\displaystyle \textbf{Question 1: } \text{The degree of the differential equation } \frac{d^2 y}{dx^2}+3\left(\frac{dy}{dx}\right)^2=x^2 \log\left(\frac{d^2 y}{dx^2}\right) \text{ is}   
\displaystyle \text{(a) } 2 \quad \text{(b) } 1 \quad \text{(c) } 3 \quad \text{(d) Not defined}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, } \frac{d^{2}y}{dx^{2}}+3\left(\frac{dy}{dx}\right)^{2}=x^{2}\log\left(\frac{d^{2}y}{dx^{2}}\right)
\displaystyle \text{Here, degree is not defined since equation is not polynomial in derivatives}
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\displaystyle \textbf{Question 2: } \text{The order and degree of the differential equation }  \\ \frac{d^3 y}{dx^3}+\frac{d^2 y}{dx^2}+\left(\frac{dy}{dx}\right)^2=3 \text{ is}
\displaystyle \text{(a) order 3 and degree 1 \quad (b) order 1 and degree 3}
\displaystyle \text{(c) order 2 and degree 1 \quad (d) order 2 and degree 2}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } \frac{d^{3}y}{dx^{3}}+\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=3
\displaystyle \text{Highest order derivative is } \frac{d^{3}y}{dx^{3}}
\displaystyle \therefore \text{Order}=3,\ \text{Degree}=1
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\displaystyle \textbf{Question 3: } \text{The degree of the differential equation } \frac{d^2 y}{dx^2}+3\left(\frac{dy}{dx}\right)^2=x^2\left(\frac{d^2 y}{dx^2}\right)^2 \text{ is} 
\displaystyle \text{(a) } 1 \quad \text{(b) } 2 \quad \text{(c) } 3 \quad \text{(d) } 4
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } \frac{d^{2}y}{dx^{2}}+3\left(\frac{dy}{dx}\right)^{2}=x^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{2} \quad (i)
\displaystyle \text{Highest order derivative is } \frac{d^{2}y}{dx^{2}} \text{ and its power is } 2
\displaystyle \therefore \text{Order}=2,\ \text{Degree}=2
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\displaystyle \textbf{Question 4: } \text{Find the sum of order and degree of the differential equation:}
\displaystyle \left(\frac{dy}{dx}\right)^5+3xy\left(\frac{d^3 y}{dx^3}\right)^2+y^2\left(\frac{d^2 y}{dx^2}\right)^3=0 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \left(\frac{dy}{dx}\right)^{5}+3x y\left(\frac{d^{3}y}{dx^{3}}\right)^{2}+y^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{3}=0
\displaystyle \text{Highest order derivative is } \frac{d^{3}y}{dx^{3}}
\displaystyle \therefore \text{Order}=3,\ \text{Degree}=2
\displaystyle \text{Required sum}=3+2=5
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\displaystyle \textbf{Question 5: } \text{Solve } \frac{dy}{dx}=\sin x-x.  
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \frac{dy}{dx}=\sin x-x
\displaystyle dy=(\sin x-x)\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dy=\int (\sin x-x)\,dx
\displaystyle y=-\cos x-\frac{x^{2}}{2}+C
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\displaystyle \textbf{Question 6: } \text{Solve } \frac{dy}{dx}+2x=e^{3x}.  
\displaystyle \text{Answer:}
\displaystyle \text{We have, } \frac{dy}{dx}+2x=e^{3x}
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x}-2x
\displaystyle dy=(e^{3x}-2x)\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dy=\int (e^{3x}-2x)\,dx
\displaystyle y=\frac{e^{3x}}{3}-x^{2}+C
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\displaystyle \textbf{Question 7: } \text{Solve the differential equation } (1+y^{2})(1+\log x)\,dx+x\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, } (1+y^{2})(1+\log x)\,dx+x\,dy=0
\displaystyle \Rightarrow \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 \text{Let } \log x=t \Rightarrow \frac{1}{x}dx=dt
\displaystyle \therefore \int (1+t)\,dt=-\tan^{-1}y+C
\displaystyle t+\frac{t^{2}}{2}=-\tan^{-1}y+C
\displaystyle \Rightarrow \log x+\frac{(\log x)^{2}}{2}+\tan^{-1}y=C
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\displaystyle \textbf{Question 8: } \text{Solve the differential equation } \mathrm{cosec}^{3}x\,dy-\mathrm{cosec}\,y\,dx=0. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \mathrm{cosec}^{3}x\,dy-\mathrm{cosec}\,y\,dx=0
\displaystyle \Rightarrow \mathrm{cosec}^{3}x\,dy=\mathrm{cosec}\,y\,dx
\displaystyle \Rightarrow \frac{dy}{\sin^{3}x}=\frac{dx}{\sin y}
\displaystyle \Rightarrow \sin y\,dy=\sin^{3}x\,dx
\displaystyle \text{Using identity } \sin^{3}x=\frac{3\sin x-\sin 3x}{4}
\displaystyle \therefore \sin y\,dy=\frac{3\sin x-\sin 3x}{4}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int \sin y\,dy=\int \frac{3\sin x-\sin 3x}{4}\,dx
\displaystyle -\cos y=-\frac{3}{4}\cos x+\frac{\cos 3x}{12}+C
\displaystyle \Rightarrow 12\cos y-9\cos x+\cos 3x+C=0
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\displaystyle \textbf{Question 9: } \text{Solve the differential equation } \frac{dy}{dx}=2^{-y}. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \frac{dy}{dx}=2^{-y} \Rightarrow 2^{y}dy=dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int 2^{y}dy=\int dx
\displaystyle \frac{2^{y}}{\log 2}=x+C
\displaystyle \Rightarrow 2^{y}=x\log 2+C
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\displaystyle \textbf{Question 10: } \text{Solve } \frac{dy}{dx}=1-xy+y-x. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \frac{dy}{dx}=1-xy+y-x
\displaystyle \Rightarrow \frac{dy}{dx}=(1-x)+y(1-x)
\displaystyle \Rightarrow \frac{dy}{dx}=(1-x)(1+y)
\displaystyle \Rightarrow \frac{dy}{1+y}=(1-x)\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int \frac{dy}{1+y}=\int (1-x)\,dx
\displaystyle \log(1+y)=x-\frac{x^{2}}{2}+C
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\displaystyle \textbf{Question 11: } \text{For what value of } n \text{ is the given a homogeneous differential equation? } \\ \frac{dy}{dx}=\frac{x^{3}-y^{3}}{x^{n}y+xy^{n}} 
\displaystyle \text{(a) } 1 \quad \text{(b) } 3 \quad \text{(c) } 0 \quad \text{(d) } 2
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, } \frac{dy}{dx}=\frac{x^{3}-y^{3}}{x^{2}y+xy^{2}}
\displaystyle \text{Let } f(x,y)=\frac{x^{3}-y^{3}}{x^{2}y+xy^{2}}
\displaystyle \text{On putting } x=\lambda x,\ y=\lambda y
\displaystyle f(\lambda x,\lambda y)=\frac{\lambda^{3}(x^{3}-y^{3})}{\lambda^{3}(x^{2}y+xy^{2})}=\lambda^{0}f(x,y)
\displaystyle \therefore \text{Equation is homogeneous}
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\displaystyle \textbf{Question 12: } \text{Solve the following differential equation } x\frac{dy}{dx}+2y=x^{2}\log x. 
\displaystyle \text{Answer:}
\displaystyle \text{We have, } x\frac{dy}{dx}+2y=x^{2}\log x
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x\log x
\displaystyle \text{This is of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=\frac{2}{x},\ Q=x\log x
\displaystyle \text{IF}=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^{2}
\displaystyle \text{Solution: } y\cdot IF=\int Q\cdot IF\,dx+C
\displaystyle \Rightarrow yx^{2}=\int x^{3}\log x\,dx+C
\displaystyle \text{Using integration by parts:}
\displaystyle \int x^{3}\log x\,dx=\log x\cdot \frac{x^{4}}{4}-\int \frac{1}{x}\cdot \frac{x^{4}}{4}dx
\displaystyle =\frac{x^{4}}{4}\log x-\int \frac{x^{3}}{4}dx
\displaystyle =\frac{x^{4}}{4}\log x-\frac{x^{4}}{16}+C
\displaystyle \Rightarrow yx^{2}=\frac{x^{4}}{4}\log x-\frac{x^{4}}{16}+C
\displaystyle \Rightarrow y=\frac{x^{2}}{4}\log x-\frac{x^{2}}{16}+\frac{C}{x^{2}}
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\displaystyle \textbf{Question 13: } \text{Solve } x^{3}dy+(xy-1)\,dx=0, \text{ given } y=3-e \text{ when }  \\ x=\frac{1}{2}, \text{ then find } y \text{ when } x=1.  
\displaystyle \text{Answer:}
\displaystyle \text{We have, } x^{3}\frac{dy}{dx}+(xy-1)dx=0
\displaystyle \Rightarrow x^{3}\frac{dy}{dx}+xy-1=0
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{x^{2}}=\frac{1}{x^{3}}
\displaystyle \text{This is linear differential equation of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=\frac{1}{x^{2}},\ Q=\frac{1}{x^{3}}
\displaystyle \therefore \text{IF}=e^{\int Pdx}=e^{\int \frac{1}{x^{2}}dx}=e^{-1/x}
\displaystyle \text{Solution is given by } y\cdot IF=\int Q\cdot IF\,dx+C \quad (i)
\displaystyle \Rightarrow ye^{-1/x}=\int \frac{1}{x^{3}}e^{-1/x}dx+C
\displaystyle \text{Let } t=-\frac{1}{x} \Rightarrow \frac{1}{x^{2}}dx=dt
\displaystyle \Rightarrow I=\int \frac{1}{x}e^{t}dt
\displaystyle =\int (-t)e^{t}dt
\displaystyle \text{Using integration by parts, we get}
\displaystyle \int te^{t}dt=te^{t}-e^{t}+C
\displaystyle \Rightarrow I=-(te^{t}-e^{t})=-te^{t}+e^{t}
\displaystyle \Rightarrow I=e^{t}(1-t)
\displaystyle \Rightarrow I=e^{-1/x}\left(1+\frac{1}{x}\right)
\displaystyle \text{From Eq. (i), we get}
\displaystyle ye^{-1/x}=e^{-1/x}\left(1+\frac{1}{x}\right)+C
\displaystyle \text{Given, } y=3-e \text{ when } x=\frac{1}{2}
\displaystyle \Rightarrow (3-e)e^{-2}=e^{-2}(1+2)+C
\displaystyle \Rightarrow 3e^{-2}-e^{-1}=3e^{-2}+C
\displaystyle \Rightarrow C=-e^{-1}
\displaystyle \text{On putting } C \text{ in solution, we get}
\displaystyle ye^{-1/x}=e^{-1/x}\left(1+\frac{1}{x}\right)-e^{-1}
\displaystyle \text{Now, on putting } x=1, \text{ we get}
\displaystyle y e^{-1}=e^{-1}(2)-e^{-1}
\displaystyle \Rightarrow y=1
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\displaystyle \textbf{Question 14: } \text{Find the particular solution of the differential equation } \\ (1+x^{2})dy=(\tan^{-1}x-y)\,dx, \text{ given that } y=1, \text{ when } x=0. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } (1+x^{2})\frac{dy}{dx}=\tan^{-1}x-y
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{1+x^{2}}=\frac{\tan^{-1}x}{1+x^{2}}
\displaystyle \text{Comparing with } \frac{dy}{dx}+P(x)y=Q(x), \text{ we get}
\displaystyle P(x)=\frac{1}{1+x^{2}},\quad Q(x)=\frac{\tan^{-1}x}{1+x^{2}}
\displaystyle \text{Integrating factor, IF}=e^{\int P(x)dx}=e^{\int \frac{1}{1+x^{2}}dx}=e^{\tan^{-1}x}
\displaystyle \text{Solution is } y\cdot IF=\int Q\cdot IF\,dx+C
\displaystyle \Rightarrow y e^{\tan^{-1}x}=\int \frac{\tan^{-1}x}{1+x^{2}}e^{\tan^{-1}x}dx+C
\displaystyle \text{Put } t=\tan^{-1}x \Rightarrow \frac{dx}{1+x^{2}}=dt
\displaystyle \Rightarrow y e^{t}=\int t e^{t}dt+C
\displaystyle \text{Using integration by parts: } \int t e^{t}dt=te^{t}-e^{t}+C
\displaystyle \Rightarrow y e^{t}=te^{t}-e^{t}+C
\displaystyle \Rightarrow y=t-1+Ce^{-t}
\displaystyle \Rightarrow y=\tan^{-1}x-1+Ce^{-\tan^{-1}x} \quad (i)
\displaystyle \text{Where, } x=0,\ y=1
\displaystyle \Rightarrow 1=0-1+C \Rightarrow C=2
\displaystyle \text{On substituting } C=2 \text{ in Eq. (i), we get}
\displaystyle y=\tan^{-1}x-1+2e^{-\tan^{-1}x}
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\displaystyle \textbf{Question 15: } \text{Write a particular solution of the differential equation } \\ \frac{dy}{dx}=\frac{y^{2}}{xy-x^{2}}, \text{ when } x=1, y=1. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \frac{dy}{dx}=\frac{y^{2}}{xy-x^{2}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}}{x(y-x)}
\displaystyle \text{Put } y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v^{2}x^{2}}{x(vx-x)}=\frac{v^{2}}{v-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}}{v-1}-v
\displaystyle =\frac{v^{2}-v(v-1)}{v-1}=\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 \log|x|=v-\log|v|+C
\displaystyle \Rightarrow \log|x|=\frac{y}{x}-\log\left|\frac{y}{x}\right|+C
\displaystyle \Rightarrow \log|x|=\frac{y}{x}-\log|y|+\log|x|+C
\displaystyle \Rightarrow \log|y|=\frac{y}{x}+C
\displaystyle \Rightarrow x\log y-y=Cx
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\displaystyle \textbf{Question 16: } \text{Write a particular solution of the differential equation } \\ (1+x^{2})\frac{dy}{dx}+2xy=\frac{1}{1+x^{2}}, \text{ when } y=0, x=0. 
\displaystyle \text{Answer:}
\displaystyle \text{Given, differential equation is } (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}}
\displaystyle \text{This is linear differential equation of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=\frac{2x}{x^{2}+1},\quad Q=\frac{1}{(x^{2}+1)^{2}}
\displaystyle \text{Now, IF}=e^{\int Pdx}=e^{\int \frac{2x}{x^{2}+1}dx}=e^{\log(x^{2}+1)}=x^{2}+1
\displaystyle \text{So, the required general solution is } y\cdot IF=\int Q\cdot IF\,dx+C
\displaystyle \Rightarrow y(x^{2}+1)=\int \frac{1}{(x^{2}+1)^{2}}(x^{2}+1)dx+C
\displaystyle \Rightarrow y(x^{2}+1)=\int \frac{dx}{x^{2}+1}+C
\displaystyle \Rightarrow y(x^{2}+1)=\tan^{-1}x+C
\displaystyle \text{When } x=0,\ \text{then } y=0
\displaystyle \Rightarrow 0=\tan^{-1}0+C \Rightarrow C=0
\displaystyle \Rightarrow y(x^{2}+1)=\tan^{-1}x
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