{e^{v}}{\sin v}
\displaystyle \Rightarrow e^{-v}\sin v\,dv=-\frac{dx}{x}
\displaystyle \Rightarrow \int e^{-v}\sin v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -\frac12 e^{-v}(\sin v+\cos v)=-\log|x|+C
\displaystyle \Rightarrow e^{-v}\!\left(\sin v+\cos v\right)=2\log|x|-2\log C\ \ ...(ii)
\displaystyle \text{It is given that }y(1)=0.\text{ Putting }x=1,y=0\text{ in (ii), we get }\log C=-\frac12.
\displaystyle \text{Putting }\log C=-\frac12\text{ in (ii), we get}
\displaystyle e^{-y/x}\!\left[\sin\!\left(\frac{y}{x}\right)+\cos\!\left(\frac{y}{x}\right)\right]=\log|x|^{2}+1.

\displaystyle \textbf{Question 25:} \text{Solve the following initial value problems:}
\displaystyle \textbf{(i)}\ x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)+x-y\sin\!\left(\frac{y}{x}\right)=0,\ y(1)=\frac{\pi}{2}
\displaystyle \textbf{(ii)}\ xe^{y/x}-y\sin\!\left(\frac{y}{x}\right)+x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)=0,\ y(1)=0\hspace{4.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)+x-y\sin\!\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x-y\sin(\tfrac{y}{x})}{x\sin(\tfrac{y}{x})}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{1-v\sin v}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v\sin v}{\sin v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1}{\sin v}
\displaystyle \Rightarrow \sin v\,dv=-\frac{dx}{x},\ x\neq0
\displaystyle \Rightarrow \int\sin v\,dv=-\int\frac{dx}{x}\ \ [\text{On integrating}]
\displaystyle \Rightarrow -\cos v=-\log|x|+C
\displaystyle \Rightarrow -\cos\!\left(\frac{y}{x}\right)+\log|x|=C\ \ ...(i)
\displaystyle \text{It is given that }y(1)=\frac{\pi}{2}.\text{ Putting }x=1,y=\frac{\pi}{2}\text{ in (i), we get }C=0.
\displaystyle \text{Putting }C=0\text{ in (i), we get }
\displaystyle \log|x|=\cos\!\left(\frac{y}{x}\right).
\displaystyle \text{Hence, }\log|x|=\cos\!\left(\frac{y}{x}\right)\text{ is the required solution.}
\displaystyle \text{(ii) We have,}
\displaystyle xe^{y/x}-y\sin\!\left(\frac{y}{x}\right)+x\frac{dy}{dx}\sin\!\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y\sin(\tfrac{y}{x})-xe^{y/x}}{x\sin(\tfrac{y}{x})}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{v\sin v-e^{v}}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\sin v-e^{v}}{\sin v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{e^{v}}{\sin v}
\displaystyle \Rightarrow e^{-v}\sin v\,dv=-\frac{dx}{x}
\displaystyle \Rightarrow \int e^{-v}\sin v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -\frac12 e^{-v}(\sin v+\cos v)=-\log|x|+C
\displaystyle \Rightarrow e^{-v}\!\left(\sin v+\cos v\right)=2\log|x|-2\log C\ \ ...(ii)
\displaystyle \text{It is given that }y(1)=0.\text{ Putting }x=1,y=0\text{ in (ii), we get }\log C=-\frac12.
\displaystyle \text{Putting }\log C=-\frac12\text{ in (ii), we get}
\displaystyle e^{-y/x}\!\left[\sin\!\left(\frac{y}{x}\right)+\cos\!\left(\frac{y}{x}\right)\right]=\log|x|^{2}+1.

\displaystyle \textbf{Question 26: } \text{Solve each of the following initial value problems:}
\displaystyle \text{(i)}\ 2x^{2}\frac{dy}{dx}-2xy+y^{2}=0,\ y(e)=e\qquad \text{[CBSE 2012]}
\displaystyle \text{(ii)}\ 2xy+y^{2}-2x^{2}\frac{dy}{dx}=0,\ y(1)=2
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle 2x^{2}\frac{dy}{dx}-2xy+y^{2}=0\Rightarrow \frac{dy}{dx}=\frac{2xy-y^{2}}{2x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{2v-v^{2}}{2}
\displaystyle \Rightarrow 2x\frac{dv}{dx}=-v^{2}
\displaystyle \Rightarrow -\frac{2}{v^{2}}\,dv=\frac{dx}{x}\ \ \ [\text{On integrating}]
\displaystyle \Rightarrow \frac{2}{v}=\log|x|+C\ \ \ ...(i)
\displaystyle \text{It is given that }y(e)=e,\ i.e.\ v=1\text{ when }x=e.
\displaystyle \text{Putting }x=e,v=1\text{ in (i), we get }2=1+C\Rightarrow C=1.
\displaystyle \text{Putting }C=1\text{ in (i), we get }\frac{2x}{y}=1+\log|x|.
\displaystyle \text{Hence, }y=\frac{2x}{1+\log|x|}\text{ gives the required solution.}
\displaystyle \text{(ii) We have,}\ 2xy+y^{2}-2x^{2}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{2v+v^{2}}{2}
\displaystyle \Rightarrow 2x\frac{dv}{dx}=v^{2}
\displaystyle \Rightarrow \frac{2}{v^{2}}\,dv=\frac{dx}{x}\ \ \ [\text{On integrating}]
\displaystyle \Rightarrow -\frac{2}{v}=\log|x|+C\ \ \ ...(ii)
\displaystyle \text{It is given that }y(1)=2,\ i.e.\ v=2\text{ when }x=1.
\displaystyle \text{Putting }x=1,v=2\text{ in (ii), we get }-1=C.
\displaystyle \text{Putting }C=-1\text{ in (ii), we get }\frac{2x}{y}=1-\log|x|.
\displaystyle \text{Hence, }y=\frac{2x}{1-\log|x|}\text{ gives the solution of the given differential equation.}

\displaystyle \textbf{Question 27: } \text{Solve each of the following initial value problems:}
\displaystyle \text{(i)}\ (x^{2}+y^{2})\,dx+xy\,dy=0,\ y(1)=1 \\ \text{(ii)}\ (xe^{y/x}+y)\,dx=x\,dy,\ y(1)=1 \\ \text{(iii)}\ (x^{2}-2y^{2})\,dx+2xy\,dy=0,\ y(1)=1\hspace{4.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle (x^{2}+y^{2})\,dx+xy\,dy=0\Rightarrow \frac{dy}{dx}=-\frac{x^{2}+y^{2}}{xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=-\frac{1+v^{2}}{v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{1+2v^{2}}{v}
\displaystyle \Rightarrow \frac{v\,dv}{2v^{2}+1}=-\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{4v}{2v^{2}+1}\,dv=-\int\frac{4}{x}\,dx\ \ \ [\text{On integrating}]
\displaystyle \Rightarrow \log(2v^{2}+1)=-4\log|x|+\log C
\displaystyle \Rightarrow (2y^{2}+x^{2})x^{2}=|C|\ \ \ ...(i)
\displaystyle \text{Putting }x=1,y=1\text{ in (i), we get }|C|=3.
\displaystyle \text{Hence, }(2y^{2}+x^{2})x^{2}=3\text{ is the required solution.}
\displaystyle \text{(ii) We have,}\ (xe^{y/x}+y)\,dx=x\,dy\Rightarrow \frac{dy}{dx}=e^{y/x}+\frac{y}{x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=e^{v}+v
\displaystyle \Rightarrow x\frac{dv}{dx}=e^{v}
\displaystyle \Rightarrow e^{-v}\,dv=\frac{dx}{x}
\displaystyle \Rightarrow \int e^{-v}\,dv=\int\frac{dx}{x}
\displaystyle \Rightarrow -e^{-v}=\log|x|+C\ \ \ ...(ii)
\displaystyle \text{Putting }x=1,y=1\text{ i.e.\ }v=1\text{ in (ii), we get }C=-\frac{1}{e}.
\displaystyle \Rightarrow e^{-y/x}=\frac{1}{e}-\log|x|
\displaystyle \Rightarrow y=x-x\log(1-e\log|x|)\text{ is the required solution.}
\displaystyle \text{(iii) We have,}\ (x^{2}-2y^{2})\,dx+2xy\,dy=0\Rightarrow \frac{dy}{dx}=\frac{2y^{2}-x^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx},\text{ it reduces to}
\displaystyle v+x\frac{dv}{dx}=\frac{2v^{2}-1}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{1}{2v}
\displaystyle \Rightarrow 2v\,dv=-\frac{dx}{x}
\displaystyle \Rightarrow \int2v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow v^{2}=-\log|x|+C
\displaystyle \Rightarrow y^{2}=-x^{2}\log|x|+Cx^{2}\ \ \ ...(iii)
\displaystyle \text{Putting }x=1,y=1\text{ in (iii), we get }C=1.
\displaystyle \text{Hence, }y^{2}=-x^{2}\log|x|+x^{2}\text{ is the required solution.}

\displaystyle \textbf{Question 28: } \text{Solve the differential equation } \frac{dy}{dx}-\frac{y}{x}=2x^{2}\hspace{0.5cm} \text{[CBSE 2004, 2007, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle \frac{dy}{dx}-\frac{1}{x}y=2x^{2}\ \ \ ...(i)
\displaystyle \text{Clearly, it is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=-\frac{1}{x}\text{ and }Q=2x^{2}.
\displaystyle \text{Now, I.F.}=e^{\int P\,dx} =e^{\int -\frac{1}{x}\,dx} =e^{-\log x} =\frac{1}{x}.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=\frac{1}{x},\text{ we get}
\displaystyle \frac{1}{x}\frac{dy}{dx}-\frac{y}{x^{2}}=2x.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y\!\left(\frac{1}{x}\right)=\int 2x\,dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow \frac{y}{x}=x^{2}+C
\displaystyle \Rightarrow y=x^{3}+Cx,\text{ which is the required solution.}

\displaystyle \textbf{Question 29: } \text{Solve the differential equation: } x\log x\,\frac{dy}{dx}+y=\frac{2}{x}\log x\hspace{1.0cm} \text{[CBSE 2010, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle x\log x\,\frac{dy}{dx}+y=\frac{2}{x}\log x
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x\log x}\,y=\frac{2}{x^{2}}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{1}{x\log x}\text{ and }Q=\frac{2}{x^{2}}.
\displaystyle \text{I.F.}=e^{\int P\,dx} =e^{\int \frac{1}{x\log x}\,dx} =e^{\int \frac{1}{t}\,dt},\ \text{where }t=\log x
\displaystyle \text{I.F.}=e^{\log t}=t=\log x.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=\log x,\text{ we get}
\displaystyle \log x\,\frac{dy}{dx}+\frac{y}{x}=\frac{2}{x^{2}}\log x.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y\log x=\int\frac{2}{x^{2}}\log x\,dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y\log x =2\int\log x\cdot x^{-2}\,dx+C
\displaystyle \Rightarrow y\log x =2\left[\log x\!\int x^{-2}dx-\int x^{-2}\cdot\frac{1}{x}dx\right]+C
\displaystyle \Rightarrow y\log x =2\left[-\frac{\log x}{x}-\frac{1}{x}\right]+C
\displaystyle \Rightarrow y\log x =-\frac{2}{x}(1+\log x)+C,\text{ which gives the required solution.}

\displaystyle \textbf{Question 30: } \text{Solve the differential equation: }  (x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}\hspace{1.0cm} \text{[CBSE 2010, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle (x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{x^{2}-1}y=\frac{1}{(x^{2}-1)^{2}}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{2x}{x^{2}-1}\text{ and }Q=\frac{1}{(x^{2}-1)^{2}}.
\displaystyle \text{I.F.}=e^{\int Pdx} =e^{\int\frac{2x}{x^{2}-1}dx} =e^{\log(x^{2}-1)} =x^{2}-1.
\displaystyle \text{Multiplying both sides of (i) by I.F. }(x^{2}-1),\text{ we get}
\displaystyle (x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y(x^{2}-1)=\int\frac{1}{x^{2}-1}dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y(x^{2}-1)=\frac12\log\!\left|\frac{x-1}{x+1}\right|+C,\text{ which is the required solution.}

\displaystyle \textbf{Question 31: } \text{Solve: }  \frac{dy}{dx}+y\sec x=\tan x\hspace{4.0cm} \text{[CBSE 2008, 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle \frac{dy}{dx}+(\sec x)y=\tan x\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\sec x\text{ and }Q=\tan x.
\displaystyle \text{I.F.}=e^{\int\sec x\,dx} =e^{\log(\sec x+\tan x)} =\sec x+\tan x.
\displaystyle \text{Multiplying both sides of (i) by I.F., we get}
\displaystyle (\sec x+\tan x)\frac{dy}{dx}+y\sec x(\sec x+\tan x)=\tan x(\sec x+\tan x).
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y(\sec x+\tan x)=\int\tan x(\sec x+\tan x)\,dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y(\sec x+\tan x)=\int(\tan x\sec x+\tan^{2}x)\,dx+C
\displaystyle \Rightarrow y(\sec x+\tan x)=\int(\tan x\sec x+\sec^{2}x-1)\,dx+C
\displaystyle \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C,\text{ which is the required solution.}

\displaystyle \textbf{Question 32: } \text{Solve: }  \cos^{2}x\,\frac{dy}{dx}+y=\tan x\hspace{4.0cm} \text{[CBSE 2008, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that}
\displaystyle \cos^{2}x\,\frac{dy}{dx}+y=\tan x
\displaystyle \Rightarrow \frac{dy}{dx}+(\sec^{2}x)y=\tan x\sec^{2}x\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\sec^{2}x\text{ and }Q=\tan x\sec^{2}x.
\displaystyle \text{I.F.}=e^{\int\sec^{2}x\,dx}=e^{\tan x}.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=e^{\tan x},\text{ we get}
\displaystyle e^{\tan x}\frac{dy}{dx}+\sec^{2}x\,e^{\tan x}y =e^{\tan x}\tan x\sec^{2}x.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y e^{\tan x} =\int e^{\tan x}\tan x\sec^{2}x\,dx+C \ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y e^{\tan x} =\int t e^{t}\,dt+C,\text{ where }t=\tan x.
\displaystyle \Rightarrow y e^{\tan x} =t e^{t}-\int e^{t}\,dt+C.
\displaystyle \Rightarrow y e^{\tan x} =t e^{t}-e^{t}+C.
\displaystyle \Rightarrow y e^{\tan x} =e^{\tan x}(\tan x-1)+C,\text{ which is the required solution.}

\displaystyle \textbf{Question 33: } \text{Solve: }(x^{2}+1)\frac{dy}{dx}+2xy=\sqrt{x^{2}+4}\hspace{4.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{We are given that} 
\displaystyle (x^{2}+1)\frac{dy}{dx}+2xy=\sqrt{x^{2}+4} \Rightarrow \frac{dy}{dx}+\frac{2x}{x^{2}+1}y=\frac{\sqrt{x^{2}+4}}{x^{2}+1}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{2x}{x^{2}+1}\text{ and }Q=\frac{\sqrt{x^{2}+4}}{x^{2}+1}.
\displaystyle \text{I.F.}=e^{\int P\,dx} =e^{\int\frac{2x}{x^{2}+1}dx} =e^{\log(x^{2}+1)} =(x^{2}+1).
\displaystyle \text{Multiplying both sides of (i) by I.F. }=(x^{2}+1),\text{ we get}
\displaystyle (x^{2}+1)\frac{dy}{dx}+2xy=\sqrt{x^{2}+4}.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we obtain}
\displaystyle y(x^{2}+1)=\int\sqrt{x^{2}+4}\,dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y(x^{2}+1)=\frac{1}{2}x\sqrt{x^{2}+4}+2\log\!\left|x+\sqrt{x^{2}+4}\right|+C,
\displaystyle \text{which is the required solution.}

\displaystyle \textbf{Question 34: } \text{Solve:} \frac{dy}{dx}+\frac{y}{x}=\cos x+\frac{\sin x}{x}\hspace{4.0cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle \frac{dy}{dx}+\frac{1}{x}y=\cos x+\frac{\sin x}{x}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{1}{x}\text{ and }Q=\cos x+\frac{\sin x}{x}.
\displaystyle \text{I.F.}=e^{\int \frac{1}{x}\,dx} =e^{\log x} =x.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=x,\text{ we get}
\displaystyle x\frac{dy}{dx}+y=x\cos x+\sin x.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle xy=\int(x\cos x+\sin x)\,dx+C \ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow xy=\int x\cos x\,dx+\int\sin x\,dx+C
\displaystyle \Rightarrow xy=x\sin x-\int\sin x\,dx+\int\sin x\,dx+C \ \ \ [\text{Integrating first integral by parts}]
\displaystyle \Rightarrow xy=x\sin x+C
\displaystyle \Rightarrow y=\sin x+\frac{C}{x},\text{ which gives the required solution.}

\displaystyle \textbf{Question 35:} \text{Solve:}
\displaystyle \textbf{(i)}\ x\frac{dy}{dx}+y-x+xy\cot x=0 \hspace{4.0cm} \text{[CBSE 2011, 2012]}
\displaystyle \textbf{(ii)}\ (1+x^{2})\,dy+2xy\,dx=\cot x\,dx \hspace{4.0cm} \text{[CBSE 2012]}
\displaystyle \textbf{(iii)}\ y+\frac{d}{dx}(xy)=x(\sin x+\log x)
\displaystyle \text{Answer:}
\displaystyle \text{ (i) We have,}
\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0
\displaystyle \Rightarrow \frac{dy}{dx}+\left(\frac{1}{x}+\cot x\right)y=1\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation with }P=\frac{1}{x}+\cot x\text{ and }Q=1.
\displaystyle \text{I.F.}=e^{\int(\frac{1}{x}+\cot x)\,dx} =e^{\log x+\log\sin x}=x\sin x.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=x\sin x,\text{ we get}
\displaystyle x\sin x\,\frac{dy}{dx}+(\sin x+x\cos x)y=x\sin x.
\displaystyle \text{Integrating with respect to }x,\text{ we get}
\displaystyle y(x\sin x)=\int x\sin x\,dx+C
\displaystyle \Rightarrow xy\sin x=-x\cos x+\sin x+C,\text{ which is the required solution.}
\displaystyle \text{(ii) We have,}
\displaystyle (1+x^{2})\,dy+2xy\,dx=\cot x\,dx
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^{2}}\,y=\frac{\cot x}{1+x^{2}}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation with }P=\frac{2x}{1+x^{2}}\text{ and }Q=\frac{\cot x}{1+x^{2}}.
\displaystyle \text{I.F.}=e^{\int\frac{2x}{1+x^{2}}dx}=e^{\log(1+x^{2})}=1+x^{2}.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=1+x^{2},\text{ we get}
\displaystyle (1+x^{2})\frac{dy}{dx}+2xy=\cot x.
\displaystyle \text{Integrating with respect to }x,\text{ we get}
\displaystyle y(1+x^{2})=\int\cot x\,dx+C
\displaystyle \Rightarrow y(1+x^{2})=\log|\sin x|+C,\text{ which is the required solution.}
\displaystyle \text{(iii) We have,}
\displaystyle y+\frac{d}{dx}(xy)=x(\sin x+\log x)
\displaystyle \Rightarrow x\frac{dy}{dx}+2y=x(\sin x+\log x)
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=\sin x+\log x\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation with }P=\frac{2}{x}\text{ and }Q=\sin x+\log x.
\displaystyle \text{I.F.}=e^{\int\frac{2}{x}dx}=e^{2\log x}=x^{2}.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=x^{2},\text{ we obtain}
\displaystyle x^{2}\frac{dy}{dx}+2xy=x^{2}(\sin x+\log x).
\displaystyle \text{Integrating with respect to }x,\text{ we get}
\displaystyle yx^{2}=\int x^{2}\sin x\,dx+\int x^{2}\log x\,dx+C
\displaystyle \Rightarrow yx^{2}=-x^{2}\cos x+2x\sin x+2\cos x+\frac{x^{3}}{3}\log x-\frac{x^{3}}{9}+C.

\displaystyle \textbf{Question 36:} \text{ Solve: }y\,dx+(x-y^{3})\,dy=0 \hspace{4.0cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is} 
\displaystyle y\,dx+(x-y^{3})\,dy=0
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{y}=y^{2}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation of the form }\frac{dx}{dy}+Rx=S,
\displaystyle \text{where }R=\frac{1}{y}\text{ and }S=y^{2}.
\displaystyle \text{I.F.}=e^{\int R\,dy} =e^{\int \frac{1}{y}\,dy} =e^{\log y} =y.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=y,\text{ we obtain}
\displaystyle y\frac{dx}{dy}+x=y^{3}.
\displaystyle \text{Integrating both sides with respect to }y,\text{ we get}
\displaystyle xy=\int y^{3}\,dy+C\ \ \ [\text{Using: }x(\text{I.F.})=\int S(\text{I.F.})dy+C]
\displaystyle \Rightarrow xy=\frac{y^{4}}{4}+C,\text{ which is the required solution.}

\displaystyle \textbf{Question 37: } \text{Solve: } \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1 \hspace{2.0cm} \text{[CBSE 2012, CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{\sqrt{x}}y=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\ \ \ ...(i)
\displaystyle \text{This is a linear differential equation with }P=\frac{1}{\sqrt{x}}\text{ and }Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int \frac{1}{\sqrt{x}}\,dx}=e^{2\sqrt{x}}.
\displaystyle \text{Multiplying both sides of (i) by I.F. }=e^{2\sqrt{x}},\text{ we get}
\displaystyle e^{2\sqrt{x}}\frac{dy}{dx}+\frac{y}{\sqrt{x}}e^{2\sqrt{x}}=\frac{1}{\sqrt{x}}.
\displaystyle \text{Integrating both sides with respect to }x,\text{ we get}
\displaystyle y e^{2\sqrt{x}}=\int\frac{1}{\sqrt{x}}\,dx+C\ \ \ [\text{Using: }y(\text{I.F.})=\int Q(\text{I.F.})dx+C]
\displaystyle \Rightarrow y e^{2\sqrt{x}}=2\sqrt{x}+C
\displaystyle \Rightarrow y=(2\sqrt{x}+C)e^{-2\sqrt{x}},\text{ which gives the required solution.}

\displaystyle \textbf{Question 38. }\text{The order and degree (if defined) of the differential equation,}
\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)^{2}+\left(\frac{dy}{dx}\right)^{3}=x\sin\left(\frac{dy}{dx}\right)\text{ respectively are} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }2,2 \qquad \text{(b) }1,3
\displaystyle \text{(c) }2,3 \qquad \text{(d) }2,\text{ degree not defined}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, differential equation is } \left(\frac{d^{2}y}{dx^{2}}\right)^{2}+\left(\frac{dy}{dx}\right)^{3}=x\sin\left(\frac{dy}{dx}\right)
\displaystyle \text{The highest order derivative occurring in differential equation is }\frac{d^{2}y}{dx^{2}}, \\ \text{so its order is }2.
\displaystyle \text{The given differential equation is not a polynomial equation in derivatives of }y, \\ \text{so its degree is not defined.}
\\

\displaystyle \textbf{Question 39. }\text{The order and degree of the differential equation}
\displaystyle \left(1+3\frac{dy}{dx}\right)^{2}=4\frac{d^{3}y}{dx^{3}}\text{ respectively are} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }1,\frac{2}{3} \qquad \text{(b) }3,1
\displaystyle \text{(c) }3,3 \qquad \text{(d) }1,2
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, differential equation is }  \left(1+3\frac{dy}{dx}\right)^{2}=4\frac{d^{3}y}{dx^{3}}
\displaystyle \text{The highest order derivative occurring in differential equation is }\frac{d^{3}y}{dx^{3}}, \\ \text{so its order is }3.
\displaystyle \text{Also, it is a polynomial equation in derivatives and highest power raised to }\frac{d^{3}y}{dx^{3}}\text{ is }1, \\ \text{so its degree is }1.
\\

\displaystyle \textbf{Question 40. }\text{What is the product of the order and degree of the}
\displaystyle \text{differential equation }\frac{d^{2}y}{dx^{2}}\sin y+\left(\frac{dy}{dx}\right)\cos y=\sqrt{y}? \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }3 \qquad \text{(b) }2
\displaystyle \text{(c) }6 \qquad \text{(d) }\text{not defined}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, differential equation is } \frac{d^{2}y}{dx^{2}}\sin y+\left(\frac{dy}{dx}\right)^{3}\cos y=\sqrt{y}
\displaystyle \text{The highest order derivative occurring in differential equation is }\frac{d^{2}y}{dx^{2}}, \\ \text{so its order is }2.
\displaystyle \text{Also, it is a polynomial equation in derivatives and highest power raised to }\frac{d^{2}y}{dx^{2}}\text{ is }1, \\ \text{so its degree is }1.
\displaystyle \text{Hence, the product of the order and degree of the above differential} \\ \text{equation }=2\times1=2.
\\

\displaystyle \textbf{Question 41. }\text{Degree of the differential equation }   \sin x+\cos\left(\frac{dy}{dx}\right)=y^{2}\text{ is} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }2 \qquad \text{(b) }1
\displaystyle \text{(c) }\text{not defined} \qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{(c) The given differential equation is not a polynomial equation in }\frac{dy}{dx}.
\displaystyle \text{Therefore, its degree is not defined.}
\\

\displaystyle \textbf{Question 42. }\text{Find the degree of the differential equation }  1+\left(\frac{dy}{dx}\right)^{2}=x. \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }1+\left(\frac{dy}{dx}\right)^2=x
\displaystyle \text{Degree}=2
\\

\displaystyle \textbf{Question 43. }\text{Find the order and the degree of the differential equation}
\displaystyle  x^{2}\frac{d^{2}y}{dx^{2}}=\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{4}. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is }  x^2\frac{d^2y}{dx^2}=1+\left[\left(\frac{dy}{dx}\right)^2\right]^4
\displaystyle \text{Since, highest order derivative occurring in the differential equation is }\frac{d^2y}{dx^2}
\displaystyle \text{therefore order is }2\text{ and as given equation can be expressed as a polynomial in derivatives,}
\displaystyle \text{so its degree is }1,\text{ which is the power of }\frac{d^2y}{dx^2}.
\\

\displaystyle \textbf{Question 44. }\text{Write the order and degree of the differential equation } \\ x^{3}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}+x\left(\frac{dy}{dx}\right)^{4}=0\hspace{2.2cm}\text{[CBSE 2019, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The highest order derivative present is }\frac{d^{2}y}{dx^{2}}.
\displaystyle \therefore \text{ Order of the differential equation }=2.
\displaystyle \text{The power of the highest order derivative }\frac{d^{2}y}{dx^{2}}\text{ is }2.
\displaystyle \therefore \text{ Degree of the differential equation }=2.
\\

\displaystyle \textbf{Question 45. }\text{Find the order and degree (if defined) of the } \text{differential equation }
\displaystyle  \frac{d^{2}y}{dx^{2}}+x\left(\frac{dy}{dx}\right)^{2}=2x^{2}\log\left(\frac{d^{2}y}{dx^{2}}\right). \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Since, highest order derivative occurring in the differential equation is }\frac{d^2y}{dx^2},
\displaystyle \text{therefore order is }2\text{ and as the differential equation is not a polynomial in derivatives,}
\displaystyle \text{therefore its degree is not defined.}
\\

\displaystyle \textbf{Question 46. }\text{Write the sum of the order and degree of the } \text{differential equation }
\displaystyle  \frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{3}\right]=0. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is }\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^3\right]=0
\displaystyle \Rightarrow 3\left(\frac{dy}{dx}\right)^{3-1}\frac{d}{dx}\left(\frac{dy}{dx}\right)=0
\displaystyle \Rightarrow 3\left(\frac{dy}{dx}\right)^2\frac{d^2y}{dx^2}=0
\displaystyle \text{Here, order = 2 and degree = 1}
\displaystyle \therefore \text{Sum of the order and degree}=2+1=3
\\

\displaystyle \textbf{Question 47. }\text{Write the sum of the order and degree of the } \text{differential equation }
\displaystyle  \left(\frac{d^{2}y}{dx^{2}}\right)^{2}+\left(\frac{dy}{dx}\right)^{3}+x^{4}=0. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is }  \left(\frac{d^2y}{dx^2}\right)^2+\left(\frac{dy}{dx}\right)^3+x^4=0
\displaystyle \text{Here, we see that the highest order derivative is }\frac{d^2y}{dx^2}\text{ whose degree is }2.
\displaystyle \text{Here, order}=2\text{ and degree}=2
\displaystyle \therefore \text{Sum of the order and degree}=2+2=4
\\

\displaystyle \textbf{Question 48. }\text{Write the degree of the differential equation }   \\ \left(\frac{dy}{dx}\right)^{4}+3x\frac{d^{2}y}{dx^{2}}=0. \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is }  \left(\frac{dy}{dx}\right)^4+3x\frac{d^2y}{dx^2}=0
\displaystyle \text{Here, highest order derivative is }\frac{d^2y}{dx^2},\text{ whose degree is one.}
\displaystyle \text{So, the degree of differential equation is }1.
\\

\displaystyle \textbf{Question 49. }\text{Write the degree of the differential equation } \\ x\left(\frac{d^{2}y}{dx^{2}}\right)^{3}+y\left(\frac{dy}{dx}\right)^{4}+x^{3}=0\hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The highest order derivative is }\frac{d^{2}y}{dx^{2}}.  \text{ Its power is }3.
\displaystyle \therefore \text{ Degree of the differential equation }=3.
\\

\displaystyle \textbf{Question 50. }\text{Write the degree of the differential equation } \\ \left(\frac{dy}{dx}\right)^{4}+3y\frac{d^{2}y}{dx^{2}}=0\hspace{2.2cm}\text{[CBSE 2013 C]}
\displaystyle \text{Answer:}
\displaystyle \text{The highest order derivative is }\frac{d^{2}y}{dx^{2}}.  \text{ Its power is }1.
\displaystyle \therefore \text{ Degree of the differential equation }=1.
\\

\displaystyle \textbf{Question 51. }\text{The integrating factor of the differential equation}
\displaystyle (1-y^{2})\frac{dx}{dy}+x=ay \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{y^{2}-1}\qquad \text{(b) }\frac{1}{\sqrt{y^{2}-1}}
\displaystyle \text{(c) }\frac{1}{1-y^{2}}\qquad \text{(d) }\frac{1}{\sqrt{1-y^{2}}}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, differential equation is}
\displaystyle (1-y^2)\frac{dx}{dy}+xy=ay
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{y}{1-y^2}x=\frac{ay}{1-y^2}
\displaystyle \text{On comparing with the linear differential equation}
\displaystyle \frac{dx}{dy}+Px=Q,\text{ we get }P=\frac{y}{1-y^2}\text{ and }Q=\frac{ay}{1-y^2}
\displaystyle \therefore \text{Integrating factor (IF)}=e^{\int P\,dy}=e^{\int \frac{y}{1-y^2}\,dy}
\displaystyle \text{On putting }1-y^2=t\Rightarrow -2y\,dy=dt\Rightarrow y\,dy=-\frac{dt}{2}
\displaystyle \text{IF}=e^{-\frac{1}{2}\int \frac{dt}{t}}=e^{-\frac{1}{2}\log t}
\displaystyle =\frac{1}{\sqrt{t}}=\frac{1}{\sqrt{1-y^2}}
\\

\displaystyle \textbf{Question 52. }\text{The solution of the differential equation }   \frac{dx}{x}+\frac{dy}{y}=0\text{ is} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{x}+\frac{1}{y}=C \qquad \text{(b) }\log x-\log y=C
\displaystyle \text{(c) }xy=C \qquad \text{(d) }x+y=C
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, differential equation is}
\displaystyle \frac{dx}{x}+\frac{dy}{y}=0
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int \frac{dx}{x}+\int \frac{dy}{y}=0
\displaystyle \Rightarrow \log x+\log y=k
\displaystyle \Rightarrow \log(xy)=k
\displaystyle \Rightarrow xy=e^k
\displaystyle \Rightarrow xy=C,\text{ where }C=e^k
\\

\displaystyle \textbf{Question 53. }\text{The number of solutions of the differential equation}
\displaystyle \frac{dy}{dx}=\frac{y+1}{x-1}\text{ when }y(1)=2,\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) zero} \qquad \text{(b) one}
\displaystyle \text{(c) two} \qquad \text{(d) infinite}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, differential equation is}
\displaystyle \frac{dy}{dx}=\frac{y+1}{x-1}
\displaystyle \Rightarrow \frac{dy}{y+1}=\frac{dx}{x-1}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int \frac{dy}{y+1}=\int \frac{dx}{x-1}
\displaystyle \Rightarrow \log(y+1)=\log(x-1)-\log C
\displaystyle \Rightarrow C(y+1)=(x-1)
\displaystyle \Rightarrow C=\frac{x-1}{y+1}
\displaystyle \text{When }x=1\text{ and }y=2,\text{ then }C=0,\text{ so required solution is}
\displaystyle x-1=0\text{ hence, only one solution exists.}
\\

\displaystyle \textbf{Question 54. }\text{Solve the differential equation given by}
\displaystyle x\,dy-y\,dx-\sqrt{x^{2}+y^{2}}\,dx=0 \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle xdy-ydx=\sqrt{x^2+y^2}\,dx
\displaystyle \Rightarrow y+\sqrt{x^2+y^2}=x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation of the form}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)\text{ on putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{In Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v+\sqrt{1+v^2}\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}
\displaystyle \Rightarrow \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{dv}{\sqrt{1+v^2}}=\int\frac{dx}{x}
\displaystyle \Rightarrow \log\left|v+\sqrt{1+v^2}\right|=\log|x|+C
\displaystyle \Rightarrow \log\left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|-\log|x|=C
\displaystyle \Rightarrow \log\left|\frac{y+\sqrt{x^2+y^2}}{x^2}\right|=C
\displaystyle \left[\because \log m-\log n=\log\left(\frac{m}{n}\right)\right]
\displaystyle \Rightarrow \frac{y+\sqrt{x^2+y^2}}{x^2}=e^C
\displaystyle \Rightarrow y+\sqrt{x^2+y^2}=Ax^2
\displaystyle \text{where }A=e^C
\\

\displaystyle \textbf{Question 55. }\text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}+\sec^{2}x\cdot y=\tan x\sec^{2}x,\text{ given that }y(0)=0. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}+\sec^2x\cdot y=\tan x\,\sec^2x
\displaystyle \text{On comparing with the linear differential equation,}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ we get}
\displaystyle P=\sec^2x\text{ and }Q=\tan x\cdot\sec^2x
\displaystyle \therefore \text{Integrating factor (IF)}=e^{\int Pdx}=e^{\int \sec^2x\,dx}=e^{\tan x}
\displaystyle \text{Now, the solution of given differential equation is}
\displaystyle y(\text{IF})=\int (Q\times \text{IF})\,dx+C
\displaystyle \Rightarrow ye^{\tan x}=\int \tan x\cdot\sec^2x\cdot e^{\tan x}\,dx+C\qquad ...(i)
\displaystyle \text{In RHS of Eq. (i), on putting}
\displaystyle \tan x=t\Rightarrow \sec^2x\,dx=dt
\displaystyle \therefore ye^{\tan x}=\int te^t\,dt+C
\displaystyle =\left[\int e^t\,dt-\int \frac{d}{dt}(t)\left(\int e^t\,dt\right)dt\right]+C
\displaystyle \text{[using integration by parts]}
\displaystyle \Rightarrow ye^{\tan x}=te^t-\int 1\cdot e^t\,dt+C
\displaystyle \Rightarrow ye^{\tan x}=te^t-e^t+C
\displaystyle \Rightarrow ye^{\tan x}=e^t(t-1)+C
\displaystyle \Rightarrow ye^{\tan x}=e^{\tan x}(\tan x-1)+C\qquad ...(ii)
\displaystyle \text{Also, given that }y(0)=0
\displaystyle \text{On putting }x=0\text{ and }y=0\text{ in Eq. (ii), we get}
\displaystyle 0=-1+C\Rightarrow C=1
\displaystyle \therefore ye^{\tan x}=e^{\tan x}(\tan x-1)+1
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 56. }\text{Find the general solution of the differential equation}
\displaystyle (xy-x^{2})\,dy=y^{2}\,dx \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=\frac{y^2}{xy-x^2}\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{From Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^2x^2}{vx^2-x^2}=\frac{v^2}{v-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^2-v(v-1)}{v-1}=\frac{v}{v-1}
\displaystyle \Rightarrow \frac{v-1}{v}\,dv=\frac{dx}{x}
\displaystyle \Rightarrow \left(1-\frac{1}{v}\right)dv=\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{v-1}{v}\,dv=\int\frac{dx}{x}
\displaystyle \Rightarrow v-\log v=\log x+C
\displaystyle \Rightarrow \frac{y}{x}-\log\left(\frac{y}{x}\right)=\log x+C
\displaystyle \Rightarrow \frac{y}{x}-\log y+\log x=\log x+C
\displaystyle \Rightarrow \frac{y}{x}-\log y=C
\displaystyle \text{which is the required general solution.}
\\

\displaystyle \textbf{Question 57. }\text{Solve the following differential equation.}
\displaystyle xe^{\frac{y}{x}}-y+x\frac{dy}{dx}=0 \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle xe^{y/x}-y+x\frac{dy}{dx}=0
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}=-e^{y/x}\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \therefore \text{From Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}-v=-e^v
\displaystyle \Rightarrow x\frac{dv}{dx}=-e^v
\displaystyle \Rightarrow e^{-v}dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int e^{-v}dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -e^{-v}=-\log x+C
\displaystyle \Rightarrow e^{-v}=\log x-C
\displaystyle \Rightarrow e^{-y/x}=\log x-C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 58. }\text{Find the general solution of the differential equation}
\displaystyle (x^{2}+1)\frac{dy}{dx}+2xy=\sqrt{x^{2}+4}. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (x^2+1)\frac{dy}{dx}+2xy=\sqrt{x^2+4}
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{\sqrt{x^2+4}}{x^2+1}
\displaystyle \text{On comparing with the linear differential equation }
\displaystyle \frac{dy}{dx}+Py=Q,\text{ we get}
\displaystyle P=\frac{2x}{x^2+1}\text{ and }Q=\frac{\sqrt{x^2+4}}{x^2+1}
\displaystyle \therefore \text{Integrating factor (IF)}=e^{\int Pdx}=e^{\int \frac{2x}{x^2+1}dx}
\displaystyle =e^{\log(x^2+1)}=x^2+1
\displaystyle \text{Now, the solution of given differential equation is}
\displaystyle y(\text{IF})=\int(Q\times \text{IF})dx+C
\displaystyle \Rightarrow y(x^2+1)=\int\frac{\sqrt{x^2+4}}{x^2+1}(x^2+1)\,dx+C
\displaystyle \Rightarrow (x^2+1)y=\int\sqrt{x^2+4}\,dx+C
\displaystyle \Rightarrow (x^2+1)y=\frac{x}{2}\sqrt{x^2+4}+2\log\left|x+\sqrt{x^2+4}\right|+C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 59. }\text{Find the general solution of the differential equation:}
\displaystyle \frac{d}{dx}(xy^{2})=2y(1+x^{2}) \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{d(xy^2)}{dx}=2y(1+x^2)
\displaystyle \Rightarrow x\cdot2y\frac{dy}{dx}+y^2=2y(1+x^2)
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{2x}y=\frac{1+x^2}{x}
\displaystyle \text{On comparing with the linear differential equation}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ we get}
\displaystyle P=\frac{1}{2x}\text{ and }Q=\frac{1+x^2}{x}
\displaystyle \therefore \text{Integrating factor}
\displaystyle (\text{IF})=e^{\int Pdx}=e^{\frac{1}{2}\int\frac{dx}{x}}
\displaystyle =e^{\frac{1}{2}\log x}=\sqrt{x}
\displaystyle \text{Now, the solution of given differential equation is}
\displaystyle y(\text{IF})=\int(Q\times \text{IF})dx+C
\displaystyle \Rightarrow \sqrt{x}\,y=\int\left(\frac{1+x^2}{x}\right)\sqrt{x}\,dx+C
\displaystyle =\int\left(\frac{1+x^2}{\sqrt{x}}\right)dx+C
\displaystyle =\int\left(\frac{1}{\sqrt{x}}+x\sqrt{x}\right)dx+C
\displaystyle =\int\left(x^{-1/2}+x^{3/2}\right)dx+C
\displaystyle \Rightarrow \sqrt{x}\,y=2\sqrt{x}+\frac{2}{5}x^{5/2}+C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 60. }\text{Find the general solution of the differential equation}
\displaystyle \log\left(\frac{dy}{dx}\right)=ax+by. \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation}
\displaystyle \log\left(\frac{dy}{dx}\right)=ax+by
\displaystyle \Rightarrow \frac{dy}{dx}=e^{ax+by}\qquad \left[\because \log m=n\Rightarrow e^n=m\right]
\displaystyle \Rightarrow \frac{dy}{dx}=e^{ax}\cdot e^{by}
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{1}{e^{by}}\,dy=e^{ax}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int e^{-by}dy=\int e^{ax}dx
\displaystyle \Rightarrow \frac{e^{-by}}{-b}=\frac{e^{ax}}{a}+C
\displaystyle \therefore \text{General solution is }\frac{e^{-by}}{-b}-\frac{e^{ax}}{a}=C.
\\

\displaystyle \textbf{Question 61. }\text{Find the particular solution of the differential equation}
\displaystyle x\frac{dy}{dx}-y=x^{2}e^{\frac{y}{x}},\text{ given, }y(1)=0. \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation is }x\frac{dy}{dx}-y=x^2e^x
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}=xe^x\qquad ...(i)
\displaystyle \text{Eq. (i) is a linear differential equation.}
\displaystyle \text{On comparing Eq. (i) with }\frac{dy}{dx}+Py=Q,\text{ we get}
\displaystyle P=-\frac{1}{x}\text{ and }Q=xe^x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int -\frac{1}{x}dx}=e^{-\log|x|}=\frac{1}{x}
\displaystyle \text{Particular solution is}
\displaystyle y\cdot \text{IF}=\int Q\cdot \text{IF}\,dx+C
\displaystyle \Rightarrow y\cdot \frac{1}{x}=\int xe^x\cdot \frac{1}{x}\,dx+C
\displaystyle \Rightarrow \frac{y}{x}=\int e^x\,dx+C
\displaystyle \Rightarrow \frac{y}{x}=e^x+C
\displaystyle \Rightarrow y=x(e^x+C)\qquad ...(i)
\displaystyle \text{At }x=1,\text{ then }y=0
\displaystyle \Rightarrow 0=e^1+C
\displaystyle \Rightarrow C=-e
\displaystyle \therefore y=xe^x-xe
\\

\displaystyle \textbf{Question 62. }\text{Find the general solution of the differential equation}
\displaystyle x\frac{dy}{dx}=y(\log y-\log x+1). \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}=y(\log y-\log x+1)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left(\log\frac{y}{x}+1\right)\qquad ...(i)
\displaystyle \text{On putting }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v(\log v+1)
\displaystyle \Rightarrow x\frac{dv}{dx}=v\log v
\displaystyle \Rightarrow \frac{1}{v\log v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1}{v\log v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log|\log v|=\log|x|+C
\displaystyle \therefore \text{Required solution is}
\displaystyle \log\left|\log\left(\frac{y}{x}\right)\right|=\log|x|+C
\\

\displaystyle \textbf{Question 63. }\text{Find the integrating factor of the differential equation }
\displaystyle x\frac{dy}{dx}+2y=x^{2}. \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x\frac{dy}{dx}+2y=x^2
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, }P=\frac{2}{x}\text{ and }Q=x
\displaystyle \therefore \text{Integrating factor}=e^{\int \frac{2}{x}dx}=e^{2\log x}=e^{\log x^2}=x^2
\\

\displaystyle \textbf{Question 64. }\text{Solve the differential equation }
\displaystyle x\sin\left(\frac{y}{x}\right)\frac{dy}{dx}+x-y\sin\left(\frac{y}{x}\right)=0.
\displaystyle \text{Given that }x=1,\text{ when }y=\frac{\pi}{2}. \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation can be written as}
\displaystyle \frac{dy}{dx}=\frac{y}{x}-\frac{1}{\sin\frac{y}{x}}\qquad ...(i)
\displaystyle \text{Let }F(x,y)=\frac{y}{x}-\frac{1}{\sin\frac{y}{x}}
\displaystyle \text{Now, }F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}-\frac{1}{\sin\left(\frac{\lambda y}{\lambda x}\right)}=\lambda^0\left[\frac{y}{x}-\frac{1}{\sin\frac{y}{x}}\right]
\displaystyle =\lambda^0F(x,y)
\displaystyle \text{It is a homogeneous differential equation.}
\displaystyle \text{Now, on putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \therefore \text{From Eq. (i), we get }v+x\frac{dv}{dx}=v-\frac{1}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{1}{\sin v}\Rightarrow \sin v\,dv=-\frac{1}{x}dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle -\cos v=-\log|x|-C
\displaystyle \Rightarrow -\cos\left(\frac{y}{x}\right)=-\log|x|-C
\displaystyle \Rightarrow \cos\left(\frac{y}{x}\right)=\log|x|+C\qquad ...(ii)
\displaystyle \text{Given that }x=1,\text{ when }y=\frac{\pi}{2}
\displaystyle \therefore \cos\frac{\pi}{2}=\log|1|+C\Rightarrow 0=0+C
\displaystyle \Rightarrow C=0
\displaystyle \text{On putting }C=0\text{ in Eq. (ii), we get}
\displaystyle \cos\left(\frac{y}{x}\right)=\log|x|
\\

\displaystyle \textbf{Question 65. }\text{For the differential equation given below, find a}
\displaystyle \text{particular solution satisfying the given condition}
\displaystyle (x+1)\frac{dy}{dx}-2e^{-y}+1;\ y=0\text{ when }x=0. \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (x+1)\frac{dy}{dx}=2e^{-y}+1
\displaystyle \Rightarrow (x+1)\frac{dy}{dx}=\frac{2+e^y}{e^y}
\displaystyle \Rightarrow \frac{e^y}{e^y+2}\,dy=\frac{dx}{x+1}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{e^y}{e^y+2}\,dy=\int\frac{dx}{x+1}
\displaystyle \Rightarrow \log(e^y+2)=\log(x+1)+\log C
\displaystyle \Rightarrow \log(e^y+2)=\log C(x+1)
\displaystyle \Rightarrow e^y+2=C(x+1)\qquad ...(i)
\displaystyle \text{Also, given }y=0,\text{ when }x=0
\displaystyle \text{On putting }x=0\text{ and }y=0\text{ in Eq. (i), we get}
\displaystyle e^0+2=C(0+1)\Rightarrow C=1+2=3
\displaystyle \text{On putting }C\text{ in Eq. (i), we get}
\displaystyle e^y+2=3(x+1)
\displaystyle \Rightarrow e^y=3x+3-2
\displaystyle \Rightarrow e^y=3x+1\Rightarrow y=\log(3x+1)
\\

\displaystyle \textbf{Question 66. }\text{Find the general solution of the differential equation}
\displaystyle \frac{dy}{dx}+\frac{1}{x}=\frac{e^{y}}{x^{2}}. \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{dy}{dx}+\frac{1}{x}=\frac{e^y}{x}\Rightarrow \frac{dy}{dx}=\frac{e^y-1}{x}
\displaystyle \Rightarrow \frac{dy}{e^y-1}=\frac{dx}{x}
\displaystyle \text{On integrating, we get}
\displaystyle \int\frac{dy}{e^y-1}=\int\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{e^{-y}dy}{1-e^{-y}}=\int\frac{dx}{x}
\displaystyle \Rightarrow -\log(1-e^{-y})=\log x+C
\displaystyle \Rightarrow \log(1-e^{-y})=\log Cx
\displaystyle \Rightarrow 1-e^{-y}=Cx
\displaystyle \Rightarrow e^{-y}=1-Cx
\displaystyle \Rightarrow -y=\log(1-Cx)
\displaystyle \Rightarrow y+\log(1-Cx)=0
\\

\displaystyle \textbf{Question 67. }\text{Find the general solution of the differential equation}
\displaystyle \frac{dy}{dx}=e^{x+y}. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is}
\displaystyle \frac{dy}{dx}=e^{x+y}\Rightarrow \frac{dy}{dx}=e^x\cdot e^y
\displaystyle \Rightarrow e^{-y}dy=e^xdx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int e^{-y}dy=\int e^xdx
\displaystyle \Rightarrow -e^{-y}=e^x+C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 68. }\text{Solve the differential equation}
\displaystyle (x+1)\frac{dy}{dx}=2e^{-y}-1;\ y(0)=0 \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }(x+1)\frac{dy}{dx}=2e^{-y}-1
\displaystyle \Rightarrow (x+1)dy=(2e^{-y}-1)dx
\displaystyle \Rightarrow \frac{1}{x+1}dx=\frac{1}{2e^{-y}-1}dy
\displaystyle \text{[separating the variables]}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1}{x+1}dx=\int\frac{1}{2e^{-y}-1}dy
\displaystyle \Rightarrow \int\frac{1}{x+1}dx=\int\frac{e^y}{2-e^y}dy
\displaystyle \Rightarrow \int\frac{1}{x+1}dx=-\int\frac{e^y}{e^y-2}dy
\displaystyle \Rightarrow \log|x+1|=-\log|e^y-2|+\log C
\displaystyle \Rightarrow \log|x+1|+\log|e^y-2|=\log C
\displaystyle \Rightarrow \log|(x+1)(e^y-2)|=\log C
\displaystyle \Rightarrow |(x+1)(e^y-2)|=C\qquad ...(i)
\displaystyle \text{It is given that }y(0)=0,\text{ i.e. }y=0\text{ when }x=0.
\displaystyle \text{On putting }x=0\text{ and }y=0\text{ in Eq. (i), we get}
\displaystyle |(0+1)(1-2)|=C
\displaystyle \Rightarrow C=1
\displaystyle \text{Hence, }(x+1)(e^y-2)=\pm1
\displaystyle \Rightarrow e^y-2=\frac{1}{x+1}
\displaystyle \Rightarrow e^y=2+\frac{1}{x+1}
\displaystyle \Rightarrow y=\log\left(2+\frac{1}{x+1}\right)
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 69. }\text{Solve the following differential equation.}
\displaystyle x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx,\text{ given that }y=0\text{ when }x=1. \hspace{0.2cm}\text{[CBSE 2019; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx
\displaystyle \Rightarrow (y+\sqrt{x^2+y^2})\,dx=x\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v+\sqrt{1+v^2}
\displaystyle \Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}
\displaystyle \Rightarrow \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{dv}{\sqrt{1+v^2}}=\int\frac{dx}{x}
\displaystyle \Rightarrow \log\left|v+\sqrt{1+v^2}\right|=\log|x|+C
\displaystyle \Rightarrow \log\left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log|x|+C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow \log\left|y+\sqrt{x^2+y^2}\right|-\log|x|=\log|x|+C
\displaystyle \Rightarrow \log\left(\frac{y+\sqrt{x^2+y^2}}{x^2}\right)=C
\displaystyle \Rightarrow \frac{y+\sqrt{x^2+y^2}}{x^2}=e^C
\displaystyle \Rightarrow y+\sqrt{x^2+y^2}=Ax^2\qquad ...(ii)
\displaystyle \text{where }A=e^C
\displaystyle \text{Now, as }y=0,\text{ when }x=1
\displaystyle \Rightarrow 0+\sqrt{1^2+0^2}=A\cdot1
\displaystyle \Rightarrow A=1
\displaystyle \text{On putting the value of }A\text{ in Eq. (ii), we get}
\displaystyle y+\sqrt{x^2+y^2}=x^2
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 70. }\text{Solve the differential equation }   (1+x^{2})\frac{dy}{dx}+2xy-4x^{2}=0,
\displaystyle \text{subject to the initial condition }y(0)=0. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1+x^2)\frac{dy}{dx}+2xy-4x^2=0
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{4x^2}{1+x^2}
\displaystyle \text{which is the equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{2x}{1+x^2}\text{ and }Q=\frac{4x^2}{1+x^2}
\displaystyle \text{Now, IF}=e^{\int\frac{2x}{1+x^2}dx}=e^{\log(1+x^2)}=1+x^2
\displaystyle \text{The general solution is}
\displaystyle y(1+x^2)=\int(1+x^2)\frac{4x^2}{1+x^2}\,dx+C
\displaystyle \Rightarrow (1+x^2)y=\int4x^2\,dx+C
\displaystyle \Rightarrow (1+x^2)y=\frac{4x^3}{3}+C
\displaystyle \Rightarrow y=\frac{4x^3}{3(1+x^2)}+C(1+x^2)^{-1}\qquad ...(i)
\displaystyle \text{Now, }y(0)=0
\displaystyle \Rightarrow 0=\frac{4\cdot0^3}{3(1+0^2)}+C(1+0^2)^{-1}
\displaystyle \Rightarrow C=0
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle y=\frac{4x^3}{3(1+x^2)}
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 71. }\text{Solve the differential equation }   \frac{dy}{dx}-\frac{2x}{1+x^{2}}y=x^{2}+2. \\ \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2+2\qquad ...(i)
\displaystyle \text{This is a linear differential equation with}
\displaystyle P=-\frac{2x}{1+x^2}\text{ and }Q=x^2+2
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{-\int\frac{2x}{1+x^2}dx}
\displaystyle =e^{-\log(1+x^2)}=\frac{1}{1+x^2}
\displaystyle \therefore \frac{y}{1+x^2}=\int\frac{x^2+2}{1+x^2}\,dx+C
\displaystyle =\int\left(1+\frac{1}{1+x^2}\right)dx+C
\displaystyle \Rightarrow \frac{y}{1+x^2}=x+\tan^{-1}x+C
\displaystyle \Rightarrow y=(1+x^2)(x+\tan^{-1}x)+C(1+x^2)
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\displaystyle \textbf{Question 72. }\text{Solve the following differential equation. }   x\frac{dy}{dx}=y-x\tan\left(\frac{y}{x}\right). \\ \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}=y-x\tan\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\tan\left(\frac{y}{x}\right)\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v-\tan v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\tan v
\displaystyle \Rightarrow \frac{dv}{\tan v}=-\frac{dx}{x}
\displaystyle \Rightarrow \cot v\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\cot v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \log|\sin v|=-\log|x|+C
\displaystyle \Rightarrow \log|x\sin v|=C
\displaystyle \Rightarrow x\sin v=A
\displaystyle \Rightarrow x\sin\left(\frac{y}{x}\right)=A
\displaystyle \Rightarrow y=x\sin^{-1}\left(\frac{A}{x}\right)
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 73. }\text{Solve the differential equation }   \frac{dy}{dx}=\left[\frac{x+y\cos x}{1+\sin x}\right]. \\ \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{dy}{dx}=-\frac{x}{1+\sin x}-\frac{y\cos x}{1+\sin x}
\displaystyle \text{or, }\frac{dy}{dx}+\frac{y\cos x}{1+\sin x}=-\frac{x}{1+\sin x}\qquad ...(i)
\displaystyle \text{which is in the linear form, }\frac{dy}{dx}+Py=Q,
\displaystyle \text{where }P=\frac{\cos x}{1+\sin x}\text{ and }Q=-\frac{x}{1+\sin x}
\displaystyle \text{Now, IF}=e^{\int\frac{\cos x}{1+\sin x}dx}=e^{\log(1+\sin x)}=1+\sin x
\displaystyle \text{and the solution is}
\displaystyle y(1+\sin x)=\int(-x)\,dx+C
\displaystyle \Rightarrow y(1+\sin x)=-\frac{x^2}{2}+C
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\displaystyle \textbf{Question 74. }\text{Solve the differential equation }  \cos\left(\frac{dy}{dx}\right)=a,\ (a\in R). \hspace{0.2cm}\text{[CBSE 2018 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation is }\cos\left(\frac{dy}{dx}\right)=a
\displaystyle \text{which can be rewritten as }\frac{dy}{dx}=\cos^{-1}a
\displaystyle \Rightarrow dy=\cos^{-1}a\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dy=\int \cos^{-1}a\,dx
\displaystyle \Rightarrow y=\cos^{-1}a\cdot x+C
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 75. }\text{Find the particular solution of the differential equation}
\displaystyle e^{x}\tan y\,dx+(2-e^{x})\sec^{2}y\,dy=0,   \text{ given that }y=\frac{\pi}{4}\text{ when }x=0. \\ \hspace{2.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle e^x\tan y\,dx+(2-e^x)\sec^2y\,dy=0
\displaystyle \text{which can be rewritten as}
\displaystyle e^x\tan y\,dx=(e^x-2)\sec^2y\,dy
\displaystyle \Rightarrow \frac{\sec^2y}{\tan y}\,dy=\frac{e^x}{e^x-2}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{\sec^2y}{\tan y}\,dy=\int\frac{e^x}{e^x-2}\,dx
\displaystyle \Rightarrow \log|\tan y|=\log|e^x-2|+C
\displaystyle \Rightarrow \log|\tan y|-\log|e^x-2|=C
\displaystyle \Rightarrow \log\left|\frac{\tan y}{e^x-2}\right|=C
\displaystyle \Rightarrow \frac{\tan y}{e^x-2}=e^C
\displaystyle \Rightarrow \tan y=e^C(e^x-2)
\displaystyle \text{Now, it is given that }y=\frac{\pi}{4}\text{ when }x=0
\displaystyle \therefore \tan\frac{\pi}{4}=e^C(e^0-2)
\displaystyle \Rightarrow 1=e^C(1-2)\Rightarrow e^C=-1
\displaystyle \text{Thus, the particular solution of the given differential equation is }\tan y=2-e^x.
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\displaystyle \textbf{Question 76. }\text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}+2y\tan x=\sin x,\text{ given that }y=0,\text{ when }x=\frac{\pi}{3}. \\ \hspace{2.2cm}\text{[CBSE 2018; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}+2y\tan x=\sin x
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, }P=2\tan x\text{ and }Q=\sin x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{2\int\tan xdx}=e^{2\log|\sec x|}
\displaystyle =e^{\log\sec^2x}=\sec^2x
\displaystyle \text{The general solution is given by}
\displaystyle y\times \text{IF}=\int(Q\times \text{IF})\,dx+C\qquad ...(i)
\displaystyle \Rightarrow y\sec^2x=\int(\sin x\cdot\sec^2x)\,dx+C
\displaystyle \Rightarrow y\sec^2x=\int\sin x\cdot\frac{1}{\cos^2x}\,dx+C
\displaystyle \Rightarrow y\sec^2x=\int\tan x\sec x\,dx+C
\displaystyle \Rightarrow y\sec^2x=\sec x+C\qquad ...(ii)
\displaystyle \text{Also, given that }y=0,\text{ when }x=\frac{\pi}{3}
\displaystyle \text{On putting }y=0\text{ and }x=\frac{\pi}{3}\text{ in Eq. (ii), we get}
\displaystyle 0\times\sec^2\frac{\pi}{3}=\sec\frac{\pi}{3}+C
\displaystyle \Rightarrow 0=2+C\Rightarrow C=-2
\displaystyle \text{On putting the value of }C\text{ in Eq. (ii), we get}
\displaystyle y\sec^2x=\sec x-2
\displaystyle \Rightarrow y=\cos x-2\cos^2x
\displaystyle \text{which is the required particular solution of the given differential equation.}
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\displaystyle \textbf{Question 77. }\text{Solve the differential equation }  (x^{2}-y^{2})\,dx+2xydy=0. \\ \hspace{2.2cm}\text{[CBSE 2018 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation}
\displaystyle (x^2-y^2)\,dx+2xy\,dy=0
\displaystyle \text{which can be rewritten as}
\displaystyle (x^2-y^2)\,dx=-2xy\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^2-y^2}{-2xy}=\frac{y^2-x^2}{2xy}=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)}\qquad ...(i)
\displaystyle \therefore \text{In RHS, degree of numerator and denominator is same.}
\displaystyle \therefore \text{It is a homogeneous differential equation and can be written as}
\displaystyle \frac{dy}{dx}=f\left(\frac{y}{x}\right)
\displaystyle \text{Now, put }y=vx\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^2-1}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=-\frac{v^2+1}{2v}
\displaystyle \Rightarrow \frac{2v}{v^2+1}\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{2v}{v^2+1}\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \log|v^2+1|=-\log|x|+\log C
\displaystyle \Rightarrow \log\left|\frac{y^2}{x^2}+1\right|=-\log|x|+\log C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow \log\left|\frac{y^2+x^2}{x^2}\cdot x\right|=\log C
\displaystyle \Rightarrow \frac{y^2+x^2}{x}=C
\displaystyle \Rightarrow y^2+x^2=Cx,
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 78. }\text{Find the particular solution of the differential equation}
\displaystyle (1+x^{2})\frac{dy}{dx}+2xy=\frac{1}{1+x^{2}},\text{ given that }y=0,\text{ when }x=1.
\displaystyle \hspace{2.2cm}\text{[CBSE 2018 C; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (x^2+1)\frac{dy}{dx}+2xy=\frac{1}{x^2+1}
\displaystyle \text{On dividing both sides by }(x^2+1),\text{ we get}
\displaystyle \frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{1}{(x^2+1)^2}\qquad ...(i)
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ where }P=\frac{2x}{x^2+1}\text{ and }Q=\frac{1}{(x^2+1)^2}
\displaystyle \text{Now, integrating factor, IF}=e^{\int Pdx}
\displaystyle =e^{\int\frac{2x}{x^2+1}dx}
\displaystyle =e^{\log|x^2+1|}=x^2+1
\displaystyle \left[\text{put }x^2+1=t\Rightarrow 2x\,dx=dt;\ \text{then }\int\frac{2x}{x^2+1}dx=\int\frac{1}{t}dt=\log|t|=\log|x^2+1|\right]
\displaystyle \text{So, the required general solution is}
\displaystyle y\times IF=\int(Q\times IF)\,dx+C
\displaystyle \Rightarrow y(x^2+1)=\int\frac{1}{(x^2+1)^2}\times(x^2+1)\,dx+C
\displaystyle \Rightarrow y(x^2+1)=\int\frac{1}{x^2+1}\,dx+C
\displaystyle \Rightarrow y(x^2+1)=\tan^{-1}x+C\qquad ...(ii)
\displaystyle \text{when }x=1,\text{ then }y=0
\displaystyle \therefore 0=\tan^{-1}1+C\Rightarrow C=-\frac{\pi}{4}
\displaystyle \text{Now, }y(x^2+1)=\tan^{-1}x-\frac{\pi}{4}\qquad [\text{from Eq. (ii)}]
\displaystyle \text{which is the required differential equation.}
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\displaystyle \textbf{Question 79. }\text{Show that the family of curves for which }\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}\text{ is given by }x^{2}-y^{2}=cx.\hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{1+v^{2}}{2v}
\displaystyle x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v
\displaystyle x\frac{dv}{dx}=\frac{1-v^{2}}{2v}
\displaystyle \frac{2v}{1-v^{2}}\,dv=\frac{dx}{x}
\displaystyle -\log(1-v^{2})=\log x+C
\displaystyle \log(1-v^{2})=-\log x+C
\displaystyle 1-v^{2}=\frac{C}{x}
\displaystyle 1-\frac{y^{2}}{x^{2}}=\frac{C}{x}
\displaystyle \frac{x^{2}-y^{2}}{x^{2}}=\frac{C}{x}
\displaystyle x^{2}-y^{2}=Cx
\displaystyle \therefore x^{2}-y^{2}=cx
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\displaystyle \textbf{Question 80. }\text{Prove that }x^{2}-y^{2}=c(x^{2}+y^{2})^{2}\text{ is the general}
\displaystyle \text{solution of the differential equation }(x^{3}-3xy^{2})dx=(y^{3}-3x^{2}y)dy,
\displaystyle \text{where }c\text{ is a parameter.} \hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation can be rewritten as}
\displaystyle \frac{dy}{dx}=\frac{x^3-3xy^2}{y^3-3x^2y}\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation, so on putting }y=vx
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Then, Eq. (i) becomes}
\displaystyle v+x\frac{dv}{dx}=\frac{x^3-3x(vx)^2}{(vx)^3-3x^2(vx)}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v=\frac{1-v^4}{v^3-3v}
\displaystyle \Rightarrow \left(\frac{v^3-3v}{1-v^4}\right)dv=\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\left(\frac{v^3-3v}{1-v^4}\right)dv=\int\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{v^3}{1-v^4}dv-3\int\frac{v}{1-v^4}dv=\log x+\log C
\displaystyle \Rightarrow -\frac{1}{4}\log(1-v^4)-\frac{3}{4}\log\left|\frac{1+v^2}{1-v^2}\right|=\log x+\log C
\displaystyle \Rightarrow -\frac{1}{4}\log\left[(1-v^4)\left(\frac{1+v^2}{1-v^2}\right)^3\right]=\log(Cx)
\displaystyle \Rightarrow \log\left[\frac{(1+v^2)^4}{(1-v^2)^2}\right]^{-1/4}=\log(Cx)
\displaystyle \Rightarrow \frac{(1-v^2)^2}{(1+v^2)^4}=(Cx)^4
\displaystyle \Rightarrow \frac{\left(1-\frac{y^2}{x^2}\right)^2}{\left(1+\frac{y^2}{x^2}\right)^4}=\frac{1}{C_1^4x^4}\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow \frac{x^4(x^2-y^2)^2}{(x^2+y^2)^4}=\frac{1}{C_1^4x^4}
\displaystyle \Rightarrow (x^2-y^2)^2=C^2(x^2+y^2)^2
\displaystyle \Rightarrow x^2-y^2=C_1(x^2+y^2),\qquad [\text{taking square root}]
\displaystyle \text{where }C_1=C^2
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 81. }\text{Solve the differential equation } x\frac{dy}{dx}+y=x\cos x+\sin x,
\displaystyle \text{ given that }y=1\text{ when }x=\frac{\pi}{2}.   \hspace{2.2cm}\text{[CBSE 2017; CBSE 2014 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}+y=x\cos x+\sin x
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{x}=\cos x+\frac{\sin x}{x}
\displaystyle \text{which is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{here }P=\frac{1}{x}\text{ and }Q=\cos x+\frac{\sin x}{x}
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\frac{1}{x}dx}=e^{\log|x|}=x
\displaystyle \text{The general solution is given by}
\displaystyle y\times\text{IF}=\int(Q\times\text{IF})\,dx+C
\displaystyle \Rightarrow yx=\int\left(\cos x+\frac{\sin x}{x}\right)x\,dx+C
\displaystyle \Rightarrow xy=\int(x\cos x+\sin x)\,dx+C
\displaystyle \Rightarrow xy=\int x\cos x\,dx+\int\sin x\,dx+C
\displaystyle \Rightarrow xy=x\sin x-\int\left[\frac{d}{dx}(x)\int\cos x\,dx\right]dx+\int\sin x\,dx+C
\displaystyle \text{[using integration by parts]}
\displaystyle \Rightarrow xy=x\sin x-\int\sin x\,dx-\cos x+C
\displaystyle \Rightarrow xy=x\sin x+\cos x-\cos x+C
\displaystyle \Rightarrow xy=x\sin x+C
\displaystyle \Rightarrow y=\sin x+\frac{C}{x}\qquad ...(i)
\displaystyle \text{Also, given that at }x=\frac{\pi}{2},\ y=1
\displaystyle \text{On putting }x=\frac{\pi}{2}\text{ and }y=1\text{ in Eq. (i), we get}
\displaystyle 1=1+C\cdot\frac{2}{\pi}\Rightarrow C=0
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle y=\sin x
\displaystyle \text{which is the required solution of given differential equation.}
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\displaystyle \textbf{Question 82. }\text{Solve the differential equation }   (\tan^{-1}x-y)\,dx=(1+x^{2})\,dy.  \\ \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(\tan^{-1}x-y)\,dx=(1+x^2)\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{\tan^{-1}x-y}{1+x^2}
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{\tan^{-1}x}{1+x^2}\qquad ...(i)
\displaystyle \text{which is a linear differential equation of the form }\frac{dy}{dx}+Py=Q,
\displaystyle \text{here }P=\frac{1}{1+x^2}\text{ and }Q=\frac{\tan^{-1}x}{1+x^2}
\displaystyle \text{Now, IF}=e^{\int Pdx}=e^{\int\frac{1}{1+x^2}dx}=e^{\tan^{-1}x}
\displaystyle \text{The general solution is given by}
\displaystyle y\cdot\text{IF}=\int Q\cdot\text{IF}\,dx+C
\displaystyle \Rightarrow ye^{\tan^{-1}x}=\int\frac{\tan^{-1}x}{1+x^2}\cdot e^{\tan^{-1}x}dx+C
\displaystyle \text{On putting }\tan^{-1}x=t\Rightarrow \frac{1}{1+x^2}dx=dt
\displaystyle \therefore ye^{\tan^{-1}x}=\int t\cdot e^t\,dt+C
\displaystyle =t\cdot e^t-\int e^t\,dt+C
\displaystyle \text{[using integration by parts]}
\displaystyle \Rightarrow ye^{\tan^{-1}x}=te^t-e^t+C
\displaystyle \Rightarrow ye^{\tan^{-1}x}=\tan^{-1}x\cdot e^{\tan^{-1}x}-e^{\tan^{-1}x}+C
\displaystyle \Rightarrow ye^{\tan^{-1}x}=(\tan^{-1}x-1)e^{\tan^{-1}x}+C
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\displaystyle \textbf{Question 83. }\text{Find the general solution of the differential equation}
\displaystyle \frac{dy}{dx}-y=\sin x. \hspace{2.2cm}\text{[CBSE 2017, 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{dy}{dx}-y=\sin x,\text{ which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ here }P=-1\text{ and }Q=\sin x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int(-1)dx}=e^{-x}
\displaystyle \text{Now, the general solution of given differential equation is given by}
\displaystyle y\cdot(\text{IF})=\int(\text{IF})\cdot Q\,dx+C
\displaystyle \Rightarrow ye^{-x}=\int e^{-x}\sin x\,dx+C\qquad ...(i)
\displaystyle \text{Let }I=\int e^{-x}\sin x\,dx\qquad ...(ii)
\displaystyle \text{By using the method of integration by parts, we get}
\displaystyle I=\sin x\cdot\frac{e^{-x}}{-1}-\int\cos x\cdot\frac{e^{-x}}{-1}\,dx
\displaystyle =-\sin x\,e^{-x}+\int e^{-x}\cos x\,dx
\displaystyle \text{Again, by using integration by parts, we get}
\displaystyle I=-\sin x\,e^{-x}+\cos x\cdot\frac{e^{-x}}{-1}-\int(-\sin x)\cdot\frac{e^{-x}}{-1}\,dx
\displaystyle =-\sin x\,e^{-x}-\cos x\,e^{-x}-I\qquad [\text{from Eq. (ii)}]
\displaystyle \Rightarrow 2I=-e^{-x}(\sin x+\cos x)
\displaystyle \Rightarrow I=-\frac{e^{-x}}{2}(\sin x+\cos x)
\displaystyle \text{Then, from Eq. (i), we get}
\displaystyle ye^{-x}=-\frac{e^{-x}}{2}(\sin x+\cos x)+C
\displaystyle \Rightarrow y=-\frac{1}{2}(\sin x+\cos x)+Ce^x
\\

\displaystyle \textbf{Question 84. }\text{Find the particular solution of the differential equation }(x-y)\frac{dy}{dx}=x+2y,\text{ given that when }x=1,\ y=0.\hspace{2.2cm}\text{[CBSE 2017, 2013 C]}
\displaystyle \text{Answer:}
\displaystyle (x-y)\frac{dy}{dx}=x+2y
\displaystyle \frac{dy}{dx}=\frac{x+2y}{x-y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{1+2v}{1-v}
\displaystyle x\frac{dv}{dx}=\frac{1+2v}{1-v}-v
\displaystyle x\frac{dv}{dx}=\frac{v^{2}+v+1}{1-v}
\displaystyle \frac{1-v}{v^{2}+v+1}\,dv=\frac{dx}{x}
\displaystyle \int \frac{1-v}{v^{2}+v+1}\,dv=\int \frac{dx}{x}
\displaystyle -\frac{1}{2}\log(v^{2}+v+1)+\sqrt{3}\tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)=\log x+C
\displaystyle \text{Putting }v=\frac{y}{x},
\displaystyle -\frac{1}{2}\log\left(\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right)+\sqrt{3}\tan^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log x+C
\displaystyle \text{Given, }x=1,\ y=0
\displaystyle \therefore v=0
\displaystyle -\frac{1}{2}\log1+\sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=0+C
\displaystyle C=\frac{\sqrt{3}\pi}{6}
\displaystyle \therefore -\frac{1}{2}\log\left(\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right)+\sqrt{3}\tan^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log x+\frac{\sqrt{3}\pi}{6}
\\

\displaystyle \textbf{Question 85. }\text{Find the general solution of the differential equation}
\displaystyle y\,dx-(x+2y^{2})\,dy=0. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }ydx-(x+2y^2)dy=0
\displaystyle \Rightarrow y\frac{dx}{dy}-x-2y^2=0
\displaystyle \Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q,\text{ here }P=-\frac{1}{y}\text{ and }Q=2y
\displaystyle \therefore \text{IF}=e^{\int Pdy}=e^{\int-\frac{1}{y}dy}=e^{-\log y}=\frac{1}{y}
\displaystyle \text{Hence, required general solution of the differential equation is}
\displaystyle x\cdot\text{IF}=\int(Q\cdot\text{IF})\,dy+C
\displaystyle \Rightarrow x\times\frac{1}{y}=\int2y\times\frac{1}{y}\,dy+C
\displaystyle \Rightarrow \frac{x}{y}=2y+C
\displaystyle \Rightarrow x=2y^2+Cy
\\

\displaystyle \textbf{Question 86. }\text{Find the particular solution of the differential equation}
\displaystyle (1-y^{2})(1+\log|x|)dx+2xydy=0\text{ given that }y=0,   \text{when }x=1. \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1-y^2)(1+\log|x|)\,dx+2xydy=0
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{(1+\log|x|)}{x}\,dx+\frac{2y}{1-y^2}\,dy=0
\displaystyle \text{[dividing both sides by }x(1-y^2)\text{]}
\displaystyle \text{On integrating, we get}
\displaystyle \int\left(\frac{1}{x}+\frac{\log|x|}{x}\right)dx+\int\frac{2y}{1-y^2}dy=0
\displaystyle \Rightarrow \log|x|+\frac{(\log|x|)^2}{2}-\log|1-y^2|=\log C\qquad ...(i)
\displaystyle \text{Also, given }y=0\text{ and }x=1
\displaystyle \Rightarrow \log1+\frac{(\log1)^2}{2}-\log|1-0|=\log C
\displaystyle \Rightarrow 0+0-0=\log C\Rightarrow \log C=0
\displaystyle \text{On putting }\log C=0\text{ in Eq. (i), we get}
\displaystyle \log|x|+\frac{(\log|x|)^2}{2}-\log|1-y^2|=0
\\

\displaystyle \textbf{Question 87. }\text{Find the general solution of the following differential}
\displaystyle (1+y^{2})+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0
\displaystyle \text{It can be rewritten as}
\displaystyle (1+y^2)\frac{dx}{dy}+x=e^{\tan^{-1}y}
\displaystyle \text{or}
\displaystyle \frac{dx}{dy}+\frac{1}{1+y^2}x=\frac{e^{\tan^{-1}y}}{1+y^2}
\displaystyle \text{[dividing both sides by }(1+y^2)\text{]}
\displaystyle \text{It is a linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q
\displaystyle \text{Here, }P=\frac{1}{1+y^2}\text{ and }Q=\frac{e^{\tan^{-1}y}}{1+y^2}
\displaystyle \text{Now, integrating factor, IF}=e^{\int Pdy}
\displaystyle =e^{\int\frac{1}{1+y^2}dy}=e^{\tan^{-1}y}
\displaystyle \text{The general solution of linear differential equation is given by}
\displaystyle x\cdot \text{IF}=\int(Q\times \text{IF})dy+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\int\frac{e^{\tan^{-1}y}\cdot e^{\tan^{-1}y}}{1+y^2}dy+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^2}dy+C\qquad ...(i)
\displaystyle \text{On putting }\tan^{-1}y=t\Rightarrow \frac{1}{1+y^2}dy=dt\text{ in Eq. (i), we get}
\displaystyle xe^{\tan^{-1}y}=\int e^{2t}dt+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\frac{e^{2t}}{2}+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\frac{e^{2\tan^{-1}y}}{2}+C
\\

\displaystyle \textbf{Question 88. }\text{Find the particular solution of the differential equation}
\displaystyle 2ye^{x/y}dx+(y-2xe^{x/y})dy=0,\text{ given that }x=0,\text{ when}
\displaystyle y=1. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Or}
\displaystyle \text{Show that the differential equation }   2ye^{x/y}dx+(y-2xe^{x/y})dy=0\text{ is homogeneous.}
\displaystyle \text{Find the particular solution of this differential equation, given that }x=0,
\displaystyle \text{when }y=1. \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle 2ye^{x/y}dx+(y-2xe^{x/y})dy=0
\displaystyle \text{It can be written as}
\displaystyle \frac{dx}{dy}=\frac{2xe^{x/y}-y}{2ye^{x/y}}\qquad ...(i)
\displaystyle \text{Let }F(x,y)=\frac{2xe^{x/y}-y}{2ye^{x/y}}
\displaystyle \text{On replacing }x\text{ by }\lambda x\text{ and }y\text{ by }\lambda y\text{ both sides, we get}
\displaystyle F(\lambda x,\lambda y)=\frac{2\lambda xe^{\lambda x/\lambda y}-\lambda y}{2\lambda ye^{\lambda x/\lambda y}}
\displaystyle =\lambda^0F(x,y)
\displaystyle \text{Thus, }F(x,y)\text{ is a homogeneous function of degree zero.}
\displaystyle \therefore \text{the given differential equation is a homogeneous differential equation.}
\displaystyle \text{To solve it, put }x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}\text{ in Eq. (i), we get}
\displaystyle v+y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}
\displaystyle \Rightarrow y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}-v=-\frac{1}{2e^v}
\displaystyle \Rightarrow 2e^v\,dv=-\frac{dy}{y}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int2e^v\,dv=-\int\frac{dy}{y}
\displaystyle \Rightarrow 2e^v+\log|y|=C\qquad ...(ii)
\displaystyle \text{Also, given that }x=0,\text{ when }y=1
\displaystyle \text{On substituting }x=0\text{ and }y=1\text{ in Eq. (ii), we get}
\displaystyle 2e^0+\log|1|=C\Rightarrow C=2
\displaystyle \text{On substituting the value of }C\text{ in Eq. (ii), we get}
\displaystyle 2e^{x/y}+\log|y|=2
\displaystyle \text{which is the particular solution of the given differential equation.}
\\

\displaystyle \textbf{Question 89. }\text{Solve the differential equation }   y+y\frac{dx}{dy}=x-y\frac{dx}{dy}. \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }y+x\frac{dy}{dx}=x-y\frac{dy}{dx}
\displaystyle \Rightarrow x\frac{dy}{dx}+y\frac{dy}{dx}=x-y
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x-y}{x+y}\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{1-v}{1+v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v-v-v^2}{1+v}=\frac{1-2v-v^2}{1+v}
\displaystyle \Rightarrow \frac{1+v}{v^2+2v-1}dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1+v}{v^2+2v-1}dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\log|v^2+2v-1|=-\log|x|+\log C
\displaystyle \Rightarrow \log|v^2+2v-1|+2\log|x|=\log C^2
\displaystyle \Rightarrow \log\left(\frac{y^2}{x^2}+\frac{2y}{x}-1\right)+\log x^2=\log C^2
\displaystyle \Rightarrow \log(y^2+2xy-x^2)=\log C^2
\displaystyle \Rightarrow y^2+2xy-x^2=C_1,\text{ where }C_1=C^2
\\

\displaystyle \textbf{Question 90. }\text{Solve the following differential equation}
\displaystyle y^{2}dx+(x^{2}-xy+y^{2})dy=0. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }y^2dx+(x^2-xy+y^2)dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-y^2}{x^2-xy+y^2}
\displaystyle \text{This is homogeneous differential equation.}
\displaystyle \text{Now, on putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{-v^2x^2}{x^2-vx^2+v^2x^2}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{-v^2}{1-v+v^2}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{-v-v^3}{1-v+v^2}
\displaystyle \Rightarrow \frac{1-v+v^2}{v(1+v^2)}\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1+v^2}{v(1+v^2)}\,dv-\int\frac{v}{1+v^2}dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{1}{v}dv-\int\frac{1}{1+v^2}dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \log|v|-\tan^{-1}v=-\log|x|+\log C
\displaystyle \Rightarrow \log\left|\frac{vx}{C}\right|=\tan^{-1}v
\displaystyle \Rightarrow \frac{vx}{C}=e^{\tan^{-1}v}
\displaystyle \Rightarrow |y|=Ce^{\tan^{-1}(y/x)}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 91. }\text{Solve the following differential equation}
\displaystyle (\cot^{-1}y+x)dy-(1+y^{2})dx=0. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }(\cot^{-1}y+x)dy=(1+y^2)dx
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{\cot^{-1}y+x}{1+y^2}
\displaystyle \Rightarrow \frac{dx}{dy}+\left(\frac{-1}{1+y^2}\right)x=\frac{\cot^{-1}y}{1+y^2}
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q
\displaystyle \text{Here, }P=\frac{-1}{1+y^2}\text{ and }Q=\frac{\cot^{-1}y}{1+y^2}
\displaystyle \therefore \text{IF}=e^{\int Pdy}=e^{-\int\frac{1}{1+y^2}dy}=e^{-\cot^{-1}y}
\displaystyle \text{The solution of given linear differential equation is given by}
\displaystyle x\cdot\text{IF}=\int(Q\cdot\text{IF})dy+C
\displaystyle \Rightarrow xe^{-\cot^{-1}y}=\int\frac{\cot^{-1}y}{1+y^2}e^{-\cot^{-1}y}dy+C\qquad ...(i)
\displaystyle \text{On putting }\cot^{-1}y=t\Rightarrow \frac{1}{1+y^2}dy=-dt\text{ in Eq. (i), we get}
\displaystyle xe^{-t}=\int(-t)e^{-t}dt+C
\displaystyle \Rightarrow xe^{-t}=e^{-t}(t+1)+C
\displaystyle \Rightarrow xe^{-\cot^{-1}y}=e^{-\cot^{-1}y}(1+\cot^{-1}y)+C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 92. }\text{Find the integrating factor of the differential equation}
\displaystyle \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation can be rewritten as}
\displaystyle \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ here }P=\frac{1}{\sqrt{x}}\text{ and }Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}
\displaystyle \therefore \text{Integrating Factor, IF}=e^{\int Pdx}=e^{\int \frac{1}{\sqrt{x}}dx}
\displaystyle =e^{2\sqrt{x}}
\\

\displaystyle \textbf{Question 93. }\text{Write the integrating factor of the following } \text{differential equation }
\displaystyle (1+y^{2})+(2xy-\cot y)\frac{dy}{dx}=0. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1+y^2)+(2xy-\cot y)\frac{dy}{dx}=0
\displaystyle \text{The above equation can be rewritten as}
\displaystyle (\cot y-2xy)\frac{dy}{dx}=1+y^2
\displaystyle \Rightarrow \frac{\cot y-2xy}{1+y^2}\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{2y}{1+y^2}x=\frac{\cot y}{1+y^2}
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q,\text{ here }P=\frac{2y}{1+y^2}\text{ and }Q=\frac{\cot y}{1+y^2}
\displaystyle \text{Now, integrating factor}=e^{\int Pdy}=e^{\int \frac{2y}{1+y^2}dy}
\displaystyle \text{On putting }1+y^2=t\Rightarrow 2y\,dy=dt
\displaystyle \therefore \text{IF}=e^{\int \frac{dt}{t}}=e^{\log t}=t=1+y^2
\\

\displaystyle \textbf{Question 94. }\text{Write the solution of the differential equation }   \frac{dy}{dx}=2^{-y}.  \\ \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{ Given, differential equation is}
\displaystyle \frac{dy}{dx}=2^{-y}
\displaystyle \text{On separating the variables, we get}
\displaystyle 2^y\,dy=dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int 2^y\,dy=\int dx
\displaystyle \Rightarrow \frac{2^y}{\log 2}=x+C_1
\displaystyle \Rightarrow 2^y=x\log 2+C_1\log 2
\displaystyle \therefore 2^y=x\log 2+C,\text{ where }C=C_1\log 2
\\

\displaystyle \textbf{Question 95. }\text{Find the solution of the differential equation }   \frac{dy}{dx}=x^{3}e^{-2y}.  \\ \hspace{0.2cm}\text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=x^3e^{-2y}
\displaystyle \text{On separating the variables, we get}
\displaystyle e^{2y}dy=x^3dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int e^{2y}dy=\int x^3dx
\displaystyle \Rightarrow \frac{e^{2y}}{2}=\frac{x^4}{4}+C_1
\displaystyle \Rightarrow 2e^{2y}=\frac{x^4}{2}+4C_1
\displaystyle \therefore 4e^{2y}=x^4+C,\text{ where }C=8C_1
\\

\displaystyle \textbf{Question 96. }\text{Solve the following differential equation.}
\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0,\ x\neq0.   \hspace{0.2cm}\text{[CBSE 2015C, CBSE 2014C, CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0,\ x\neq0
\displaystyle \text{Above equation can be written as}
\displaystyle x\frac{dy}{dx}+y(1+x\cot x)=x
\displaystyle \text{On dividing both sides by }x,\text{ we get}
\displaystyle \frac{dy}{dx}+y\left(\frac{1}{x}+\cot x\right)=1
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, }P=\frac{1}{x}+\cot x\text{ and }Q=1
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\left(\frac{1}{x}+\cot x\right)dx}
\displaystyle =e^{\log|x|+\log|\sin x|}
\displaystyle =x\sin x
\displaystyle \text{The solution of given linear differential equation is}
\displaystyle y\times\text{IF}=\int(Q\times\text{IF})dx+C
\displaystyle \Rightarrow yx\sin x=\int x\sin x\,dx+C
\displaystyle \Rightarrow yx\sin x=x\int\sin x\,dx-\int\left[\frac{d}{dx}(x)\int\sin x\,dx\right]dx+C
\displaystyle \text{[using integration by parts]}
\displaystyle \Rightarrow yx\sin x=-x\cos x+\int\cos x\,dx+C
\displaystyle \Rightarrow yx\sin x=-x\cos x+\sin x+C
\displaystyle \text{On dividing both sides by }x\sin x,\text{ we get}
\displaystyle y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 97. }\text{Find the particular solution of the differential equation}
\displaystyle x^{2}dy+(xy+y^{2})dx=0,\text{ when }y(1)=1.   \hspace{0.2cm}\text{[CBSE 2015C, 2013C; CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x^2dy+(xy+y^2)dx=0
\displaystyle \Rightarrow x^2\frac{dy}{dx}=-(xy+y^2)
\displaystyle \Rightarrow \frac{dy}{dx}=-\left(\frac{y}{x}+\frac{y^2}{x^2}\right)\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=-(v+v^2)
\displaystyle \Rightarrow x\frac{dv}{dx}=-v^2-2v
\displaystyle \Rightarrow \frac{dv}{v^2+2v}=-\frac{dx}{x}
\displaystyle \Rightarrow \int\frac{dv}{v(v+2)}=-\int\frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\log\left|\frac{v}{v+2}\right|=-\log|x|+C\qquad ...(ii)
\displaystyle \text{Also, given at }x=1\text{ and }y=1,\text{ so }v=1
\displaystyle \text{On putting }v=1\text{ in Eq. (ii), we get}
\displaystyle \frac{1}{2}\log\frac{1}{3}=-\log1+C
\displaystyle \Rightarrow C=\frac{1}{2}\log\frac{1}{3}
\displaystyle \text{On putting the value of }C\text{ in Eq. (ii), we get}
\displaystyle \frac{1}{2}\log\left|\frac{v}{v+2}\right|=-\log|x|+\frac{1}{2}\log\frac{1}{3}
\displaystyle \Rightarrow \log\left|\frac{v}{v+2}\right|=\log\frac{1}{3x^2}
\displaystyle \Rightarrow \frac{v}{v+2}=\frac{1}{3x^2}
\displaystyle \Rightarrow \frac{y}{y+2x}=\frac{1}{3x^2}
\displaystyle \Rightarrow y(3x^2-1)=2x
\displaystyle \Rightarrow y=\frac{2x}{3x^2-1}
\displaystyle \text{which is the required particular solution.}
\\

\displaystyle \textbf{Question 98. }\text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}},\text{ given that }y=1,\text{ when }x=0. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^2+y^2}=\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{Now, put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ from Eq. (i),}
\displaystyle \text{we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v}{1+v^2}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v-v-v^3}{1+v^2}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^3}{1+v^2}
\displaystyle \Rightarrow \frac{1+v^2}{v^3}dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\left(\frac{1}{v^3}+\frac{1}{v}\right)dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{2v^2}+\log|v|=-\log|x|+C
\displaystyle \Rightarrow -\frac{x^2}{2y^2}+\log\left|\frac{y}{x}\right|=-\log|x|+C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow -\frac{x^2}{2y^2}+\log|y|-\log|x|=-\log|x|+C
\displaystyle \Rightarrow -\frac{x^2}{2y^2}+\log|y|=C\qquad ...(ii)
\displaystyle \text{Also, it is given that }y=1,\text{ when }x=0.
\displaystyle \text{From Eq. (ii), we have}
\displaystyle 0+\log|1|=C\Rightarrow C=0
\displaystyle \text{On putting }C=0\text{ in Eq. (ii), we get}
\displaystyle -\frac{x^2}{2y^2}+\log|y|=0\Rightarrow \log|y|=\frac{x^2}{2y^2}
\displaystyle \therefore y=e^{\frac{x^2}{2y^2}}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 99. }\text{Show that the differential eq. }x\sin^{2}\left(\frac{y}{x}\right) -y\,dx+x\,dy=0
\displaystyle \text{ is homogeneous. Find the particular solution of this differential equation,}
\displaystyle \text{given that }y=\frac{\pi}{4}\text{ when }x=1.   \hspace{0.2cm}\text{[CBSE 2015C, 2014C, 2013; CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\sin^2\left(\frac{y}{x}\right)-y+x\frac{dy}{dx}=0
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}+\sin^2\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\sin^2\left(\frac{y}{x}\right)\qquad ...(i)
\displaystyle \text{Let }F(x,y)=\frac{y}{x}-\sin^2\frac{y}{x}
\displaystyle \text{On replacing }x\text{ by }\lambda x\text{ and }y\text{ by }\lambda y\text{ both sides, we get}
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}-\sin^2\frac{\lambda y}{\lambda x}=\lambda^0\left[\frac{y}{x}-\sin^2\frac{y}{x}\right]
\displaystyle =\lambda^0F(x,y)
\displaystyle \text{So, the given differential equation is homogeneous.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v-\sin^2v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\sin^2v
\displaystyle \Rightarrow \mathrm{cosec}^2v\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\mathrm{cosec}^2v\,dv+\int\frac{dx}{x}=0
\displaystyle \Rightarrow -\cot v+\log|x|=C
\displaystyle \Rightarrow -\cot\left(\frac{y}{x}\right)+\log|x|=C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \text{The general solution of differential equation is}
\displaystyle -\cot\left(\frac{y}{x}\right)+\log|x|=C\qquad ...(ii)
\displaystyle \text{Also, given that }y=\frac{\pi}{4},\text{ when }x=1
\displaystyle \text{On putting }x=1\text{ and }y=\frac{\pi}{4}\text{ in Eq. (ii), we get}
\displaystyle -\cot\left(\frac{\pi}{4}\right)+\log1=C\Rightarrow C=-1
\displaystyle \text{On putting this value of }C\text{ in Eq. (ii), we get}
\displaystyle -\cot\left(\frac{y}{x}\right)+\log|x|=-1
\displaystyle \therefore 1+\log|x|-\cot\left(\frac{y}{x}\right)=0
\displaystyle \text{which is the required particular solution of given differential equation.}
\\

\displaystyle \textbf{Question 100. }\text{Solve the differential equation } \frac{dy}{dx}-3y\cot x=\sin2x,
\displaystyle \text{ given }y=2\text{ when }x=\frac{\pi}{2}.   \hspace{2.2cm}\text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{dy}{dx}-3y\cot x=\sin2x
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ where }P=-3\cot x\text{ and }Q=\sin2x.
\displaystyle \text{IF}=e^{\int Pdx}=e^{-3\int\cot xdx}
\displaystyle =e^{-3\log|\sin x|}=e^{\log|\sin x|^{-3}}=(\sin x)^{-3}
\displaystyle \text{The general solution of differential equation is given by}
\displaystyle y(\sin x)^{-3}=\int(\sin x)^{-3}(\sin2x)\,dx+C
\displaystyle \Rightarrow y(\sin x)^{-3}=\int\frac{2\sin x\cos x}{\sin^3x}\,dx+C
\displaystyle \Rightarrow y(\sin x)^{-3}=\int\frac{2\cos x}{\sin^2x}\,dx+C\qquad ...(i)
\displaystyle \text{On putting }\sin x=t\Rightarrow \cos x\,dx=dt\text{ in Eq. (i), we get}
\displaystyle y(\sin x)^{-3}=2\int\frac{1}{t^2}dt+C=2\frac{t^{-1}}{-1}+C
\displaystyle \Rightarrow y(\sin x)^{-3}=-\frac{2}{\sin x}+C
\displaystyle \Rightarrow y=-2\sin^2x+C\sin^3x\qquad ...(ii)
\displaystyle \text{On putting }x=\frac{\pi}{2}\text{ and }y=2\text{ in Eq. (ii), we get}
\displaystyle 2=-2\sin^2\frac{\pi}{2}+C\sin^3\frac{\pi}{2}
\displaystyle \Rightarrow 2=-2+1\cdot C
\displaystyle \Rightarrow C=4
\displaystyle \therefore y=-2\sin^2x+4\sin^3x
\displaystyle \text{which is the required particular solution.}
\\

\displaystyle \textbf{Question 101. }\text{Show that the differential equation } \frac{dy}{dx}=\frac{y^{2}}{xy-x^{2}}
\displaystyle \text{is homogeneous and also solve it.}   \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=\frac{y^2}{xy-x^2}\qquad ...(i)
\displaystyle \text{Let }F(x,y)=\frac{y^2}{xy-x^2}
\displaystyle \text{Now, on replacing }x\text{ by }\lambda x\text{ and }y\text{ by }\lambda y\text{, we get}
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda^2y^2}{\lambda^2(xy-x^2)}=\lambda^0\frac{y^2}{xy-x^2}=\lambda^0F(x,y)
\displaystyle \text{Thus, the given differential equation is homogeneous.}
\displaystyle \text{Now, to solve it, put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{From Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^2x^2}{vx^2-x^2}=\frac{v^2}{v-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^2}{v-1}-v=\frac{v}{v-1}
\displaystyle \Rightarrow \frac{v-1}{v}dv=\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\left(1-\frac{1}{v}\right)dv=\int\frac{dx}{x}
\displaystyle \Rightarrow v-\log|v|=\log|x|+C
\displaystyle \Rightarrow \frac{y}{x}-\log\left|\frac{y}{x}\right|=\log|x|+C
\displaystyle \Rightarrow \frac{y}{x}-\log|y|+\log|x|=\log|x|+C
\displaystyle \therefore \frac{y}{x}-\log|y|=C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 102. }\text{Find the particular solution of the differential}
\displaystyle \text{equation }(\tan^{-1}y-x)\,dy+(1+y^{2})\,dx,\text{ given that }x=1   \text{ when }y=0. \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(\tan^{-1}y-x)\,dy=(1+y^2)\,dx
\displaystyle \Rightarrow (1+y^2)\frac{dx}{dy}+x=\tan^{-1}y
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{\tan^{-1}y}{1+y^2}
\displaystyle \text{Here, }P(y)=\frac{1}{1+y^2},\quad Q(y)=\frac{\tan^{-1}y}{1+y^2}
\displaystyle \text{Integrating factor }=e^{\int P(y)\,dy}=e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}
\displaystyle \text{Solution of equation is}
\displaystyle x\times \text{IF}=\int Q\times \text{IF}\,dy+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\int\frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}dy
\displaystyle \text{On putting }\tan^{-1}y=t\Rightarrow \frac{1}{1+y^2}dy=dt
\displaystyle \Rightarrow xe^{\tan^{-1}y}=\int te^t\,dt
\displaystyle \Rightarrow xe^{\tan^{-1}y}=te^t-e^t+C
\displaystyle \Rightarrow xe^{\tan^{-1}y}=e^{\tan^{-1}y}(\tan^{-1}y-1)+C
\displaystyle \text{Also, it is given that }x=1,\text{ when }y=0.
\displaystyle \therefore \text{From above, we have }1\cdot e^0=e^0(0-1)+C
\displaystyle \Rightarrow 1=-1+C\Rightarrow C=2
\displaystyle \therefore xe^{\tan^{-1}y}=e^{\tan^{-1}y}(\tan^{-1}y-1)+2
\displaystyle \text{Hence, the required solution is obtained.}
\\

\displaystyle \textbf{Question 103. }\text{Solve the following differential equation }
\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0.   \hspace{2.2cm}\text{[CBSE 2015, CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \sqrt{1+x^2+y^2+x^2y^2}+xy\frac{dy}{dx}=0
\displaystyle \Rightarrow \sqrt{(1+x^2)+y^2(1+x^2)}=-xy\frac{dy}{dx}
\displaystyle \Rightarrow \sqrt{(1+x^2)(1+y^2)}=-xy\frac{dy}{dx}
\displaystyle \Rightarrow \sqrt{1+x^2}\sqrt{1+y^2}=-xy\frac{dy}{dx}
\displaystyle \Rightarrow \frac{y}{\sqrt{1+y^2}}\,dy=-\frac{\sqrt{1+x^2}}{x}\,dx\qquad ...(i)
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{y}{\sqrt{1+y^2}}\,dy=-\int\frac{\sqrt{1+x^2}}{x}\,dx
\displaystyle \text{On putting }1+y^2=t\text{ and }1+x^2=u^2
\displaystyle \Rightarrow 2y\,dy=dt\text{ and }2x\,dx=2u\,du
\displaystyle \Rightarrow y\,dy=\frac{dt}{2}\text{ and }x\,dx=u\,du
\displaystyle \therefore \frac{1}{2}\int\frac{dt}{\sqrt t}=-\int\frac{u}{u^2-1}\cdot u\,du
\displaystyle \Rightarrow \frac{1}{2}\int t^{-1/2}dt=-\int\frac{u^2}{u^2-1}\,du
\displaystyle \Rightarrow t^{1/2}=-\int\frac{(u^2-1)+1}{u^2-1}\,du
\displaystyle \Rightarrow \sqrt{1+y^2}=-\int\frac{u^2-1}{u^2-1}\,du-\int\frac{1}{u^2-1}\,du
\displaystyle \Rightarrow \sqrt{1+y^2}=-u-\frac{1}{2}\log\left|\frac{u-1}{u+1}\right|+C
\displaystyle \therefore \sqrt{1+y^2}=-\sqrt{1+x^2}-\frac{1}{2}\log\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 105. }\text{Solve the differential equation}
\displaystyle \left[y-x\cos\left(\frac{y}{x}\right)\right]dy+\left[y\cos\left(\frac{y}{x}\right)-2x\sin\left(\frac{y}{x}\right)\right]dx=0.   \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \left[y-x\cos\left(\frac{y}{x}\right)\right]dy+\left[y\cos\left(\frac{y}{x}\right)-2x\sin\left(\frac{y}{x}\right)\right]dx=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2x\sin\left(\frac{y}{x}\right)-y\cos\left(\frac{y}{x}\right)}{y-x\cos\left(\frac{y}{x}\right)}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\sin\left(\frac{y}{x}\right)-\frac{y}{x}\cos\left(\frac{y}{x}\right)}{\frac{y}{x}-\cos\left(\frac{y}{x}\right)}\qquad ...(i)
\displaystyle \text{[divide numerator and denominator by }x]
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{2\sin v-v\cos v}{v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2\sin v-v\cos v-v^2+v\cos v}{v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2\sin v-v^2}{v-\cos v}
\displaystyle \Rightarrow \left(\frac{v-\cos v}{v^2-2\sin v}\right)dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{v-\cos v}{v^2-2\sin v}dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\log|v^2-2\sin v|=-\log|x|+\log C_1
\displaystyle \left[\because \frac{d}{dv}(v^2-2\sin v)=2(v-\cos v)\right]
\displaystyle \Rightarrow \log\sqrt{v^2-2\sin v}=\log\left|\frac{C_1}{x}\right|
\displaystyle \Rightarrow \sqrt{v^2-2\sin v}=\frac{C_1}{x}
\displaystyle \Rightarrow \sqrt{\frac{y^2}{x^2}-2\sin\left(\frac{y}{x}\right)}=\frac{C_1}{x}\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow \sqrt{y^2-2x^2\sin\left(\frac{y}{x}\right)}=C_1
\displaystyle \Rightarrow y^2-2x^2\sin\left(\frac{y}{x}\right)=C_1^2
\displaystyle \therefore y^2-2x^2\sin\left(\frac{y}{x}\right)=C,\text{ where }C=C_1^2
\\

\displaystyle \textbf{Question 106. }\text{If }y(x)\text{ is a solution of the differential equation}
\displaystyle \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x\text{ and }y(0)=1,\text{ then find the}   \text{value of }y\left(\frac{\pi}{2}\right). \hspace{0.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \left(\frac{2+\sin x}{1+y}\right)dy=\cos x\,dx
\displaystyle \Rightarrow \frac{1}{1+y}dy=\frac{\cos x}{2+\sin x}dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1}{1+y}dy=\int\frac{\cos x}{2+\sin x}dx
\displaystyle \Rightarrow \log|1+y|=\log|2+\sin x|+C
\displaystyle \text{[put }2+\sin x=t\Rightarrow \cos x\,dx=dt]
\displaystyle \Rightarrow \log|1+y|-\log|2+\sin x|=\log C
\displaystyle \Rightarrow \log\left|\frac{1+y}{2+\sin x}\right|=\log C
\displaystyle \Rightarrow (1+y)=C(2+\sin x)\qquad ...(i)
\displaystyle \text{Also, given that at }x=0,\ y=1
\displaystyle \text{On putting }x=0\text{ and }y=1\text{ in Eq. (i), we get}
\displaystyle (1+1)=C(2+\sin0)\Rightarrow C=1
\displaystyle \text{On putting }C=1\text{ in Eq. (i), we get}
\displaystyle 1+y=2+\sin x
\displaystyle \Rightarrow y=1+\sin x
\displaystyle \text{Now, at }x=\frac{\pi}{2},\ y\left(\frac{\pi}{2}\right)=1+\sin\frac{\pi}{2}=2
\\

\displaystyle \textbf{Question 107. }\text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}\times\frac{x(2\log|x|+1)}{\sin y+y\cos y},   \text{given that }y=\frac{\pi}{2},\text{ when }x=1. \hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=\frac{x(2\log|x|+1)}{\sin y+y\cos y}
\displaystyle \text{On separating the variables, we get}
\displaystyle (\sin y+y\cos y)\,dy=x(2\log|x|+1)\,dx
\displaystyle \Rightarrow \sin y\,dy+y\cos y\,dy=2x\log|x|\,dx+x\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\sin y\,dy+\int y\cos y\,dy=\int2x\log|x|\,dx+\int x\,dx
\displaystyle \Rightarrow -\cos y+\int y\cos y\,dy=2\left[\log|x|\int x\,dx-\int\frac{d}{dx}(\log|x|)\left(\int x\,dx\right)dx\right]+\frac{x^2}{2}
\displaystyle \text{[by using integration by parts]}
\displaystyle \Rightarrow -\cos y+y\sin y-\int\sin y\,dy
\displaystyle =2\left[\frac{x^2}{2}\log|x|-\int\frac{1}{x}\cdot\frac{x^2}{2}\,dx\right]+\frac{x^2}{2}
\displaystyle \Rightarrow -\cos y+y\sin y+\cos y=x^2\log|x|-\int x\,dx+\frac{x^2}{2}
\displaystyle \Rightarrow y\sin y=x^2\log|x|-\frac{x^2}{2}+\frac{x^2}{2}+C
\displaystyle \Rightarrow y\sin y=x^2\log|x|+C\qquad ...(i)
\displaystyle \text{Also, given that }y=\frac{\pi}{2}\text{ when }x=1
\displaystyle \text{On putting }y=\frac{\pi}{2}\text{ and }x=1\text{ in Eq. (i), we get}
\displaystyle \frac{\pi}{2}\sin\frac{\pi}{2}=1^2\log(1)+C
\displaystyle \Rightarrow C=\frac{\pi}{2}\qquad \left[\because \sin\frac{\pi}{2}=1,\ \log1=0\right]
\displaystyle \text{On substituting the value of }C\text{ in Eq. (i), we get}
\displaystyle y\sin y=x^2\log|x|+\frac{\pi}{2}
\displaystyle \text{which is the required particular solution.}
\\

\displaystyle \textbf{Question 107. }\text{Solve the following differential equation } \\ (x^{2}-1)\frac{dy}{dx}+2xy=\frac{2}{x^{2}-1},\ x\neq1.\hspace{0.2cm}\text{[CBSE 2014; CBSE 2014 C, 2010]}
\displaystyle \text{Answer:}
\displaystyle (x^{2}-1)\frac{dy}{dx}+2xy=\frac{2}{x^{2}-1}
\displaystyle \frac{dy}{dx}+\frac{2x}{x^{2}-1}y=\frac{2}{(x^{2}-1)^{2}}
\displaystyle \text{This is a linear differential equation.}
\displaystyle \text{I.F.}=e^{\int \frac{2x}{x^{2}-1}\,dx}=e^{\log(x^{2}-1)}=x^{2}-1
\displaystyle \text{Multiplying by I.F.,}
\displaystyle (x^{2}-1)\frac{dy}{dx}+2xy=\frac{2}{x^{2}-1}
\displaystyle \frac{d}{dx}\left[y(x^{2}-1)\right]=\frac{2}{x^{2}-1}
\displaystyle y(x^{2}-1)=\int \frac{2}{x^{2}-1}\,dx
\displaystyle y(x^{2}-1)=\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right)\,dx
\displaystyle y(x^{2}-1)=\log|x-1|-\log|x+1|+C
\displaystyle y(x^{2}-1)=\log\left|\frac{x-1}{x+1}\right|+C
\displaystyle \therefore y=\frac{1}{x^{2}-1}\left[\log\left|\frac{x-1}{x+1}\right|+C\right]

\displaystyle \textbf{Question 109. }\text{Find the particular solution of the differential equation } \\ e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0,\text{ given that }y=1,\text{ when }x=0.\hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0
\displaystyle \frac{y}{\sqrt{1-y^{2}}}\,dy=-xe^{x}\,dx
\displaystyle \int \frac{y}{\sqrt{1-y^{2}}}\,dy=-\int xe^{x}\,dx
\displaystyle -\sqrt{1-y^{2}}=-e^{x}(x-1)+C
\displaystyle \sqrt{1-y^{2}}=e^{x}(x-1)+C
\displaystyle \text{Given, }y=1\text{ when }x=0
\displaystyle 0=e^{0}(0-1)+C
\displaystyle C=1
\displaystyle \therefore \sqrt{1-y^{2}}=e^{x}(x-1)+1

\displaystyle \textbf{Question 110. }\text{Solve the following differential equation}
\displaystyle \mathrm{cosec}\,x\log|y|\frac{dy}{dx}+x^{2}y^{2}=0. \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \mathrm{cosec} x\log|y|\frac{dy}{dx}+x^2y^2=0
\displaystyle \text{It can be rewritten as}
\displaystyle \mathrm{cosec} x\log|y|\frac{dy}{dx}=-x^2y^2
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{\log|y|}{y^2}\,dy=-\frac{x^2}{\mathrm{cosec} x}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{\log|y|}{y^2}\,dy=-\int\frac{x^2}{\mathrm{cosec} x}\,dx
\displaystyle \Rightarrow I_1=-I_2
\displaystyle \text{where, }I_1=\int\frac{\log|y|}{y^2}\,dy
\displaystyle \text{and }I_2=\int x^2\sin x\,dx
\displaystyle \text{Consider, }I_1=\int\frac{\log|y|}{y^2}\,dy
\displaystyle \text{On putting }\log y=t\Rightarrow y=e^t,\ \frac{1}{y}=e^{-t}
\displaystyle \Rightarrow I_1=\int te^{-t}\,dt
\displaystyle =-te^{-t}-\int(-e^{-t})dt
\displaystyle =-te^{-t}-e^{-t}+C_1
\displaystyle =-\frac{\log|y|}{y}-\frac{1}{y}+C_1\qquad ...(iii)
\displaystyle \text{And }I_2=\int x^2\sin x\,dx
\displaystyle =x^2(-\cos x)-\int2x(-\cos x)\,dx
\displaystyle =-x^2\cos x+2\int x\cos x\,dx
\displaystyle =-x^2\cos x+2(x\sin x+\cos x)+C_2\qquad ...(iv)
\displaystyle \text{On putting the values of }I_1\text{ and }I_2\text{ from Eqs. (iii) and (iv), we get}
\displaystyle -\frac{\log|y|}{y}-\frac{1}{y}+C_1=x^2\cos x-2x\sin x-2\cos x-C_2
\displaystyle \Rightarrow \frac{1+\log|y|}{y}=x^2\cos x-2x\sin x-2\cos x+C
\displaystyle \text{where, }C=-C_2-C_1
\displaystyle \text{which is the required solution of given differential equation.}
\\

\displaystyle \textbf{Question 111. }\text{Solve the differential equation }\frac{dy}{dx}+y\cot x=2\cos x,
\displaystyle \text{given that }y=0,\text{ when }x=\frac{\pi}{2}. \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}+y\cot x=2\cos x
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, }P=\cot x\text{ and }Q=2\cos x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\cot x\,dx}=e^{\log|\sin x|}=\sin x
\displaystyle \text{The general solution is given by}
\displaystyle y\times\text{IF}=\int(\text{IF}\times Q)\,dx+C
\displaystyle \Rightarrow y\sin x=\int2\sin x\cos x\,dx+C
\displaystyle \Rightarrow y\sin x=-\frac{\cos2x}{2}+C\qquad ...(i)
\displaystyle \text{Also, given that }y=0,\text{ when }x=\frac{\pi}{2}
\displaystyle \text{On putting }x=\frac{\pi}{2}\text{ and }y=0\text{ in Eq. (i), we get}
\displaystyle 0=-\frac{\cos\pi}{2}+C
\displaystyle \Rightarrow C=-\frac{1}{2}
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle y\sin x=-\frac{\cos2x}{2}-\frac{1}{2}
\displaystyle \Rightarrow 2y\sin x+\cos2x+1=0
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 112. }\text{Find the particular solution of the differential equation}
\displaystyle x\frac{dy}{dx}-y+x\,\mathrm{cosec}\left(\frac{y}{x}\right)=0\text{ given that }y=0\text{ when }x=1.   \hspace{0.2cm}\text{[CBSE 2014C, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}-y+x\mathrm{cosec}\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}+\mathrm{cosec}\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\mathrm{cosec}\left(\frac{y}{x}\right)\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as }
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=v-\mathrm{cosec} v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\mathrm{cosec} v
\displaystyle \Rightarrow \sin v\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle -\cos v=-\log|x|+C
\displaystyle \Rightarrow \cos v=\log|x|-C\qquad ...(ii)
\displaystyle \text{Also, given that }x=1\text{ and }y=0
\displaystyle \text{On putting above values in Eq. (ii), we get}
\displaystyle \cos0=\log|1|-C
\displaystyle \Rightarrow 1=0-C
\displaystyle \Rightarrow C=-1
\displaystyle \therefore \cos\left(\frac{y}{x}\right)=\log|x|+1
\displaystyle \text{which is the required particular solution of the differential equation.}
\\

\displaystyle \textbf{Question 113. }\text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}=1+x+y+xy,   \text{given that }y=0\text{ when }x=1. \hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=1+x+y+xy
\displaystyle \Rightarrow \frac{dy}{dx}=1(1+x)+y(1+x)
\displaystyle \Rightarrow \frac{dy}{dx}=(1+x)(1+y)\qquad ...(i)
\displaystyle \text{On separating variables, we get}
\displaystyle \frac{1}{1+y}\,dy=(1+x)\,dx\qquad ...(ii)
\displaystyle \text{On integrating both sides of Eq. (ii), we get}
\displaystyle \int\frac{1}{1+y}\,dy=\int(1+x)\,dx
\displaystyle \Rightarrow \log|1+y|=x+\frac{x^2}{2}+C\qquad ...(iii)
\displaystyle \text{Also, given that }y=0,\text{ when }x=1.
\displaystyle \text{On substituting }x=1,\ y=0\text{ in Eq. (iii), we get}
\displaystyle \log|1+0|=1+\frac{1}{2}+C
\displaystyle \Rightarrow C=-\frac{3}{2}\qquad [\because \log1=0]
\displaystyle \text{Now, on substituting the value of }C\text{ in Eq. (iii), we get}
\displaystyle \log|1+y|=x+\frac{x^2}{2}-\frac{3}{2}
\displaystyle \text{which is the required particular solution of given differential equation.}
\\

\displaystyle \textbf{Question 114. }\text{Solve the differential equation }(1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}.\hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle (1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}
\displaystyle \frac{dy}{dx}+\frac{1}{1+x^{2}}y=\frac{e^{\tan^{-1}x}}{1+x^{2}}
\displaystyle \text{This is a linear differential equation.}
\displaystyle \text{I.F.}=e^{\int \frac{1}{1+x^{2}}\,dx}=e^{\tan^{-1}x}
\displaystyle \text{Multiplying by I.F.,}
\displaystyle \frac{d}{dx}\left(ye^{\tan^{-1}x}\right)=\frac{e^{2\tan^{-1}x}}{1+x^{2}}
\displaystyle ye^{\tan^{-1}x}=\int \frac{e^{2\tan^{-1}x}}{1+x^{2}}\,dx
\displaystyle \text{Let }\tan^{-1}x=t,\text{ then }\frac{dx}{1+x^{2}}=dt
\displaystyle ye^{\tan^{-1}x}=\int e^{2t}\,dt
\displaystyle ye^{\tan^{-1}x}=\frac{1}{2}e^{2t}+C
\displaystyle ye^{\tan^{-1}x}=\frac{1}{2}e^{2\tan^{-1}x}+C
\displaystyle \therefore y=\frac{1}{2}e^{\tan^{-1}x}+Ce^{-\tan^{-1}x}

\displaystyle \textbf{Question 115. }\text{Find the particular solution of the differential equation}
\displaystyle \log\left(\frac{dy}{dx}\right)=3x+4y,\text{ given that }y=0,\text{ when }x=0.   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \log\left(\frac{dy}{dx}\right)=3x+4y
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x+4y}\qquad [\because \log m=n\Rightarrow e^n=m]
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x}\cdot e^{4y}
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{1}{e^{4y}}\,dy=e^{3x}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int e^{-4y}\,dy=\int e^{3x}\,dx\Rightarrow \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C\qquad ...(i)
\displaystyle \text{Also, given that }y=0,\text{ when }x=0.
\displaystyle \text{On putting }y=0\text{ and }x=0\text{ in Eq. (i), we get}
\displaystyle \frac{e^{-4(0)}}{-4}=\frac{e^{3(0)}}{3}+C
\displaystyle \Rightarrow -\frac{1}{4}=\frac{1}{3}+C\qquad [\because e^0=e^0=1]
\displaystyle \Rightarrow C=-\frac{1}{4}-\frac{1}{3}
\displaystyle \therefore C=-\frac{7}{12}
\displaystyle \text{On substituting the value of }C\text{ in Eq. (i), we get}
\displaystyle \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}-\frac{7}{12}
\displaystyle \Rightarrow 4e^{3x}+3e^{-4y}-7=0
\displaystyle \text{which is the required particular solution of given differential equation.}
\\

\displaystyle \textbf{Question 116. }\text{Find the particular solution of the differential equation}
\displaystyle x(1+y^{2})\,dx-y(1+x^{2})\,dy=0,\text{ given that }y=1,\text{ when}   x=0. \hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x(1+y^2)\,dx-y(1+x^2)\,dy=0
\displaystyle \Rightarrow x(1+y^2)\,dx=y(1+x^2)\,dy
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{y}{1+y^2}\,dy=\frac{x}{1+x^2}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{y}{1+y^2}\,dy=\int\frac{x}{1+x^2}\,dx
\displaystyle \Rightarrow \frac{1}{2}\log|1+y^2|=\frac{1}{2}\log|1+x^2|+C\qquad ...(i)
\displaystyle \text{Also, given that }y=1,\text{ when }x=0.
\displaystyle \text{On substituting the values of }x\text{ and }y\text{ in Eq. (i), we get}
\displaystyle \frac{1}{2}\log|1+(1)^2|=\frac{1}{2}\log|1+(0)^2|+C
\displaystyle \Rightarrow \frac{1}{2}\log2=C\qquad [\because \log1=0]
\displaystyle \text{On putting }C=\frac{1}{2}\log2\text{ in Eq. (i), we get}
\displaystyle \frac{1}{2}\log|1+y^2|=\frac{1}{2}\log|1+x^2|+\frac{1}{2}\log2
\displaystyle \Rightarrow \log|1+y^2|=\log|1+x^2|+\log2
\displaystyle \Rightarrow \log|1+y^2|-\log|1+x^2|=\log2
\displaystyle \Rightarrow \log\left|\frac{1+y^2}{1+x^2}\right|=\log2
\displaystyle \Rightarrow \frac{1+y^2}{1+x^2}=2\Rightarrow 1+y^2=2+2x^2
\displaystyle \Rightarrow y^2-2x^2-1=0
\displaystyle \text{which is the required particular solution of given differential equation.}
\\

\displaystyle \textbf{Question 117. }\text{Solve the differential equation }   x\log|x|\frac{dy}{dx}+y=\frac{2}{x}\log|x|. \\ \hspace{0.2cm}\text{[CBSE 2014; CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (x\log|x|)\frac{dy}{dx}+y=\frac{2}{x}\log|x|
\displaystyle \text{On dividing both sides by }x\log|x|,\text{ we get}
\displaystyle \frac{dy}{dx}+\frac{y}{x\log|x|}=\frac{2\log|x|}{x^2\log|x|}=\frac{2}{x^2}
\displaystyle \text{which is a linear differential equation of first order}
\displaystyle \text{and is of the form }\frac{dy}{dx}+Py=Q.
\displaystyle \text{Here, }P=\frac{1}{x\log|x|}\text{ and }Q=\frac{2}{x^2}
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\frac{1}{x\log|x|}dx}=e^{\log|\log x|}
\displaystyle \left[\because I=\int\frac{1}{x\log|x|}dx,\text{ put }\log|x|=t\Rightarrow \frac{1}{x}dx=dt\right]
\displaystyle \therefore I=\int\frac{1}{t}dt=\log|t|=\log|\log x|
\displaystyle \Rightarrow \text{IF}=\log|x|\qquad [\because e^{\log x}=x]
\displaystyle \text{Now, solution of above equation is given by}
\displaystyle y\times\text{IF}=\int(Q\times\text{IF})dx+C
\displaystyle y\log|x|=\int\frac{2}{x^2}\log|x|dx
\displaystyle \Rightarrow y\log|x|=2\left[\log|x|\int\frac{1}{x^2}dx-\int\left(\frac{d}{dx}(\log|x|)\int\frac{1}{x^2}dx\right)dx\right]
\displaystyle \text{[by using integration by parts]}
\displaystyle \Rightarrow y\log|x|=2\left[\log|x|\left(-\frac{1}{x}\right)-\int\frac{1}{x}\left(-\frac{1}{x}\right)dx\right]
\displaystyle \therefore y\log|x|=-\frac{2}{x}\log|x|-\frac{2}{x}+C,
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 118. }\text{Solve the following differential equation } \\ x\cos\left(\frac{y}{x}\right)dy=\left[y\cos\left(\frac{y}{x}\right)+x\right]dx,\ x\neq0.\hspace{0.2cm}\text{[CBSE 2014 C]}
\displaystyle \text{Answer:}
\displaystyle x\cos\left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos\left(\frac{y}{x}\right)+x
\displaystyle \frac{dy}{dx}=\frac{y}{x}+\sec\left(\frac{y}{x}\right)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=v+\sec v
\displaystyle x\frac{dv}{dx}=\sec v
\displaystyle \cos v\,dv=\frac{dx}{x}
\displaystyle \int \cos v\,dv=\int \frac{dx}{x}
\displaystyle \sin v=\log x+C
\displaystyle \therefore \sin\left(\frac{y}{x}\right)=\log x+C
\\

\displaystyle \textbf{Question 119. }\text{Solve the differential equation}
\displaystyle (x^{2}-yx^{2})\,dy+(y^{2}+x^{2}y^{2})\,dx=0,\text{ given that }y=1,   \text{when }x=1. \hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (x^2-yx^2)dy+(y^2+x^2y^2)dx=0
\displaystyle \Rightarrow x^2(1-y)dy+y^2(1+x^2)dx=0
\displaystyle \Rightarrow -x^2(1-y)dy=y^2(1+x^2)dx
\displaystyle \Rightarrow x^2(y-1)dy=y^2(1+x^2)dx
\displaystyle \Rightarrow \frac{y-1}{y^2}dy=\frac{1+x^2}{x^2}dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{y-1}{y^2}dy=\int\frac{1+x^2}{x^2}dx
\displaystyle \Rightarrow \int\frac{1}{y}dy-\int\frac{1}{y^2}dy=\int\frac{1}{x^2}dx+\int1dx
\displaystyle \Rightarrow \log|y|+\frac{1}{y}=-\frac{1}{x}+x+C\qquad ...(i)
\displaystyle \text{Also, given that }y=1,\text{ when }x=1
\displaystyle \text{On putting }y=1\text{ and }x=1\text{ in Eq. (i), we get}
\displaystyle \log|1|+1=-1+1+C
\displaystyle \Rightarrow C=1
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle \log|y|+\frac{1}{y}=-\frac{1}{x}+x+1
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 120. }\text{Solve the following differential equation :}
\displaystyle x\cos\left(\frac{y}{x}\right)(x\,dy+y\,dx)=y\sin\left(\frac{y}{x}\right)(x\,dy-y\,dx).   \hspace{0.2cm}\text{[CBSE 2013C, 2010C]}
\displaystyle \text{Or}
\displaystyle \text{Solve the following differential equation.}
\displaystyle \left(x\cos\frac{y}{x}+y\sin\frac{y}{x}\right)y-\left(y\sin\frac{y}{x}-x\cos\frac{y}{x}\right)\frac{dy}{dx}=0.   \hspace{0.2cm}\text{[CBSE 2010C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is rewritten as}
\displaystyle \left(x\cos\frac{y}{x}+y\sin\frac{y}{x}\right)y-\left(y\sin\frac{y}{x}-x\cos\frac{y}{x}\right)x\frac{dy}{dx}=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{\left(x\cos\frac{y}{x}+y\sin\frac{y}{x}\right)y}{\left(y\sin\frac{y}{x}-x\cos\frac{y}{x}\right)x}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{(x\cos v+vx\sin v)\cdot vx}{(vx\sin v-x\cos v)\cdot x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v-v^2\sin v+v\cos v}{v\sin v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}
\displaystyle \Rightarrow \frac{v\sin v-\cos v}{v\cos v}\,dv=\frac{2dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{v\sin v-\cos v}{v\cos v}\,dv=2\int\frac{dx}{x}
\displaystyle \Rightarrow \int\left(\tan v-\frac{1}{v}\right)dv=2\int\frac{dx}{x}
\displaystyle \Rightarrow \log|\sec v|-\log|v|=2\log|x|+C
\displaystyle \Rightarrow \log|\sec v|-\log|v|-\log|x|^2=C
\displaystyle \Rightarrow \log|\sec v|-\log|vx^2|=C
\displaystyle \Rightarrow \log\left|\frac{\sec v}{vx^2}\right|=C
\displaystyle \Rightarrow \log\left|\frac{\sec\frac{y}{x}}{xy}\right|=C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow \frac{\sec\frac{y}{x}}{xy}=e^C
\displaystyle \Rightarrow \sec\frac{y}{x}=Axy\qquad [\because e^C=A]
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 121. }\text{Find the particular solution of the differential equation}
\displaystyle (3xy+y^{2})\,dx+(x^{2}+xy)\,dy=0,\text{ for }x=1\text{ and }y=1.   \hspace{2.2cm}\text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (3xy+y^2)dx+(x^2+xy)dy=0
\displaystyle \text{It can be rewritten as}
\displaystyle \frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}=-\frac{3\frac{y}{x}+\frac{y^2}{x^2}}{1+\frac{y}{x}}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as }
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}=-\frac{3v+v^2}{1+v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{3v+v^2+v+v^2}{1+v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{2(v^2+2v)}{1+v}
\displaystyle \Rightarrow \frac{(1+v)dv}{2(v^2+2v)}=-\frac{dx}{x}\qquad ...(ii)
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1+v}{2(v^2+2v)}dv=-\int\frac{dx}{x}
\displaystyle \text{Again, put }z=v^2+2v\Rightarrow (2v+2)dv=dz
\displaystyle \Rightarrow (1+v)dv=\frac{dz}{2}
\displaystyle \text{Then, Eq. (ii) becomes}
\displaystyle \int\frac{1}{2}\times\frac{dz}{2z}=-\int\frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{4}\log|z|=-\log|x|+\log|C|
\displaystyle \Rightarrow \frac{1}{4}\left[\log|z|+4\log|x|\right]=\log|C|
\displaystyle \Rightarrow \log|zx^4|=4\log|C|
\displaystyle \Rightarrow zx^4=C^4
\displaystyle \Rightarrow x^4(v^2+2v)=C_1,\text{ where }C_1=C^4
\displaystyle \Rightarrow x^4\left(\frac{y^2}{x^2}+\frac{2y}{x}\right)=C_1\qquad \left[\because v=\frac{y}{x}\right]\qquad ...(iii)
\displaystyle \text{Also, given that }y=1\text{ for }x=1.
\displaystyle \text{On putting }x=1\text{ and }y=1\text{ in Eq. (iii), we get}
\displaystyle 1\left(\frac{1^2}{1^2}+\frac{2}{1}\right)=C_1
\displaystyle \Rightarrow C_1=3
\displaystyle \text{So, on putting }C_1=3\text{ in Eq. (iii), we get}
\displaystyle x^4\left(\frac{y^2}{x^2}+\frac{2y}{x}\right)=3
\displaystyle \therefore x^2y^2+2x^3y=3
\displaystyle \text{which is the required particular solution.}
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\displaystyle \textbf{Question 122. }\text{Find the particular solution of the following differential equation}
\displaystyle \text{ given that }y=0,\text{ when }x=1:   (x^{2}+xy)\,dy=(x^{2}+y^{2})\,dx. \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }(x^2+xy)dy=(x^2+y^2)dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i),}
\displaystyle \text{we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^2+v^2x^2}{x^2+x\cdot vx}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{1-v}{1+v}
\displaystyle \therefore \left(\frac{1+v}{1-v}\right)dv=\frac{1}{x}dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\left(\frac{1+v}{1-v}\right)dv=\int\frac{1}{x}dx
\displaystyle \Rightarrow \int\left[-1+\frac{2}{1-v}\right]dv=\log|x|+\log C
\displaystyle \Rightarrow -v-2\log(1-v)=\log|x|+\log C
\displaystyle \Rightarrow -v=\log(1-v)^2+\log|x|+\log C
\displaystyle \Rightarrow -v=\log\{C|x|(1-v)^2\}
\displaystyle \Rightarrow C|x|(1-v)^2=e^{-v}
\displaystyle \Rightarrow C|x|\left(1-\frac{y}{x}\right)^2=e^{-y/x}\qquad \left[\because v=\frac{y}{x}\right]\qquad ...(ii)
\displaystyle \text{On putting }x=1\text{ and }y=0\text{ in Eq. (ii), we get}
\displaystyle C\cdot1(1-0)=e^0\Rightarrow C=1
\displaystyle \text{Thus, the required solution is}
\displaystyle |x|\left(1-\frac{y}{x}\right)^2=e^{-y/x}\Rightarrow (x-y)^2=|x|e^{-y/x}
\displaystyle \text{which is the required particular solution.}
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\displaystyle \textbf{Question 123. }\text{Show that the differential equation}
\displaystyle x\frac{dy}{dx}\sin\left(\frac{y}{x}\right)+x-y\sin\left(\frac{y}{x}\right)=0\text{ is homogeneous. Find the particular solution of this }
\displaystyle \text{differential equation, given}   \text{that }x=1,\text{ when }y=\frac{\pi}{2}. \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x\frac{dy}{dx}\sin\left(\frac{y}{x}\right)=y\sin\left(\frac{y}{x}\right)-x
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}=-\frac{1}{\sin\frac{y}{x}}\qquad ...(i)
\displaystyle \text{[dividing both sides by }x\sin\frac{y}{x}]
\displaystyle \text{Let }F(x,y)=\frac{y}{x}-\frac{1}{\sin\frac{y}{x}}
\displaystyle \text{On replacing }x\text{ by }\lambda x\text{ and }y\text{ by }\lambda y\text{ both sides, we get}
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}-\frac{1}{\sin\frac{\lambda y}{\lambda x}}=\lambda^0\left(\frac{y}{x}-\frac{1}{\sin\frac{y}{x}}\right)
\displaystyle =\lambda^0F(x,y)
\displaystyle \text{So, given differential equation is homogeneous.}
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i),}
\displaystyle \text{we get}
\displaystyle v+x\frac{dv}{dx}=v-\frac{1}{\sin v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{1}{\sin v}
\displaystyle \Rightarrow \sin v\,dv=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\sin v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -\cos v=-\log|x|+C
\displaystyle \Rightarrow -\cos\frac{y}{x}=-\log|x|+C\qquad \left[\text{put }v=\frac{y}{x}\right]\qquad ...(ii)
\displaystyle \text{Also, given that }x=1,\text{ when }y=\frac{\pi}{2}
\displaystyle \text{On putting }x=1\text{ and }y=\frac{\pi}{2}\text{ in Eq. (ii), we get}
\displaystyle -\cos\frac{\pi}{2}=-\log|1|+C
\displaystyle \Rightarrow 0=0+C\Rightarrow C=0
\displaystyle \text{On putting the value of }C\text{ in Eq. (ii), we get}
\displaystyle \cos\frac{y}{x}=\log|x|
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 124. }\text{Find the particular solution of the differential } \text{equation }
\displaystyle \frac{dx}{dy}+x\cot y=2y+y^{2}\cot y,\ (y\neq0),\text{ given that } x=0, \text{ when }y=\frac{\pi}{2}.    \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dx}{dy}+x\cot y=2y+y^2\cot y,\quad (y\neq0)
\displaystyle \text{which is a linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q,\text{ here }P=\cot y\text{ and }Q=2y+y^2\cot y
\displaystyle \therefore \text{IF}=e^{\int Pdy}=e^{\int\cot y\,dy}=e^{\log|\sin y|}=\sin y
\displaystyle \text{The solution of the differential equation is given by}
\displaystyle x\times\text{IF}=\int Q\times\text{IF}\,dy+C
\displaystyle \therefore x\sin y=\int(2y+y^2\cot y)\sin y\,dy+C
\displaystyle =2\int y\sin y\,dy+\int y^2\cos y\,dy+C
\displaystyle =2\int y\sin y\,dy+y^2\sin y-\int2y\sin y\,dy+C
\displaystyle \text{[using integration by parts]}
\displaystyle =2\int y\sin y\,dy+y^2\sin y-2\int y\sin y\,dy+C
\displaystyle \Rightarrow x\sin y=y^2\sin y+C\qquad ...(i)
\displaystyle \text{Also, given that }x=0,\text{ when }y=\frac{\pi}{2}.
\displaystyle \text{On putting }x=0\text{ and }y=\frac{\pi}{2}\text{ in Eq. (i), we get}
\displaystyle 0=\left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2}+C
\displaystyle \Rightarrow C=-\frac{\pi^2}{4}\qquad \left[\because \sin\frac{\pi}{2}=1\right]
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle x\sin y=y^2\sin y-\frac{\pi^2}{4}
\displaystyle \Rightarrow x=y^2-\frac{\pi^2}{4}\mathrm{cosec} y
\displaystyle \text{which is required particular solution of given differential equation.}
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\displaystyle \textbf{Question 125. }\text{Find the particular solution of the differential equation } \\ (\tan^{-1}y-x)\,dy=(1+y^{2})\,dx,\text{ given that }x=0,\text{ when }y=0.\hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle (\tan^{-1}y-x)\,dy=(1+y^{2})\,dx
\displaystyle \frac{dx}{dy}=\frac{\tan^{-1}y-x}{1+y^{2}}
\displaystyle \frac{dx}{dy}+\frac{x}{1+y^{2}}=\frac{\tan^{-1}y}{1+y^{2}}
\displaystyle \text{This is a linear differential equation in }x.
\displaystyle \text{I.F.}=e^{\int \frac{1}{1+y^{2}}\,dy}=e^{\tan^{-1}y}
\displaystyle \text{Hence, }x e^{\tan^{-1}y}=\int e^{\tan^{-1}y}\frac{\tan^{-1}y}{1+y^{2}}\,dy
\displaystyle \text{Let }\tan^{-1}y=t,\text{ then }\frac{dy}{1+y^{2}}=dt
\displaystyle x e^{\tan^{-1}y}=\int te^{t}\,dt
\displaystyle x e^{\tan^{-1}y}=e^{t}(t-1)+C
\displaystyle x e^{\tan^{-1}y}=e^{\tan^{-1}y}(\tan^{-1}y-1)+C
\displaystyle x=\tan^{-1}y-1+Ce^{-\tan^{-1}y}
\displaystyle \text{Given, }x=0\text{ when }y=0
\displaystyle 0=0-1+C
\displaystyle C=1
\displaystyle \therefore x=\tan^{-1}y-1+e^{-\tan^{-1}y}
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\displaystyle \textbf{Question 126. }\text{Solve the following differential equation}
\displaystyle \frac{dy}{dx}-y=\cos x,\text{ given that if }x=0,\ y=1. \hspace{2.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{dy}{dx}-y=\cos x
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\text{ here }P=-1\text{ and }Q=\cos x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int(-1)dx}=e^{-x}
\displaystyle \text{The general solution is given by}
\displaystyle y\times\text{IF}=\int(\text{IF}\times Q)\,dx+C
\displaystyle \Rightarrow ye^{-x}=\int e^{-x}\cos x\,dx+C\qquad ...(i)
\displaystyle \text{Now, }\int e^{-x}\cos x\,dx=e^{-x}\sin x+\int e^{-x}\sin x\,dx
\displaystyle \text{[integrating by parts]}
\displaystyle =e^{-x}\sin x-e^{-x}\cos x-\int e^{-x}\cos x\,dx
\displaystyle \Rightarrow 2\int e^{-x}\cos x\,dx=e^{-x}(\sin x-\cos x)
\displaystyle \Rightarrow \int e^{-x}\cos x\,dx=\frac{1}{2}e^{-x}(\sin x-\cos x)\qquad ...(ii)
\displaystyle \text{On substituting this value in Eq. (i), we get}
\displaystyle ye^{-x}=\frac{1}{2}e^{-x}(\sin x-\cos x)+C\qquad ...(iii)
\displaystyle \text{On putting }x=0,\ y=1\text{ in Eq. (iii), we get}
\displaystyle 1\cdot e^0=\frac{1}{2}e^0(\sin0-\cos0)+C
\displaystyle \Rightarrow 1=\frac{1}{2}(-1)+C\Rightarrow C=\frac{3}{2}
\displaystyle \text{On putting }C=\frac{3}{2}\text{ in Eq. (iii), we get}
\displaystyle ye^{-x}=\frac{1}{2}e^{-x}(\sin x-\cos x)+\frac{3}{2}
\displaystyle \Rightarrow y=\frac{1}{2}(\sin x-\cos x)+\frac{3}{2}e^x
\displaystyle \text{which is the required particular solution.}
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\displaystyle \textbf{Question 127. }\text{Find the particular solution of the following differential equation,} \\ \text{given that }x=2,\ y=1.\ x\frac{dy}{dx}+2y=x^{2},\ (x\neq0).\hspace{2.2cm}\text{[CBSE 2012 C]}
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}+2y=x^{2}
\displaystyle \frac{dy}{dx}+\frac{2}{x}y=x
\displaystyle \text{This is a linear differential equation.}
\displaystyle \text{I.F.}=e^{\int \frac{2}{x}\,dx}=e^{2\log x}=x^{2}
\displaystyle \text{Multiplying by I.F.,}
\displaystyle x^{2}\frac{dy}{dx}+2xy=x^{3}
\displaystyle \frac{d}{dx}(x^{2}y)=x^{3}
\displaystyle x^{2}y=\int x^{3}\,dx
\displaystyle x^{2}y=\frac{x^{4}}{4}+C
\displaystyle y=\frac{x^{2}}{4}+\frac{C}{x^{2}}
\displaystyle \text{Given, }x=2,\ y=1
\displaystyle 1=\frac{2^{2}}{4}+\frac{C}{2^{2}}
\displaystyle 1=1+\frac{C}{4}
\displaystyle C=0
\displaystyle \therefore y=\frac{x^{2}}{4}
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\displaystyle \textbf{Question 128. }\text{Find the particular solution of differential equation}
\displaystyle \frac{dy}{dx}+y\cot x=2x+x^{2}\cot x,  x\neq0,\text{ given that }y=0,\text{ when }x=\frac{\pi}{2}. \hspace{0.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{dy}{dx}+y\cot x=2x+x^2\cot x,\ (x\neq0)
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q.
\displaystyle \text{Here, }P=\cot x\text{ and }Q=2x+x^2\cot x.
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\cot xdx}=e^{\log|\sin x|}=\sin x
\displaystyle \text{The general solution is given by}
\displaystyle y\cdot\text{IF}=\int(\text{IF}\times Q)dx+C
\displaystyle \Rightarrow y\sin x=\int(2x+x^2\cot x)\sin x\,dx+C
\displaystyle =2\int x\sin x\,dx+\int x^2\cos x\,dx+C
\displaystyle =2\int x\sin x\,dx+x^2\sin x-\int2x\sin x\,dx+C
\displaystyle \Rightarrow y\sin x=x^2\sin x+C\qquad ...(i)
\displaystyle \text{On putting }x=\frac{\pi}{2}\text{ and }y=0\text{ in Eq. (i), we get}
\displaystyle 0\cdot\sin\frac{\pi}{2}=\left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2}+C\Rightarrow C=-\frac{\pi^2}{4}
\displaystyle \text{On putting }C=-\frac{\pi^2}{4}\text{ in Eq. (i), we get}
\displaystyle y\sin x=x^2\sin x-\frac{\pi^2}{4}
\displaystyle \therefore y=x^2-\frac{\pi^2}{4}\mathrm{cosec} x
\displaystyle \text{[dividing both sides by }\sin x\text{]}
\\

\displaystyle \textbf{Question 129. }\text{Solve the following differential equation } \\ \frac{dy}{dx}+y\cot x=4x\,\mathrm{cosec}\,x,\text{ given that }y=0,\text{ when }x=\frac{\pi}{2}. \\ \hspace{0.2cm}\text{[CBSE 2012 C; CBSE 2012; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}+y\cot x=4x\,\mathrm{cosec}\,x
\displaystyle \text{This is a linear differential equation.}
\displaystyle \text{I.F.}=e^{\int \cot x\,dx}=e^{\log\sin x}=\sin x
\displaystyle \text{Multiplying by I.F.,}
\displaystyle \sin x\frac{dy}{dx}+y\sin x\cot x=4x
\displaystyle \sin x\frac{dy}{dx}+y\cos x=4x
\displaystyle \frac{d}{dx}(y\sin x)=4x
\displaystyle y\sin x=\int 4x\,dx
\displaystyle y\sin x=2x^{2}+C
\displaystyle \text{Given, }y=0\text{ when }x=\frac{\pi}{2}
\displaystyle 0=2\left(\frac{\pi}{2}\right)^{2}+C
\displaystyle C=-\frac{\pi^{2}}{2}
\displaystyle \therefore y\sin x=2x^{2}-\frac{\pi^{2}}{2}

\displaystyle \textbf{Question 130. }\text{Find the particular solution of the following differential equation}
\displaystyle x\frac{dy}{dx}=(x+2)(y+2);\ y=-1\text{ when }x=1. \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }xy\frac{dy}{dx}=(x+2)(y+2)
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{y}{y+2}dy=\frac{x+2}{x}dx
\displaystyle \Rightarrow \left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\left(1-\frac{2}{y+2}\right)dy=\int\left(1+\frac{2}{x}\right)dx
\displaystyle \Rightarrow y-2\log|y+2|=x+2\log|x|+C\qquad ...(i)
\displaystyle \text{Given that }y=-1,\text{ when }x=1
\displaystyle \text{On putting }x=1\text{ and }y=-1\text{ in Eq. (i), we get}
\displaystyle -1-2\log(1)=1+2\log|1|+C
\displaystyle \Rightarrow -1=1+C
\displaystyle \Rightarrow C=-2
\displaystyle \text{On putting }C=-2\text{ in Eq. (i), we get}
\displaystyle y-2\log|y+2|=x+2\log|x|-2
\displaystyle \text{which is required particular solution.}
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\displaystyle \textbf{Question 131. }\text{Solve the following differential eq. }   2x^{2}\frac{dy}{dx}-2xy+y^{2}=0 \hspace{0.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle 2x^2\frac{dy}{dx}-2xy+y^2=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\frac{y^2}{2x^2}\qquad ...(i)
\displaystyle \text{which is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i),}
\displaystyle \text{we get}
\displaystyle v+x\frac{dv}{dx}=v-\frac{v^2}{2}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^2}{2}\Rightarrow \frac{2dv}{v^2}=-\frac{dx}{x}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{2dv}{v^2}=-\int\frac{dx}{x}
\displaystyle \Rightarrow 2\int v^{-2}dv=-\log|x|+C
\displaystyle \Rightarrow \frac{2v^{-1}}{-1}=-\log|x|+C
\displaystyle \Rightarrow -\frac{2}{v}=-\log|x|+C
\displaystyle \Rightarrow -\frac{2x}{y}=-\log|x|+C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \Rightarrow y=\frac{-2x}{-\log|x|+C}
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 132. }\text{Solve the following differential equation}
\displaystyle \frac{dy}{dx}=1+x^{2}+y^{2}+x^{2}y^{2},\text{ given that }y=1,\text{ when }x=0. \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}=1+x^2+y^2+x^2y^2
\displaystyle \Rightarrow \frac{dy}{dx}=1(1+x^2)+y^2(1+x^2)
\displaystyle \Rightarrow \frac{dy}{dx}=(1+x^2)(1+y^2)
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{dy}{1+y^2}=(1+x^2)dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{dy}{1+y^2}=\int(1+x^2)dx
\displaystyle \Rightarrow \tan^{-1}y=x+\frac{x^3}{3}+C\qquad ...(i)
\displaystyle \text{Also, given that }y=1,\text{ when }x=0.
\displaystyle \text{On putting }x=0\text{ and }y=1\text{ in Eq. (i), we get}
\displaystyle \tan^{-1}1=C
\displaystyle \Rightarrow C=\frac{\pi}{4}\qquad \left[\because \tan^{-1}1=\frac{\pi}{4}\right]
\displaystyle \text{On putting the value of }C\text{ in Eq. (i), we get}
\displaystyle \tan^{-1}y=x+\frac{x^3}{3}+\frac{\pi}{4}
\displaystyle \therefore y=\tan\left(x+\frac{x^3}{3}+\frac{\pi}{4}\right)
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 133. }\text{Solve the following differential equation}
\displaystyle \frac{dy}{dx}+y\sec x=\tan x,\ \left(0\leq x<\frac{\pi}{2}\right). \hspace{2.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle \frac{dy}{dx}+y\sec x=\tan x
\displaystyle \text{which is a linear differential equation of first order and is of the form}
\displaystyle \frac{dy}{dx}+Py=Q\qquad ...(i)
\displaystyle \text{Here, }P=\sec x\text{ and }Q=\tan x
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\sec xdx}=e^{\log|\sec x+\tan x|}
\displaystyle \Rightarrow \text{IF}=\sec x+\tan x
\displaystyle \text{The general solution is}
\displaystyle y\times\text{IF}=\int(Q\times\text{IF})dx+C
\displaystyle \Rightarrow y(\sec x+\tan x)=\int\tan x(\sec x+\tan x)dx+C
\displaystyle \Rightarrow y(\sec x+\tan x)=\int\sec x\tan xdx+\int\tan^2xdx+C
\displaystyle \Rightarrow y(\sec x+\tan x)=\sec x+\int(\sec^2x-1)dx
\displaystyle \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C
\displaystyle \text{On dividing both sides by }(\sec x+\tan x),\text{ we get}
\displaystyle y=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 134. }\text{Solve the following differential equation}
\displaystyle x(x^{2}-1)\frac{dy}{dx}=1,\ y=0,\text{ when }x=2. \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle x(x^2-1)\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x(x^2-1)}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x(x-1)(x+1)}\qquad [\because a^2-b^2=(a-b)(a+b)]
\displaystyle \Rightarrow dy=\frac{dx}{x(x-1)(x+1)}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dy=\int\frac{dx}{x(x-1)(x+1)}
\displaystyle \Rightarrow y=I+K\qquad ...(i)
\displaystyle \text{where, }I=\int\frac{dx}{x(x-1)(x+1)}
\displaystyle \text{By using partial fraction method,}
\displaystyle \text{let }\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}
\displaystyle \Rightarrow 1=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)
\displaystyle =A(x^2-1)+B(x^2+x)+C(x^2-x)
\displaystyle \text{On comparing the coefficients of }x^2,\ x\text{ and constant terms from both sides, we get}
\displaystyle A+B+C=0,\quad B-C=0,\quad -A=1
\displaystyle \Rightarrow A=-1
\displaystyle \text{On solving above equations, we get}
\displaystyle A=-1,\ B=\frac{1}{2}\text{ and }C=\frac{1}{2}
\displaystyle \text{then }\frac{1}{x(x-1)(x+1)}=-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}
\displaystyle \text{On integrating both sides w.r.t. }x,\text{ we get}
\displaystyle I=\int\frac{dx}{x(x-1)(x+1)}
\displaystyle =\int-\frac{1}{x}dx+\frac{1}{2}\int\frac{dx}{x-1}+\frac{1}{2}\int\frac{dx}{x+1}
\displaystyle \Rightarrow I=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|
\displaystyle \text{On putting the value of }I\text{ in Eq. (i), we get}
\displaystyle y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+K\qquad ...(ii)
\displaystyle \text{Also, given that }y=0,\text{ when }x=2.
\displaystyle \text{On putting }y=0\text{ and }x=2\text{ in Eq. (ii), we get}
\displaystyle 0=-\log2+\frac{1}{2}\log1+\frac{1}{2}\log3+K
\displaystyle \Rightarrow K=\log2-\frac{1}{2}\log1-\frac{1}{2}\log3
\displaystyle \Rightarrow K=\log2-\log\sqrt3\qquad [\because \log1=0]
\displaystyle \Rightarrow K=\log\frac{2}{\sqrt3}
\displaystyle \text{On putting the value of }K\text{ in Eq. (ii), we get}
\displaystyle y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+\log\frac{2}{\sqrt3}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 135. }\text{Solve the following differential equation}
\displaystyle (1+x^{2})dy+2xy\,dx=\cot x\,dx,\text{ where }x\neq0. \hspace{0.2cm}\text{[CBSE 2012, 2011C, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1+x^2)dy+2xydx=\cot x\,dx\qquad [x\neq0]
\displaystyle \Rightarrow (1+x^2)dy=(\cot x-2xy)dx
\displaystyle \text{On dividing both sides by }(1+x^2),\text{ we get}
\displaystyle dy=\frac{\cot x-2xy}{1+x^2}dx
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{\cot x}{1+x^2}
\displaystyle \text{which is a linear differential equation of }1\text{st order and is of the form}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, }P=\frac{2x}{1+x^2}\text{ and }Q=\frac{\cot x}{1+x^2}
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int\frac{2x}{1+x^2}dx}=e^{\log|1+x^2|}=1+x^2
\displaystyle \text{The solution of linear differential equation is given by}
\displaystyle y\times\text{IF}=\int(Q\times\text{IF})dx+C
\displaystyle \Rightarrow y(1+x^2)=\int\frac{\cot x}{1+x^2}(1+x^2)dx+C
\displaystyle \Rightarrow y(1+x^2)=\int\cot x\,dx+C
\displaystyle \Rightarrow y(1+x^2)=\log|\sin x|+C
\displaystyle \text{On dividing both sides by }(1+x^2),\text{ we get}
\displaystyle y=\frac{\log|\sin x|}{1+x^2}+\frac{C}{1+x^2}
\displaystyle \text{which is the required solution.}
\\

\displaystyle \textbf{Question 136. }\text{Find the particular solution of the following } \text{differential equation }
\displaystyle x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0,\text{ given that }   \text{when }x=2,\ y=\pi. \hspace{0.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}+\sin\left(\frac{y}{x}\right)=0
\displaystyle \text{[dividing both sides by }x]\qquad ...(i)
\displaystyle \text{This is a homogeneous differential equation as}
\displaystyle \frac{dy}{dx}=F\left(\frac{y}{x}\right)
\displaystyle \text{On putting }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in Eq. (i), we get}
\displaystyle v+x\frac{dv}{dx}-v+\sin v=0
\displaystyle \Rightarrow x\frac{dv}{dx}+\sin v=0
\displaystyle \Rightarrow \mathrm{cosec} v\,dv+\frac{dx}{x}=0
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\mathrm{cosec} v\,dv+\int\frac{dx}{x}=\log C
\displaystyle \Rightarrow \log|\mathrm{cosec} v-\cot v|+\log x=\log C
\displaystyle \Rightarrow x(\mathrm{cosec} v-\cot v)=C
\displaystyle \Rightarrow x\left[\mathrm{cosec}\left(\frac{y}{x}\right)-\cot\left(\frac{y}{x}\right)\right]=C\qquad \left[\because v=\frac{y}{x}\right]
\displaystyle \text{On putting }x=2\text{ and }y=\pi\text{ in above equation, we get}
\displaystyle 2\left[\mathrm{cosec}\left(\frac{\pi}{2}\right)-\cot\left(\frac{\pi}{2}\right)\right]=C\Rightarrow C=2
\displaystyle \text{On putting }C=2\text{ in Eq. (ii), we get}
\displaystyle x\left[\mathrm{cosec}\left(\frac{y}{x}\right)-\cot\left(\frac{y}{x}\right)\right]=2
\displaystyle \text{which is the required particular solution.}
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\displaystyle \textbf{Question 137. }\text{Solve the following differential equation}
\displaystyle (1+y^{2})(1+\log|x|)\,dx+x\,dy=0. \hspace{2.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is}
\displaystyle (1+y^2)(1+\log|x|)\,dx+xdy=0
\displaystyle \text{On separating the variables, we get}
\displaystyle \frac{1+\log|x|}{x}dx=-\frac{dy}{1+y^2}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1+\log|x|}{x}dx=-\int\frac{dy}{1+y^2}
\displaystyle \Rightarrow \log|x|+I_1+K=-\tan^{-1}y\qquad ...(i)
\displaystyle \text{where, }I_1=\int\frac{\log|x|}{x}dx
\displaystyle \text{On putting }\log|x|=t\Rightarrow \frac{1}{x}dx=dt
\displaystyle \therefore I_1=\int t\,dt=\frac{t^2}{2}+C_1=\frac{(\log|x|)^2}{2}+C_1
\displaystyle \text{On putting the value of }I_1\text{ in Eq. (i), we get}
\displaystyle \log|x|+\frac{(\log|x|)^2}{2}+C=-\tan^{-1}y
\displaystyle \Rightarrow \tan^{-1}y=-\log|x|-\frac{(\log|x|)^2}{2}-C
\displaystyle \therefore y=\tan\left[-\log|x|-\frac{(\log|x|)^2}{2}-C\right]
\displaystyle \text{which is the required solution.}
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