\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.}

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