Class 12: Differential Equations

$\displaystyle \text{Solve the following differential equations:}$

$\displaystyle \textbf{Question 1: }~\frac{dy}{dx}+2y=e^{3x}. \hspace{2.0cm} \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=e^{3x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=e^{3x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}e^{3x}$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{5x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=\int e^{5x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=\frac{e^{5x}}{5}+C$
$\displaystyle \Rightarrow y=\frac{1}{5}e^{3x}+Ce^{-2x}$
$\displaystyle \text{Hence, } y=\frac{1}{5}e^{3x}+Ce^{-2x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 2: }~4\frac{dy}{dx}+8y=5e^{-3x}. \hspace{2.0cm} \text{[CBSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } 4\frac{dy}{dx}+8y=5e^{-3x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+2y=\frac{5}{4}e^{-3x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=\frac{5}{4}e^{-3x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=\frac{5}{4}e^{2x}e^{-3x}$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=\frac{5}{4}e^{-x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=\frac{5}{4}\int e^{-x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=-\frac{5}{4}e^{-x}+C$
$\displaystyle \Rightarrow y=-\frac{5}{4}e^{-3x}+Ce^{-2x}$
$\displaystyle \text{Hence, } y=-\frac{5}{4}e^{-3x}+Ce^{-2x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 3: }~\frac{dy}{dx}+2y=6e^{x}. \hspace{2.0cm} \text{[CBSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=6e^{x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=6e^{x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=6e^{2x}e^{x}$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=6e^{3x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=6\int e^{3x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=6\left(\frac{e^{3x}}{3}\right)+C$
$\displaystyle \Rightarrow y e^{2x}=2e^{3x}+C$
$\displaystyle \text{Hence, } y e^{2x}=2e^{3x}+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 4: }~\frac{dy}{dx}+y=e^{-2x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y=e^{-2x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=1$
$\displaystyle Q=e^{-2x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 1\,dx}$
$\displaystyle =e^{x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{x}, \text{ we get}$
$\displaystyle e^{x}\left(\frac{dy}{dx}+y\right)=e^{x}e^{-2x}$
$\displaystyle \Rightarrow e^{x}\frac{dy}{dx}+e^{x}y=e^{-x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{x}=\int e^{-x}\,dx + C$
$\displaystyle \Rightarrow y e^{x}=-e^{-x}+C$
$\displaystyle \Rightarrow y=-e^{-2x}+Ce^{-x}$
$\displaystyle \text{Hence, } y=-e^{-2x}+Ce^{-x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 5: }~x\frac{dy}{dx}=x+y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}=x+y$
$\displaystyle \Rightarrow \frac{dy}{dx}=1+\frac{y}{x}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=1 \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{1}{x}$
$\displaystyle Q=1$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\frac{1}{x}\,dx}$
$\displaystyle =e^{-\log|x|}$
$\displaystyle =\frac{1}{|x|}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{x}, \text{ we get}$
$\displaystyle \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}\times 1$
$\displaystyle \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^{2}}y=\frac{1}{x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{y}{x}=\int \frac{1}{x}\,dx + C$
$\displaystyle \Rightarrow \frac{y}{x}=\log|x|+C$
$\displaystyle \text{Hence, } \frac{y}{x}=\log|x|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 6: }~\frac{dy}{dx}+2y=4x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=4x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=4x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=4xe^{2x}$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=4xe^{2x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=4\int x e^{2x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=4\left(x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}\,dx\right)+C$
$\displaystyle \Rightarrow y e^{2x}=2xe^{2x}-2\int e^{2x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=2xe^{2x}-2\left(\frac{e^{2x}}{2}\right)+C$
$\displaystyle \Rightarrow y e^{2x}=2xe^{2x}-e^{2x}+C$
$\displaystyle \Rightarrow y e^{2x}=(2x-1)e^{2x}+C$
$\displaystyle \Rightarrow y=(2x-1)+Ce^{-2x}$
$\displaystyle \text{Hence, } y=(2x-1)+Ce^{-2x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 7: }~x\frac{dy}{dx}+y=xe^{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+y=xe^{x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=e^{x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{x}$
$\displaystyle Q=e^{x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{x}\,dx}$
$\displaystyle =e^{\log|x|}$
$\displaystyle =x$
$\displaystyle \text{Multiplying both sides of (1) by } x, \text{ we get}$
$\displaystyle x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=xe^{x}$
$\displaystyle \Rightarrow x\frac{dy}{dx}+y=xe^{x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle xy=\int xe^{x}\,dx + C$
$\displaystyle \Rightarrow xy=x e^{x}-\int e^{x}\,dx + C$
$\displaystyle \Rightarrow xy=x e^{x}-e^{x}+C$
$\displaystyle \Rightarrow xy=(x-1)e^{x}+C$
$\displaystyle \Rightarrow y=\frac{x-1}{x}e^{x}+\frac{C}{x}$
$\displaystyle \text{Hence, } y=\frac{x-1}{x}e^{x}+\frac{C}{x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 8: }~\frac{dy}{dx}+\frac{4x}{x^{2}+1}\,y+\frac{1}{(x^{2}+1)^{2}}=0. \hspace{2.0cm} \text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+\frac{4x}{x^{2}+1}y+\frac{1}{(x^{2}+1)^{2}}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{4x}{x^{2}+1}y=-\frac{1}{(x^{2}+1)^{2}} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{4x}{x^{2}+1}$
$\displaystyle Q=-\frac{1}{(x^{2}+1)^{2}}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{4x}{x^{2}+1}\,dx}$
$\displaystyle =e^{2\log(x^{2}+1)}$
$\displaystyle =(x^{2}+1)^{2}$
$\displaystyle \text{Multiplying both sides of (1) by } (x^{2}+1)^{2}, \text{ we get}$
$\displaystyle (x^{2}+1)^{2}\left(\frac{dy}{dx}+\frac{4x}{x^{2}+1}y\right)=-(x^{2}+1)^{2}\frac{1}{(x^{2}+1)^{2}}$
$\displaystyle \Rightarrow (x^{2}+1)^{2}\frac{dy}{dx}+4x(x^{2}+1)y=-1$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle (x^{2}+1)^{2}y=-\int 1\,dx + C$
$\displaystyle \Rightarrow (x^{2}+1)^{2}y=-x+C$
$\displaystyle \text{Hence, } (x^{2}+1)^{2}y=-x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 9: }~x\frac{dy}{dx}+y=x\log x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+y=x\log x$
$\displaystyle \text{Dividing both sides by } x, \text{ we get}$
$\displaystyle \frac{dy}{dx}+\frac{y}{x}=\log x$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\frac{1}{x}$
$\displaystyle Q=\log x$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int \frac{1}{x}\,dx}$
$\displaystyle =e^{\log|x|}$
$\displaystyle =x$
$\displaystyle \text{So, the solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow xy=\int x\log x\,dx + C$
$\displaystyle \Rightarrow xy=x\int \log x\,dx - \int \frac{d}{dx}(x)\int \log x\,dx\,dx + C$
$\displaystyle \Rightarrow xy=x\left(x\log x-x\right)-\int \left(x\log x-x\right)\,dx + C$
$\displaystyle \Rightarrow xy=x^{2}\log x-x^{2}-\left(\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}\right)+C$
$\displaystyle \Rightarrow xy=\frac{x^{2}\log x}{2}-\frac{x^{2}}{4}+C$
$\displaystyle \Rightarrow 4xy=2x^{2}\log x-x^{2}+K$

$\displaystyle \textbf{Question 10: }~x\frac{dy}{dx}-y=(x-1)e^{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}-y=(x-1)e^{x}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\frac{x-1}{x}e^{x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{1}{x}$
$\displaystyle Q=\frac{x-1}{x}e^{x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\frac{1}{x}\,dx}$
$\displaystyle =e^{-\log|x|}$
$\displaystyle =\frac{1}{x}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{x}, \text{ we get}$
$\displaystyle \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{x-1}{x^{2}}e^{x}$
$\displaystyle \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^{2}}y=\frac{x-1}{x^{2}}e^{x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{y}{x}=\int \frac{x-1}{x^{2}}e^{x}\,dx + C$
$\displaystyle \Rightarrow \frac{y}{x}=\frac{e^{x}}{x}+C$
$\displaystyle \Rightarrow y=e^{x}+Cx$
$\displaystyle \text{Hence, } y=e^{x}+Cx \text{ is the required solution.}$

$\displaystyle \textbf{Question 11: }~\frac{dy}{dx}+\frac{y}{x}=x^{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+\frac{y}{x}=x^{3} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{x}$
$\displaystyle Q=x^{3}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{x}\,dx}$
$\displaystyle =e^{\log|x|}$
$\displaystyle =x$
$\displaystyle \text{Multiplying both sides of (1) by } x, \text{ we get}$
$\displaystyle x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x^{4}$
$\displaystyle \Rightarrow x\frac{dy}{dx}+y=x^{4}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle xy=\int x^{4}\,dx + C$
$\displaystyle \Rightarrow xy=\frac{x^{5}}{5}+C$
$\displaystyle \Rightarrow 5xy=x^{5}+5C$
$\displaystyle \Rightarrow 5xy=x^{5}+K$
$\displaystyle \text{Hence, } 5xy=x^{5}+K \text{ is the required solution.}$

$\displaystyle \textbf{Question 12: }~\frac{dy}{dx}+y=\sin x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y=\sin x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=1$
$\displaystyle Q=\sin x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 1\,dx}$
$\displaystyle =e^{x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{x}, \text{ we get}$
$\displaystyle e^{x}\left(\frac{dy}{dx}+y\right)=e^{x}\sin x$
$\displaystyle \Rightarrow e^{x}\frac{dy}{dx}+e^{x}y=e^{x}\sin x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{x}=\int e^{x}\sin x\,dx + C$
$\displaystyle \Rightarrow y e^{x}=\frac{e^{x}}{2}(\sin x-\cos x)+C$
$\displaystyle \Rightarrow y=Ce^{-x}+\frac{1}{2}(\sin x-\cos x)$
$\displaystyle \text{Hence, } y=Ce^{-x}+\frac{1}{2}(\sin x-\cos x) \text{ is the required solution.}$

$\displaystyle \textbf{Question 13: }~\frac{dy}{dx}+y=\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y=\cos x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=1$
$\displaystyle Q=\cos x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 1\,dx}$
$\displaystyle =e^{x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{x}, \text{ we get}$
$\displaystyle e^{x}\left(\frac{dy}{dx}+y\right)=e^{x}\cos x$
$\displaystyle \Rightarrow e^{x}\frac{dy}{dx}+e^{x}y=e^{x}\cos x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{x}=\int e^{x}\cos x\,dx + C$
$\displaystyle \Rightarrow y e^{x}=\frac{e^{x}}{2}(\cos x+\sin x)+C$
$\displaystyle \Rightarrow y=\frac{1}{2}(\cos x+\sin x)+Ce^{-x}$
$\displaystyle \text{Hence, } y=\frac{1}{2}(\cos x+\sin x)+Ce^{-x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 14: }~\frac{dy}{dx}+2y=\sin x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=\sin x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=\sin x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}\sin x$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}\sin x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=\int e^{2x}\sin x\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=\frac{e^{2x}}{5}(2\sin x-\cos x)+C$
$\displaystyle \Rightarrow y=\frac{1}{5}(2\sin x-\cos x)+Ce^{-2x}$
$\displaystyle \text{Hence, } y=\frac{1}{5}(2\sin x-\cos x)+Ce^{-2x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 15: }~\frac{dy}{dx}=y\tan x-2\sin x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}=y\tan x-2\sin x$
$\displaystyle \Rightarrow \frac{dy}{dx}-y\tan x=-2\sin x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\tan x$
$\displaystyle Q=-2\sin x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\tan x\,dx}$
$\displaystyle =e^{\log(\cos x)}$
$\displaystyle =\cos x$
$\displaystyle \text{Multiplying both sides of (1) by } \cos x, \text{ we get}$
$\displaystyle \cos x\left(\frac{dy}{dx}-y\tan x\right)=-2\sin x\cos x$
$\displaystyle \Rightarrow \cos x\frac{dy}{dx}-y\sin x=-\sin 2x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y\cos x=-\int \sin 2x\,dx + C$
$\displaystyle \Rightarrow y\cos x=\frac{\cos 2x}{2}+C$
$\displaystyle \Rightarrow 2y\cos x=\cos 2x+K$
$\displaystyle \text{Hence, } 2y\cos x=\cos 2x+K \text{ is the required solution.}$

$\displaystyle \textbf{Question 16: }~(1+x^{2})\frac{dy}{dx}+y=\tan^{-1}x. \hspace{2.0cm} \text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+x^{2})\frac{dy}{dx}+y=\tan^{-1}x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{1+x^{2}}=\frac{\tan^{-1}x}{1+x^{2}} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{1+x^{2}}$
$\displaystyle Q=\frac{\tan^{-1}x}{1+x^{2}}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{1+x^{2}}\,dx}$
$\displaystyle =e^{\tan^{-1}x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\tan^{-1}x}, \text{ we get}$
$\displaystyle e^{\tan^{-1}x}\left(\frac{dy}{dx}+\frac{y}{1+x^{2}}\right)=e^{\tan^{-1}x}\frac{\tan^{-1}x}{1+x^{2}}$
$\displaystyle \Rightarrow e^{\tan^{-1}x}\frac{dy}{dx}+e^{\tan^{-1}x}\frac{y}{1+x^{2}}=e^{\tan^{-1}x}\frac{\tan^{-1}x}{1+x^{2}}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle e^{\tan^{-1}x}y=\int \frac{\tan^{-1}x\,e^{\tan^{-1}x}}{1+x^{2}}\,dx + C$
$\displaystyle \text{Let } I=\int \frac{\tan^{-1}x\,e^{\tan^{-1}x}}{1+x^{2}}\,dx$
$\displaystyle \text{Putting } \tan^{-1}x=t, \text{ we get}$
$\displaystyle \frac{1}{1+x^{2}}\,dx=dt$
$\displaystyle \therefore I=\int te^{t}\,dt$
$\displaystyle =t e^{t}-\int e^{t}\,dt$
$\displaystyle =t e^{t}-e^{t}$
$\displaystyle =(t-1)e^{t}$
$\displaystyle =(\tan^{-1}x-1)e^{\tan^{-1}x}$
$\displaystyle \text{Substituting in the earlier equation, we get}$
$\displaystyle e^{\tan^{-1}x}y=(\tan^{-1}x-1)e^{\tan^{-1}x}+C$
$\displaystyle \Rightarrow y=\tan^{-1}x-1+Ce^{-\tan^{-1}x}$
$\displaystyle \text{Hence, } y=\tan^{-1}x-1+Ce^{-\tan^{-1}x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 17: }~\frac{dy}{dx}+y\tan x=\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\tan x=\cos x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\tan x$
$\displaystyle Q=\cos x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \tan x\,dx}$
$\displaystyle =e^{-\log(\cos x)}$
$\displaystyle =\sec x$
$\displaystyle \text{Multiplying both sides of (1) by } \sec x, \text{ we get}$
$\displaystyle \sec x\left(\frac{dy}{dx}+y\tan x\right)=\cos x\sec x$
$\displaystyle \Rightarrow \sec x\frac{dy}{dx}+y\sec x\tan x=1$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y\sec x=\int 1\,dx + C$
$\displaystyle \Rightarrow y\sec x=x+C$
$\displaystyle \text{Hence, } y\sec x=x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 18: }~\frac{dy}{dx}+y\cot x=x^{2}\cot x+2x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\cot x=x^{2}\cot x+2x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\cot x$
$\displaystyle Q=x^{2}\cot x+2x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}$
$\displaystyle =\sin x$
$\displaystyle \text{Multiplying both sides of (1) by } \sin x, \text{ we get}$
$\displaystyle \sin x\left(\frac{dy}{dx}+y\cot x\right)=\sin x\left(x^{2}\cot x+2x\right)$
$\displaystyle \Rightarrow \sin x\frac{dy}{dx}+y\cos x=x^{2}\cos x+2x\sin x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y\sin x=\int x^{2}\cos x\,dx + \int 2x\sin x\,dx + C$
$\displaystyle \Rightarrow y\sin x=x^{2}\sin x - \int 2x\sin x\,dx + \int 2x\sin x\,dx + C$
$\displaystyle \Rightarrow y\sin x=x^{2}\sin x + C$
$\displaystyle \text{Hence, } y\sin x=x^{2}\sin x + C \text{ is the required solution.}$

$\displaystyle \textbf{Question 19: }~\frac{dy}{dx}+y\tan x=x^{2}\cos^{2}x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\tan x=x^{2}\cos^{2}x$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\tan x$
$\displaystyle Q=x^{2}\cos^{2}x$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int \tan x\,dx}$
$\displaystyle =e^{-\log(\cos x)}$
$\displaystyle =\sec x$
$\displaystyle \text{Therefore, solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow y\sec x=\int x^{2}\cos^{2}x\sec x\,dx + C$
$\displaystyle \Rightarrow y\sec x=\int x^{2}\cos x\,dx + C$
$\displaystyle \text{Let } I=\int x^{2}\cos x\,dx$
$\displaystyle =x^{2}\sin x-\int 2x\sin x\,dx$
$\displaystyle =x^{2}\sin x-2\int x\sin x\,dx$
$\displaystyle =x^{2}\sin x-2\left(-x\cos x+\int \cos x\,dx\right)$
$\displaystyle =x^{2}\sin x+2x\cos x-2\sin x$
$\displaystyle \Rightarrow y\sec x=x^{2}\sin x+2x\cos x-2\sin x + C$
$\displaystyle \text{Hence, } y\sec x=x^{2}\sin x+2x\cos x-2\sin x + C \text{ is the required solution.}$

$\displaystyle \textbf{Question 20: }~(1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}. \hspace{2.0cm} \text{[CBSE 2002, 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{1+x^{2}}=\frac{e^{\tan^{-1}x}}{1+x^{2}} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{1+x^{2}}$
$\displaystyle Q=\frac{e^{\tan^{-1}x}}{1+x^{2}}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{1+x^{2}}\,dx}$
$\displaystyle =e^{\tan^{-1}x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\tan^{-1}x}, \text{ we get}$
$\displaystyle e^{\tan^{-1}x}\left(\frac{dy}{dx}+\frac{y}{1+x^{2}}\right)=\frac{e^{2\tan^{-1}x}}{1+x^{2}}$
$\displaystyle \Rightarrow e^{\tan^{-1}x}\frac{dy}{dx}+e^{\tan^{-1}x}\frac{y}{1+x^{2}}=\frac{e^{2\tan^{-1}x}}{1+x^{2}}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{\tan^{-1}x}=\int \frac{e^{2\tan^{-1}x}}{1+x^{2}}\,dx + C$
$\displaystyle \text{Let } I=\int \frac{e^{2\tan^{-1}x}}{1+x^{2}}\,dx$
$\displaystyle \text{Putting } \tan^{-1}x=t, \text{ we get}$
$\displaystyle \frac{1}{1+x^{2}}\,dx=dt$
$\displaystyle \therefore I=\int e^{2t}\,dt$
$\displaystyle =\frac{e^{2t}}{2}$
$\displaystyle =\frac{e^{2\tan^{-1}x}}{2}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle y e^{\tan^{-1}x}=\frac{e^{2\tan^{-1}x}}{2}+C$
$\displaystyle \Rightarrow 2y e^{\tan^{-1}x}=e^{2\tan^{-1}x}+2C$
$\displaystyle \Rightarrow 2y e^{\tan^{-1}x}=e^{2\tan^{-1}x}+k$
$\displaystyle \text{Hence, } 2y e^{\tan^{-1}x}=e^{2\tan^{-1}x}+k \text{ is the required solution.}$

$\displaystyle \textbf{Question 21: }~x\,dy=(2y+2x^{4}+x^{2})\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\,dy=(2y+2x^{4}+x^{2})\,dx$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2}{x}y+2x^{3}+x$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{2}{x}y=2x^{3}+x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{2}{x}$
$\displaystyle Q=2x^{3}+x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\frac{2}{x}\,dx}$
$\displaystyle =e^{-2\log|x|}$
$\displaystyle =\frac{1}{x^{2}}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{x^{2}}, \text{ we get}$
$\displaystyle \frac{1}{x^{2}}\left(\frac{dy}{dx}-\frac{2}{x}y\right)=\frac{1}{x^{2}}(2x^{3}+x)$
$\displaystyle \Rightarrow \frac{1}{x^{2}}\frac{dy}{dx}-\frac{2}{x^{3}}y=2x+\frac{1}{x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{y}{x^{2}}=\int \left(2x+\frac{1}{x}\right)\,dx + C$
$\displaystyle \Rightarrow \frac{y}{x^{2}}=x^{2}+\log|x|+C$
$\displaystyle \Rightarrow y=x^{4}+x^{2}\log|x|+Cx^{2}$
$\displaystyle \text{Hence, } y=x^{4}+x^{2}\log|x|+Cx^{2} \text{ is the required solution.}$

$\displaystyle \textbf{Question 22: }~(1+y^{2})+\left(x-e^{\tan^{-1}y}\right)\frac{dy}{dx}=0. \hspace{2.0cm} \text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+y^{2})+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0$
$\displaystyle \Rightarrow (x-e^{\tan^{-1}y})\frac{dy}{dx}=-(1+y^{2})$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{1+y^{2}}{x-e^{\tan^{-1}y}}$
$\displaystyle \Rightarrow \frac{dx}{dy}=-\frac{x-e^{\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^{2}}=\frac{e^{\tan^{-1}y}}{1+y^{2}} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{1+y^{2}}$
$\displaystyle Q=\frac{e^{\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int \frac{1}{1+y^{2}}\,dy}$
$\displaystyle =e^{\tan^{-1}y}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\tan^{-1}y}, \text{ we get}$
$\displaystyle e^{\tan^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^{2}}\right)=\frac{e^{2\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \Rightarrow e^{\tan^{-1}y}\frac{dx}{dy}+\frac{x e^{\tan^{-1}y}}{1+y^{2}}=\frac{e^{2\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x e^{\tan^{-1}y}=\int \frac{e^{2\tan^{-1}y}}{1+y^{2}}\,dy + C$
$\displaystyle \text{Let } I=\int \frac{e^{2\tan^{-1}y}}{1+y^{2}}\,dy$
$\displaystyle \text{Putting } \tan^{-1}y=t, \text{ we get}$
$\displaystyle \frac{1}{1+y^{2}}\,dy=dt$
$\displaystyle \therefore I=\int e^{2t}\,dt$
$\displaystyle =\frac{e^{2t}}{2}$
$\displaystyle =\frac{e^{2\tan^{-1}y}}{2}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle x e^{\tan^{-1}y}=\frac{e^{2\tan^{-1}y}}{2}+C$
$\displaystyle \Rightarrow 2x e^{\tan^{-1}y}=e^{2\tan^{-1}y}+2C$
$\displaystyle \Rightarrow 2x e^{\tan^{-1}y}=e^{2\tan^{-1}y}+k$
$\displaystyle \text{Hence, } 2x e^{\tan^{-1}y}=e^{2\tan^{-1}y}+k \text{ is the required solution.}$

$\displaystyle \textbf{Question 23: }~y^{2}\frac{dx}{dy}+x-\frac{1}{y}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } y^{2}\frac{dx}{dy}+x-\frac{1}{y}=0$
$\displaystyle \Rightarrow y^{2}\frac{dx}{dy}+x=\frac{1}{y}$
$\displaystyle \Rightarrow \frac{dx}{dy}+\frac{1}{y^{2}}x=\frac{1}{y^{3}} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{y^{2}}$
$\displaystyle Q=\frac{1}{y^{3}}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int \frac{1}{y^{2}}\,dy}$
$\displaystyle =e^{-\frac{1}{y}}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{-\frac{1}{y}}, \text{ we get}$
$\displaystyle e^{-\frac{1}{y}}\left(\frac{dx}{dy}+\frac{1}{y^{2}}x\right)=\frac{e^{-\frac{1}{y}}}{y^{3}}$
$\displaystyle \Rightarrow e^{-\frac{1}{y}}\frac{dx}{dy}+\frac{x e^{-\frac{1}{y}}}{y^{2}}=\frac{e^{-\frac{1}{y}}}{y^{3}}$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x e^{-\frac{1}{y}}=\int \frac{e^{-\frac{1}{y}}}{y^{3}}\,dy + C$
$\displaystyle \text{Let } I=\int \frac{e^{-\frac{1}{y}}}{y^{3}}\,dy$
$\displaystyle \text{Putting } t=\frac{1}{y}, \text{ we get}$
$\displaystyle dt=-\frac{1}{y^{2}}\,dy$
$\displaystyle \therefore I=-\int t e^{-t}\,dt$
$\displaystyle =- \left(-t e^{-t}-e^{-t}\right)$
$\displaystyle =(t+1)e^{-t}$
$\displaystyle =\left(\frac{1}{y}+1\right)e^{-\frac{1}{y}}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle x e^{-\frac{1}{y}}=\left(\frac{1}{y}+1\right)e^{-\frac{1}{y}}+C$
$\displaystyle \Rightarrow x=\frac{y+1}{y}+Ce^{\frac{1}{y}}$
$\displaystyle \text{Hence, } x=\frac{y+1}{y}+Ce^{\frac{1}{y}} \text{ is the required solution.}$

$\displaystyle \textbf{Question 24: }~(2x-10y^{3})\frac{dy}{dx}+y=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (2x-10y^{3})\frac{dy}{dx}+y=0$
$\displaystyle \Rightarrow (2x-10y^{3})\frac{dy}{dx}=-y$
$\displaystyle \Rightarrow \frac{dx}{dy}=-\frac{2x-10y^{3}}{y}$
$\displaystyle \Rightarrow \frac{dx}{dy}+\frac{2}{y}x=10y^{2} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{2}{y}$
$\displaystyle Q=10y^{2}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int \frac{2}{y}\,dy}$
$\displaystyle =e^{2\log|y|}$
$\displaystyle =y^{2}$
$\displaystyle \text{Multiplying both sides of (1) by } y^{2}, \text{ we get}$
$\displaystyle y^{2}\left(\frac{dx}{dy}+\frac{2}{y}x\right)=10y^{4}$
$\displaystyle \Rightarrow y^{2}\frac{dx}{dy}+2xy=10y^{4}$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x y^{2}=\int 10y^{4}\,dy + C$
$\displaystyle \Rightarrow x y^{2}=2y^{5}+C$
$\displaystyle \Rightarrow x=2y^{3}+Cy^{-2}$
$\displaystyle \text{Hence, } x=2y^{3}+Cy^{-2} \text{ is the required solution.}$

$\displaystyle \textbf{Question 25: }~(x+\tan y)\,dy=\sin 2y\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (x+\tan y)\,dy=\sin 2y\,dx$
$\displaystyle \Rightarrow \frac{dx}{dy}=x\mathrm{cosec} 2y+\frac{1}{2}\sec^{2}y$
$\displaystyle \Rightarrow \frac{dx}{dy}-x\mathrm{cosec} 2y=\frac{1}{2}\sec^{2}y \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\mathrm{cosec} 2y$
$\displaystyle Q=\frac{1}{2}\sec^{2}y$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int -\mathrm{cosec} 2y\,dy}$
$\displaystyle =e^{-\frac{1}{2}\log(\tan y)}$
$\displaystyle =\frac{1}{\sqrt{\tan y}}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{\sqrt{\tan y}}, \text{ we get}$
$\displaystyle \frac{1}{\sqrt{\tan y}}\left(\frac{dx}{dy}-x\mathrm{cosec} 2y\right)=\frac{1}{2}\frac{1}{\sqrt{\tan y}}\sec^{2}y$
$\displaystyle \Rightarrow \frac{1}{\sqrt{\tan y}}\frac{dx}{dy}-x\mathrm{cosec} 2y\frac{1}{\sqrt{\tan y}}=\frac{1}{2}\frac{1}{\sqrt{\tan y}}\sec^{2}y$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle \frac{x}{\sqrt{\tan y}}=\int \frac{1}{2}\frac{1}{\sqrt{\tan y}}\sec^{2}y\,dy + C$
$\displaystyle \text{Let } I=\int \frac{1}{2}\frac{1}{\sqrt{\tan y}}\sec^{2}y\,dy$
$\displaystyle \text{Putting } t=\tan y, \text{ we get}$
$\displaystyle dt=\sec^{2}y\,dy$
$\displaystyle \therefore I=\frac{1}{2}\int \frac{1}{\sqrt{t}}\,dt$
$\displaystyle =\sqrt{t}$
$\displaystyle =\sqrt{\tan y}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle \frac{x}{\sqrt{\tan y}}=\sqrt{\tan y}+C$
$\displaystyle \Rightarrow x=\tan y+C\sqrt{\tan y}$
$\displaystyle \text{Hence, } x=\tan y+C\sqrt{\tan y} \text{ is the required solution.}$

$\displaystyle \textbf{Question 26: }~dx+x\,dy=e^{-y}\sec^{2}y\,dy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } dx+x\,dy=e^{-y}\sec^{2}y\,dy$
$\displaystyle \Rightarrow dx=e^{-y}\sec^{2}y\,dy-x\,dy$
$\displaystyle \Rightarrow \frac{dx}{dy}=e^{-y}\sec^{2}y-x$
$\displaystyle \Rightarrow \frac{dx}{dy}+x=e^{-y}\sec^{2}y \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=1$
$\displaystyle Q=e^{-y}\sec^{2}y$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int 1\,dy}$
$\displaystyle =e^{y}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{y}, \text{ we get}$
$\displaystyle e^{y}\left(\frac{dx}{dy}+x\right)=e^{y}e^{-y}\sec^{2}y$
$\displaystyle \Rightarrow e^{y}\frac{dx}{dy}+xe^{y}=\sec^{2}y$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x e^{y}=\int \sec^{2}y\,dy + C$
$\displaystyle \Rightarrow x e^{y}=\tan y+C$
$\displaystyle \text{Hence, } x e^{y}=\tan y+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 27: }~\frac{dy}{dx}=y\tan x-2\sin x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}=y\tan x-2\sin x$
$\displaystyle \Rightarrow \frac{dy}{dx}-y\tan x=-2\sin x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\tan x$
$\displaystyle Q=-2\sin x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\tan x\,dx}$
$\displaystyle =e^{\log(\cos x)}$
$\displaystyle =\cos x$
$\displaystyle \text{Multiplying both sides of (1) by } \cos x, \text{ we get}$
$\displaystyle \cos x\left(\frac{dy}{dx}-y\tan x\right)=-2\sin x\cos x$
$\displaystyle \Rightarrow \cos x\frac{dy}{dx}-y\sin x=-\sin 2x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y\cos x=-\int \sin 2x\,dx + C$
$\displaystyle \Rightarrow y\cos x=\frac{\cos 2x}{2}+C$
$\displaystyle \Rightarrow y\cos x=\frac{1-2\sin^{2}x}{2}+C$
$\displaystyle \Rightarrow y\cos x=-\sin^{2}x+K$
$\displaystyle \Rightarrow y=\sec x(-\sin^{2}x+K)$
$\displaystyle \text{Hence, } y=\sec x(-\sin^{2}x+K) \text{ is the required solution.}$

$\displaystyle \textbf{Question 28: }~\frac{dy}{dx}+y\cos x=\sin x\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\cos x=\sin x\cos x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\cos x$
$\displaystyle Q=\sin x\cos x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \cos x\,dx}$
$\displaystyle =e^{\sin x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\sin x}, \text{ we get}$
$\displaystyle e^{\sin x}\left(\frac{dy}{dx}+y\cos x\right)=e^{\sin x}\sin x\cos x$
$\displaystyle \Rightarrow e^{\sin x}\frac{dy}{dx}+e^{\sin x}y\cos x=e^{\sin x}\sin x\cos x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{\sin x}=\int e^{\sin x}\sin x\cos x\,dx + C$
$\displaystyle \text{Let } I=\int e^{\sin x}\sin x\cos x\,dx$
$\displaystyle \text{Putting } t=\sin x, \text{ we get}$
$\displaystyle dt=\cos x\,dx$
$\displaystyle \therefore I=\int t e^{t}\,dt$
$\displaystyle =t e^{t}-\int e^{t}\,dt$
$\displaystyle =t e^{t}-e^{t}$
$\displaystyle =(t-1)e^{t}$
$\displaystyle =e^{\sin x}(\sin x-1)$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle y e^{\sin x}=e^{\sin x}(\sin x-1)+C$
$\displaystyle \Rightarrow y=\sin x-1+Ce^{-\sin x}$
$\displaystyle \text{Hence, } y=\sin x-1+Ce^{-\sin x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 29: }~(1+x^{2})\frac{dy}{dx}-2xy=(x^{2}+2)(x^{2}+1). \hspace{2.0cm} \text{[CBSE 20105]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } (1+x^{2})\frac{dy}{dx}-2xy=(x^{2}+2)(x^{2}+1)$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{2x}{1+x^{2}}y=x^{2}+2$
$\displaystyle \text{This is a linear differential equation.}$
$\displaystyle \text{I.F.}=e^{\int -\frac{2x}{1+x^{2}}\,dx}$
$\displaystyle =e^{-\log(1+x^{2})}$
$\displaystyle =\frac{1}{1+x^{2}}$
$\displaystyle y\left(\frac{1}{1+x^{2}}\right)=\int \frac{x^{2}+2}{1+x^{2}}\,dx + C$
$\displaystyle \Rightarrow y\left(\frac{1}{1+x^{2}}\right)=\int \left(1+\frac{1}{1+x^{2}}\right)\,dx + C$
$\displaystyle \Rightarrow y\left(\frac{1}{1+x^{2}}\right)=x+\tan^{-1}x + C$
$\displaystyle \Rightarrow y=(x+\tan^{-1}x + C)(1+x^{2})$

$\displaystyle \textbf{Question 30: }~(\sin x)\frac{dy}{dx}+y\cos x=2\sin^{2}x\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (\sin x)\frac{dy}{dx}+y\cos x=2\sin^{2}x\cos x$
$\displaystyle \Rightarrow \frac{dy}{dx}+y\cot x=2\sin x\cos x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\cot x$
$\displaystyle Q=2\sin x\cos x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}$
$\displaystyle =\sin x$
$\displaystyle \text{Multiplying both sides of (1) by } \sin x, \text{ we get}$
$\displaystyle \sin x\left(\frac{dy}{dx}+y\cot x\right)=2\sin^{2}x\cos x$
$\displaystyle \Rightarrow \sin x\frac{dy}{dx}+y\cos x=2\sin^{2}x\cos x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y\sin x=\int 2\sin^{2}x\cos x\,dx + C$
$\displaystyle \text{Putting } \sin x=t, \text{ we get}$
$\displaystyle dt=\cos x\,dx$
$\displaystyle \Rightarrow y\sin x=2\int t^{2}\,dt + C$
$\displaystyle \Rightarrow y\sin x=\frac{2}{3}t^{3}+C$
$\displaystyle \Rightarrow y\sin x=\frac{2}{3}\sin^{3}x+C$
$\displaystyle \text{Hence, } y\sin x=\frac{2}{3}\sin^{3}x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 31: }~(x^{2}-1)\frac{dy}{dx}+2(x+2)y=2(x+1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (x^{2}-1)\frac{dy}{dx}+2(x+2)y=2(x+1)$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2(x+2)}{x^{2}-1}y=\frac{2(x+1)}{x^{2}-1}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2(x+2)}{x^{2}-1}y=\frac{2}{x-1} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{2(x+2)}{x^{2}-1}$
$\displaystyle Q=\frac{2}{x-1}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{2(x+2)}{x^{2}-1}\,dx}$
$\displaystyle =e^{\int \frac{2x}{x^{2}-1}\,dx + \int \frac{4}{x^{2}-1}\,dx}$
$\displaystyle =e^{\log|x^{2}-1| + 2\log\left|\frac{x-1}{x+1}\right|}$
$\displaystyle =\frac{(x-1)^{3}}{x+1}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{(x-1)^{3}}{x+1}, \text{ we get}$
$\displaystyle \frac{(x-1)^{3}}{x+1}\left(\frac{dy}{dx}+\frac{2(x+2)}{x^{2}-1}y\right)=\frac{(x-1)^{3}}{x+1}\cdot\frac{2}{x-1}$
$\displaystyle \Rightarrow \frac{(x-1)^{3}}{x+1}\frac{dy}{dx}-\frac{2(x+2)(x-1)^{2}}{(x+1)^{2}}y=\frac{2(x-1)^{2}}{x+1}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{(x-1)^{3}}{x+1}y=\int \frac{2(x-1)^{2}}{x+1}\,dx + C$
$\displaystyle \Rightarrow \frac{(x-1)^{3}}{x+1}y=\int \left(2x-6+\frac{8}{x+1}\right)\,dx + C$
$\displaystyle \Rightarrow \frac{(x-1)^{3}}{x+1}y=x^{2}-6x+8\log|x+1|+C$
$\displaystyle \Rightarrow y=\frac{x+1}{(x-1)^{3}}\left(x^{2}-6x+8\log|x+1|+C\right)$
$\displaystyle \text{Hence, } y=\frac{x+1}{(x-1)^{3}}\left(x^{2}-6x+8\log|x+1|+C\right) \text{ is the required solution.}$

$\displaystyle \textbf{Question 32: }~x\frac{dy}{dx}+2y=x\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+2y=x\cos x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=\cos x$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\frac{2}{x}$
$\displaystyle Q=\cos x$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int \frac{2}{x}\,dx}$
$\displaystyle =e^{2\log|x|}$
$\displaystyle =x^{2}$
$\displaystyle \text{Solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow x^{2}y=\int x^{2}\cos x\,dx + C$
$\displaystyle \text{Let } I=\int x^{2}\cos x\,dx$
$\displaystyle =x^{2}\sin x-\int 2x\sin x\,dx$
$\displaystyle =x^{2}\sin x-2\int x\sin x\,dx$
$\displaystyle =x^{2}\sin x-2\left(-x\cos x+\int \cos x\,dx\right)$
$\displaystyle =x^{2}\sin x+2x\cos x-2\sin x$
$\displaystyle \Rightarrow x^{2}y=x^{2}\sin x+2x\cos x-2\sin x+C$
$\displaystyle \text{Hence, } x^{2}y=x^{2}\sin x+2x\cos x-2\sin x+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 33: }~\frac{dy}{dx}-y=xe^{x}. \hspace{2.0cm} \text{[CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}-y=xe^{x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-1$
$\displaystyle Q=xe^{x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -1\,dx}$
$\displaystyle =e^{-x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{-x}, \text{ we get}$
$\displaystyle e^{-x}\left(\frac{dy}{dx}-y\right)=xe^{x}e^{-x}$
$\displaystyle \Rightarrow e^{-x}\frac{dy}{dx}-e^{-x}y=x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle e^{-x}y=\int x\,dx + C$
$\displaystyle \Rightarrow e^{-x}y=\frac{x^{2}}{2}+C$
$\displaystyle \Rightarrow y=\left(\frac{x^{2}}{2}+C\right)e^{x}$
$\displaystyle \text{Hence, } y=\left(\frac{x^{2}}{2}+C\right)e^{x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 34: }~\frac{dy}{dx}+2y=xe^{4x}. \hspace{2.0cm} \text{[CBSE 20o2]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=xe^{4x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=2$
$\displaystyle Q=xe^{4x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{2x}, \text{ we get}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=xe^{6x}$
$\displaystyle \Rightarrow e^{2x}\frac{dy}{dx}+2e^{2x}y=xe^{6x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{2x}=\int x e^{6x}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=x\frac{e^{6x}}{6}-\int \frac{e^{6x}}{6}\,dx + C$
$\displaystyle \Rightarrow y e^{2x}=\frac{x e^{6x}}{6}-\frac{e^{6x}}{36}+C$
$\displaystyle \Rightarrow y=\frac{x e^{4x}}{6}-\frac{e^{4x}}{36}+Ce^{-2x}$
$\displaystyle \text{Hence, } y=\frac{x e^{4x}}{6}-\frac{e^{4x}}{36}+Ce^{-2x} \text{ is the required solution.}$

$\displaystyle \textbf{Question 35: }~\text{Solve the differential equation }(x+2y^{2})\frac{dy}{dx}=y,\ \text{given that when } \\ x=2,\ y=1. \hspace{2.0cm} \text{[CBSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (x+2y^{2})\frac{dy}{dx}=y$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{x+2y^{2}}{y}$
$\displaystyle \Rightarrow \frac{dx}{dy}-\frac{1}{y}x=2y \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{1}{y}$
$\displaystyle Q=2y$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int -\frac{1}{y}\,dy}$
$\displaystyle =e^{-\log|y|}$
$\displaystyle =\frac{1}{y}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{y}, \text{ we get}$
$\displaystyle \frac{1}{y}\left(\frac{dx}{dy}-\frac{1}{y}x\right)=\frac{1}{y}\cdot 2y$
$\displaystyle \Rightarrow \frac{1}{y}\frac{dx}{dy}-\frac{x}{y^{2}}=2$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle \frac{x}{y}=\int 2\,dy + C$
$\displaystyle \Rightarrow \frac{x}{y}=2y+C$
$\displaystyle \Rightarrow x=2y^{2}+Cy$
$\displaystyle \text{Now, } y=1 \text{ at } x=2$
$\displaystyle \Rightarrow 2=2+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Hence, } x=2y^{2} \text{ is the required solution.}$

$\displaystyle \textbf{Solve the following differential equations:}$

$\displaystyle \textbf{Question 36: }~\text{Find one-parameter families of solution curves of the following} \\ \text{differential equations (or solve the following differential equations):}$

$\displaystyle \text{(i)}~\frac{dy}{dx}+3y=e^{mx},\ m\ \text{is a given real number}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+3y=e^{mx} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=3$
$\displaystyle Q=e^{mx}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 3\,dx}$
$\displaystyle =e^{3x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{3x}, \text{ we get}$
$\displaystyle e^{3x}\left(\frac{dy}{dx}+3y\right)=e^{3x}e^{mx}$
$\displaystyle \Rightarrow e^{3x}\frac{dy}{dx}+3e^{3x}y=e^{(m+3)x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{3x}=\int e^{(m+3)x}\,dx + C \quad (m+3\neq 0)$
$\displaystyle \Rightarrow y e^{3x}=\frac{e^{(m+3)x}}{m+3}+C$
$\displaystyle \Rightarrow y=\frac{e^{mx}}{m+3}+Ce^{-3x}$
$\displaystyle \text{When } m+3=0, \text{ we have}$
$\displaystyle y e^{3x}=\int e^{0x}\,dx + C$
$\displaystyle \Rightarrow y e^{3x}=x+C$
$\displaystyle \Rightarrow y=(x+C)e^{-3x}$
$\displaystyle \text{Hence, } y=\frac{e^{mx}}{m+3}+Ce^{-3x}, \text{ where } m+3\neq 0,$
$\displaystyle \text{and } y=(x+C)e^{-3x}, \text{ where } m+3=0, \text{ are the required solutions.}$

$\displaystyle \text{(ii)}~\frac{dy}{dx}-y=\cos 2x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}-y=\cos 2x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-1$
$\displaystyle Q=\cos 2x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -1\,dx}$
$\displaystyle =e^{-x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{-x}, \text{ we get}$
$\displaystyle e^{-x}\left(\frac{dy}{dx}-y\right)=e^{-x}\cos 2x$
$\displaystyle \Rightarrow e^{-x}\frac{dy}{dx}-e^{-x}y=e^{-x}\cos 2x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{-x}=\int e^{-x}\cos 2x\,dx + C$
$\displaystyle \text{Let } I=\int e^{-x}\cos 2x\,dx$
$\displaystyle =\frac{1}{2}e^{-x}\sin 2x-\frac{1}{2}\int (-e^{-x}\sin 2x)\,dx$
$\displaystyle =\frac{1}{2}e^{-x}\sin 2x+\frac{1}{2}\int e^{-x}\sin 2x\,dx$
$\displaystyle =\frac{1}{2}e^{-x}\sin 2x-\frac{1}{4}e^{-x}\cos 2x-\frac{1}{4}\int e^{-x}\cos 2x\,dx$
$\displaystyle =\frac{1}{2}e^{-x}\sin 2x-\frac{1}{4}e^{-x}\cos 2x-\frac{1}{4}I$
$\displaystyle \Rightarrow \frac{5}{4}I=\frac{1}{2}e^{-x}\sin 2x-\frac{1}{4}e^{-x}\cos 2x$
$\displaystyle \Rightarrow I=\frac{e^{-x}}{5}(2\sin 2x-\cos 2x)$
$\displaystyle \text{Substituting in the solution, we get}$
$\displaystyle y e^{-x}=\frac{e^{-x}}{5}(2\sin 2x-\cos 2x)+C$
$\displaystyle \Rightarrow y=\frac{1}{5}(2\sin 2x-\cos 2x)+Ce^{x}$
$\displaystyle \text{Hence, } y=\frac{1}{5}(2\sin 2x-\cos 2x)+Ce^{x} \text{ is the required solution.}$

$\displaystyle \text{(iii)}~x\frac{dy}{dx}-y=(x+1)e^{-x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}-y=(x+1)e^{-x}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\frac{x+1}{x}e^{-x} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{1}{x}$
$\displaystyle Q=\frac{x+1}{x}e^{-x}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\frac{1}{x}\,dx}$
$\displaystyle =e^{-\log|x|}$
$\displaystyle =\frac{1}{x}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{x}, \text{ we get}$
$\displaystyle \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}\cdot\frac{x+1}{x}e^{-x}$
$\displaystyle \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{y}{x^{2}}=\left(\frac{1}{x}+\frac{1}{x^{2}}\right)e^{-x}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{y}{x}=\int \left(\frac{1}{x}+\frac{1}{x^{2}}\right)e^{-x}\,dx + C$
$\displaystyle \text{Let } t=\frac{e^{-x}}{x}, \text{ we get}$
$\displaystyle dt=\left(-\frac{e^{-x}}{x}-\frac{e^{-x}}{x^{2}}\right)\,dx$
$\displaystyle \therefore \left(\frac{1}{x}+\frac{1}{x^{2}}\right)e^{-x}\,dx=-dt$
$\displaystyle \Rightarrow \frac{y}{x}=-\int dt + C$
$\displaystyle \Rightarrow \frac{y}{x}=-t+C$
$\displaystyle \Rightarrow \frac{y}{x}=-\frac{e^{-x}}{x}+C$
$\displaystyle \Rightarrow y=-e^{-x}+Cx$
$\displaystyle \text{Hence, } y=-e^{-x}+Cx \text{ is the required solution.}$

$\displaystyle \text{(iv)}~x\frac{dy}{dx}+y=x^{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+y=x^{4}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=x^{3} \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\frac{1}{x}$
$\displaystyle Q=x^{3}$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{x}\,dx}$
$\displaystyle =e^{\log|x|}$
$\displaystyle =x$
$\displaystyle \text{Multiplying both sides of (1) by } x, \text{ we get}$
$\displaystyle x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x^{4}$
$\displaystyle \Rightarrow x\frac{dy}{dx}+y=x^{4}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle xy=\int x^{4}\,dx + C$
$\displaystyle \Rightarrow xy=\frac{x^{5}}{5}+C$
$\displaystyle \Rightarrow y=\frac{x^{4}}{5}+\frac{C}{x}$
$\displaystyle \text{Hence, } y=\frac{x^{4}}{5}+\frac{C}{x} \text{ is the required solution.}$

$\displaystyle \text{(v)}~(x\log x)\frac{dy}{dx}+y=\log x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (x\log x)\frac{dy}{dx}+y=\log x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{\log x}{x\log x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x\log x}y=\frac{1}{x}$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\frac{1}{x\log x}$
$\displaystyle Q=\frac{1}{x}$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{x\log x}\,dx}$
$\displaystyle =e^{\log(\log x)}$
$\displaystyle =\log x$
$\displaystyle \text{So, the solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow y\log x=\int \frac{1}{x}\log x\,dx + C$
$\displaystyle \Rightarrow y\log x=\int \log x\,d(\log x) + C$
$\displaystyle \Rightarrow y\log x=\frac{(\log x)^{2}}{2}+C$
$\displaystyle \Rightarrow y=\frac{1}{2}\log x+\frac{C}{\log x}$

$\displaystyle \text{(vi)}~\frac{dy}{dx}-\frac{2xy}{1+x^{2}}=x^{2}+2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}-\frac{2x}{1+x^{2}}y=x^{2}+2 \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=-\frac{2x}{1+x^{2}}$
$\displaystyle Q=x^{2}+2$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int -\frac{2x}{1+x^{2}}\,dx}$
$\displaystyle =e^{-\log(1+x^{2})}$
$\displaystyle =\frac{1}{1+x^{2}}$
$\displaystyle \text{Multiplying both sides of (1) by } \frac{1}{1+x^{2}}, \text{ we get}$
$\displaystyle \frac{1}{1+x^{2}}\left(\frac{dy}{dx}-\frac{2x}{1+x^{2}}y\right)=\frac{x^{2}+2}{1+x^{2}}$
$\displaystyle \Rightarrow \frac{1}{1+x^{2}}\frac{dy}{dx}-\frac{2x}{(1+x^{2})^{2}}y=\frac{x^{2}+2}{1+x^{2}}$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle \frac{y}{1+x^{2}}=\int \frac{x^{2}+2}{1+x^{2}}\,dx + C$
$\displaystyle \Rightarrow \frac{y}{1+x^{2}}=\int \left(1+\frac{1}{1+x^{2}}\right)\,dx + C$
$\displaystyle \Rightarrow \frac{y}{1+x^{2}}=x+\tan^{-1}x + C$
$\displaystyle \Rightarrow y=(1+x^{2})(x+\tan^{-1}x + C)$
$\displaystyle \text{Hence, } y=(1+x^{2})(x+\tan^{-1}x + C) \text{ is the required solution.}$

$\displaystyle \text{(vii)}~\frac{dy}{dx}+y\cos x=e^{\sin x}\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\cos x=e^{\sin x}\cos x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\cos x$
$\displaystyle Q=e^{\sin x}\cos x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \cos x\,dx}$
$\displaystyle =e^{\sin x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\sin x}, \text{ we get}$
$\displaystyle e^{\sin x}\left(\frac{dy}{dx}+y\cos x\right)=e^{\sin x}\cdot e^{\sin x}\cos x$
$\displaystyle \Rightarrow e^{\sin x}\frac{dy}{dx}+e^{\sin x}y\cos x=e^{2\sin x}\cos x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{\sin x}=\int e^{2\sin x}\cos x\,dx + C$
$\displaystyle \text{Let } I=\int e^{2\sin x}\cos x\,dx$
$\displaystyle \text{Putting } t=\sin x, \text{ we get}$
$\displaystyle dt=\cos x\,dx$
$\displaystyle \therefore I=\int e^{2t}\,dt$
$\displaystyle =\frac{e^{2t}}{2}$
$\displaystyle =\frac{e^{2\sin x}}{2}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle y e^{\sin x}=\frac{e^{2\sin x}}{2}+C$
$\displaystyle \Rightarrow y=\frac{e^{\sin x}}{2}+Ce^{-\sin x}$
$\displaystyle \text{Hence, } y=\frac{e^{\sin x}}{2}+Ce^{-\sin x} \text{ is the required solution.}$

$\displaystyle \text{(viii)}~(x+y)\frac{dy}{dx}=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (x+y)\frac{dy}{dx}=1$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x+y}$
$\displaystyle \Rightarrow \frac{dx}{dy}=x+y$
$\displaystyle \Rightarrow \frac{dx}{dy}-x=y \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=-1$
$\displaystyle Q=y$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int -1\,dy}$
$\displaystyle =e^{-y}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{-y}, \text{ we get}$
$\displaystyle e^{-y}\left(\frac{dx}{dy}-x\right)=e^{-y}y$
$\displaystyle \Rightarrow e^{-y}\frac{dx}{dy}-e^{-y}x=e^{-y}y$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x e^{-y}=\int y e^{-y}\,dy + C$
$\displaystyle \Rightarrow x e^{-y}=y(-e^{-y})-\int (-e^{-y})\,dy + C$
$\displaystyle \Rightarrow x e^{-y}=-y e^{-y}-e^{-y}+C$
$\displaystyle \Rightarrow x e^{-y}+y e^{-y}+e^{-y}=C$
$\displaystyle \Rightarrow (x+y+1)e^{-y}=C$
$\displaystyle \Rightarrow x+y+1=Ce^{y}$
$\displaystyle \text{Hence, } x+y+1=Ce^{y} \text{ is the required solution.}$

$\displaystyle \text{(ix)}~\left(\frac{dy}{dx}\right)\cos^{2}x=\tan x-y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}\cos^{2}x=\tan x-y$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{\cos^{2}x}y=\tan x\sec^{2}x$
$\displaystyle \Rightarrow \frac{dy}{dx}+y\sec^{2}x=\tan x\sec^{2}x \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where}$
$\displaystyle P=\sec^{2}x$
$\displaystyle Q=\tan x\sec^{2}x$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \sec^{2}x\,dx}$
$\displaystyle =e^{\tan x}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{\tan x}, \text{ we get}$
$\displaystyle e^{\tan x}\left(\frac{dy}{dx}+y\sec^{2}x\right)=e^{\tan x}\tan x\sec^{2}x$
$\displaystyle \Rightarrow e^{\tan x}\frac{dy}{dx}+y e^{\tan x}\sec^{2}x=e^{\tan x}\tan x\sec^{2}x$
$\displaystyle \text{Integrating both sides with respect to } x, \text{ we get}$
$\displaystyle y e^{\tan x}=\int e^{\tan x}\tan x\sec^{2}x\,dx + C$
$\displaystyle \text{Let } I=\int e^{\tan x}\tan x\sec^{2}x\,dx$
$\displaystyle \text{Putting } t=\tan x, \text{ we get}$
$\displaystyle dt=\sec^{2}x\,dx$
$\displaystyle \therefore I=\int t e^{t}\,dt$
$\displaystyle =t e^{t}-\int e^{t}\,dt$
$\displaystyle =t e^{t}-e^{t}$
$\displaystyle =(t-1)e^{t}$
$\displaystyle =( \tan x-1)e^{\tan x}$
$\displaystyle \text{Substituting the value of } I, \text{ we get}$
$\displaystyle y e^{\tan x}=(\tan x-1)e^{\tan x}+C$
$\displaystyle \Rightarrow y=\tan x-1+Ce^{-\tan x}$
$\displaystyle \text{Hence, } y=\tan x-1+Ce^{-\tan x} \text{ is the required solution.}$

$\displaystyle \text{(x)}~e^{-y}\sec^{2}y\,dy=dx+x\,dy.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } e^{-y}\sec^{2}y\,dy=dx+x\,dy$
$\displaystyle \Rightarrow dx=e^{-y}\sec^{2}y\,dy-x\,dy$
$\displaystyle \Rightarrow \frac{dx}{dy}=e^{-y}\sec^{2}y-x$
$\displaystyle \Rightarrow \frac{dx}{dy}+x=e^{-y}\sec^{2}y \quad \text{...(1)}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form}$
$\displaystyle \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where}$
$\displaystyle P=1$
$\displaystyle Q=e^{-y}\sec^{2}y$
$\displaystyle \therefore \text{I.F.}=e^{\int P\,dy}$
$\displaystyle =e^{\int 1\,dy}$
$\displaystyle =e^{y}$
$\displaystyle \text{Multiplying both sides of (1) by } e^{y}, \text{ we get}$
$\displaystyle e^{y}\left(\frac{dx}{dy}+x\right)=e^{y}e^{-y}\sec^{2}y$
$\displaystyle \Rightarrow e^{y}\frac{dx}{dy}+x e^{y}=\sec^{2}y$
$\displaystyle \text{Integrating both sides with respect to } y, \text{ we get}$
$\displaystyle x e^{y}=\int \sec^{2}y\,dy + C$
$\displaystyle \Rightarrow x e^{y}=\tan y + C$
$\displaystyle \Rightarrow x=(\tan y + C)e^{-y}$
$\displaystyle \text{Hence, } x=(\tan y + C)e^{-y} \text{ is the required solution.}$

$\displaystyle \text{(xi)}~x\log x\,\frac{dy}{dx}+y=2\log x. \hspace{2.0cm} \text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\log x\frac{dy}{dx}+y=2\log x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{2\log x}{x\log x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x}$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\frac{1}{x\log x}$
$\displaystyle Q=\frac{2}{x}$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{1}{x\log x}\,dx}$
$\displaystyle =e^{\log(\log x)}$
$\displaystyle =\log x$
$\displaystyle \text{So, the solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow y\log x=2\int \frac{1}{x}\log x\,dx + C$
$\displaystyle \text{Putting } \log x=t, \text{ we get}$
$\displaystyle \frac{1}{x}\,dx=dt$
$\displaystyle \Rightarrow y\log x=2\int t\,dt + C$
$\displaystyle \Rightarrow y\log x=t^{2}+C$
$\displaystyle \Rightarrow y\log x=(\log x)^{2}+C$
$\displaystyle \Rightarrow y=\log x+\frac{C}{\log x}$

$\displaystyle \text{(xii)}~x\frac{dy}{dx}+2y=x^{2}\log x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+2y=x^{2}\log x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x\log x$
$\displaystyle \text{Comparing with } \frac{dy}{dx}+Py=Q, \text{ we get}$
$\displaystyle P=\frac{2}{x}$
$\displaystyle Q=x\log x$
$\displaystyle \text{Now,}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int \frac{2}{x}\,dx}$
$\displaystyle =e^{2\log x}$
$\displaystyle =x^{2}$
$\displaystyle \text{So, the solution is given by}$
$\displaystyle y \times \text{I.F.}=\int Q \times \text{I.F.}\,dx + C$
$\displaystyle \Rightarrow x^{2}y=\int x^{3}\log x\,dx + C$
$\displaystyle =\int x^{3}\log x\,dx + C$
$\displaystyle =x^{4}\log x\Big/4-\int \frac{d}{dx}(\log x)\cdot\frac{x^{4}}{4}\,dx + C$
$\displaystyle =\frac{x^{4}\log x}{4}-\frac{1}{4}\int x^{3}\,dx + C$
$\displaystyle =\frac{x^{4}\log x}{4}-\frac{x^{4}}{16}+C$
$\displaystyle \Rightarrow y=\frac{x^{2}\log x}{4}-\frac{x^{2}}{16}+\frac{C}{x^{2}}$
$\displaystyle \Rightarrow y=\frac{x^{2}}{16}(4\log x-1)+\frac{C}{x^{2}}$

$\displaystyle \textbf{Question 37: }~\text{Solve each of the following initial value problems:}$
$\displaystyle \text{(i)}~y'+y=e^{x},\ y(0)=\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y=e^{x}$
$\displaystyle \text{Clearly, it is a linear differential equation of the form }$
$\displaystyle \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=1 \text{ and } Q=e^{x}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}$
$\displaystyle =e^{\int 1\,dx}$
$\displaystyle =e^{x}$
$\displaystyle \text{Multiplying both sides by } e^{x}, \text{ we get}$
$\displaystyle e^{x}\!\left(\frac{dy}{dx}+y\right)=e^{2x}$
$\displaystyle \Rightarrow e^{x}\frac{dy}{dx}+e^{x}y=e^{2x}$
$\displaystyle \text{Integrating with respect to } x, \text{ we get}$
$\displaystyle y e^{x}=\int e^{2x}\,dx + C$
$\displaystyle =\frac{e^{2x}}{2}+C$
$\displaystyle \text{Now, given } y(0)=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1}{2}=\frac{1}{2}+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \Rightarrow y e^{x}=\frac{e^{2x}}{2}$
$\displaystyle \Rightarrow y=\frac{e^{x}}{2}$
$\displaystyle \text{Hence, } y=\frac{e^{x}}{2} \text{ is the required solution.}$

$\displaystyle \text{(ii)}~x\frac{dy}{dx}-y=\log x,\ y(1)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}-y=\log x$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}=\frac{\log x}{x} \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-\frac{1}{x} \text{ and } Q=\frac{\log x}{x}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\left(-\frac{1}{x}\right)dx}$
$\displaystyle =e^{-\log|x|}=\frac{1}{|x|}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\frac{1}{|x|}$
$\displaystyle \frac{1}{|x|}\frac{dy}{dx}-\frac{1}{x|x|}y=\frac{\log x}{x|x|}$
$\displaystyle \Rightarrow \frac{d}{dx}\left(\frac{y}{|x|}\right)=\frac{\log x}{x|x|}$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle \frac{y}{|x|}=\int\frac{\log x}{x|x|}\,dx+C$
$\displaystyle \text{For } x>0,\ |x|=x,$
$\displaystyle \Rightarrow \frac{y}{x}=\int\frac{\log x}{x^{2}}\,dx+C$
$\displaystyle =\log x\int\frac{1}{x^{2}}\,dx-\int\left(\frac{d}{dx}(\log x)\int\frac{1}{x^{2}}\,dx\right)dx+C$
$\displaystyle =\log x\left(-\frac{1}{x}\right)-\int\left(\frac{1}{x}\cdot\left(-\frac{1}{x}\right)\right)dx+C$
$\displaystyle =-\frac{\log x}{x}+\int\frac{1}{x^{2}}\,dx+C$
$\displaystyle =-\frac{\log x}{x}-\frac{1}{x}+C$
$\displaystyle \Rightarrow y=-\log x-1+Cx \qquad (2)$
$\displaystyle \text{Now, } y(1)=0$
$\displaystyle \Rightarrow 0=-\log 1-1+C(1)$
$\displaystyle \Rightarrow 0=-1+C$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Putting } C=1 \text{ in (2),}$
$\displaystyle y=-\log x-1+x$
$\displaystyle \Rightarrow y=x-1-\log x$

$\displaystyle \text{(iii)}~\frac{dy}{dx}+2y=e^{-2x}\sin x,\ y(0)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y=e^{-2x}\sin x$
$\displaystyle \Rightarrow \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=2 \text{ and } Q=e^{-2x}\sin x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int 2\,dx}$
$\displaystyle =e^{2x}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{2x}$
$\displaystyle e^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}e^{-2x}\sin x$
$\displaystyle \Rightarrow e^{2x}\left(\frac{dy}{dx}+2y\right)=\sin x$
$\displaystyle \Rightarrow \frac{d}{dx}\left(ye^{2x}\right)=\sin x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle ye^{2x}=\int \sin x\,dx+C$
$\displaystyle =-\cos x+C \qquad (2)$
$\displaystyle \text{Now, } y(0)=0$
$\displaystyle \Rightarrow 0\cdot e^{0}=-\cos 0+C$
$\displaystyle \Rightarrow 0=-1+C$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Putting } C=1 \text{ in (2),}$
$\displaystyle ye^{2x}=-\cos x+1$
$\displaystyle \Rightarrow ye^{2x}=1-\cos x$

$\displaystyle \text{(iv)}~x\frac{dy}{dx}-y=(x+1)e^{-x},\ y(1)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}-y=(x+1)e^{-x}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\frac{x+1}{x}e^{-x} \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-\frac{1}{x} \text{ and } Q=\frac{x+1}{x}e^{-x}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\left(-\frac{1}{x}\right)dx}$
$\displaystyle =e^{-\log|x|}=\frac{1}{|x|}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\frac{1}{|x|}$
$\displaystyle \frac{1}{|x|}\frac{dy}{dx}-\frac{1}{x|x|}y=\frac{x+1}{x|x|}e^{-x}$
$\displaystyle \Rightarrow \frac{d}{dx}\left(\frac{y}{|x|}\right)=\frac{x+1}{x|x|}e^{-x}$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle \frac{y}{|x|}=\int\frac{x+1}{x|x|}e^{-x}\,dx+C$
$\displaystyle \text{For } x>0,\ |x|=x,$
$\displaystyle \Rightarrow \frac{y}{x}=\int\left(\frac{1}{x}+\frac{1}{x^{2}}\right)e^{-x}\,dx+C$
$\displaystyle \text{Put } t=\frac{1}{x}e^{-x}$
$\displaystyle \Rightarrow dt=\left(-\frac{1}{x}-\frac{1}{x^{2}}\right)e^{-x}dx$
$\displaystyle \Rightarrow \left(\frac{1}{x}+\frac{1}{x^{2}}\right)e^{-x}dx=-dt$
$\displaystyle \Rightarrow \frac{y}{x}=\int(-dt)+C$
$\displaystyle \Rightarrow \frac{y}{x}=-t+C$
$\displaystyle \Rightarrow \frac{y}{x}=-\frac{e^{-x}}{x}+C$
$\displaystyle \Rightarrow y=-e^{-x}+Cx \qquad (2)$
$\displaystyle \text{Now, } y(1)=0$
$\displaystyle \Rightarrow 0=-e^{-1}+C$
$\displaystyle \Rightarrow C=e^{-1}$
$\displaystyle \text{Putting } C=e^{-1} \text{ in (2),}$
$\displaystyle y=-e^{-x}+xe^{-1}$
$\displaystyle \Rightarrow y=xe^{-1}-e^{-x}$

$\displaystyle \text{(v)}~(1+y^{2})\,dx+\left(x-e^{-\tan^{-1}y}\right)dy=0,\ y(0)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (1+y^{2})\,dx+(x-e^{-\tan^{-1}y})\,dy=0$
$\displaystyle \Rightarrow (x-e^{-\tan^{-1}y})\frac{dy}{dx}=-(1+y^{2})$
$\displaystyle \Rightarrow (1+y^{2})\frac{dx}{dy}=-(x-e^{-\tan^{-1}y})$
$\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^{2}}=\frac{e^{-\tan^{-1}y}}{1+y^{2}} \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where } P=\frac{1}{1+y^{2}} \text{ and } Q=\frac{e^{-\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int\frac{1}{1+y^{2}}\,dy}$
$\displaystyle =e^{\tan^{-1}y}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{\tan^{-1}y}$
$\displaystyle e^{\tan^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^{2}}\right)=e^{\tan^{-1}y}\frac{e^{-\tan^{-1}y}}{1+y^{2}}$
$\displaystyle \Rightarrow e^{\tan^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^{2}}\right)=\frac{1}{1+y^{2}}$
$\displaystyle \Rightarrow \frac{d}{dy}\left(xe^{\tan^{-1}y}\right)=\frac{1}{1+y^{2}}$
$\displaystyle \text{Integrating both sides with respect to } y,$
$\displaystyle xe^{\tan^{-1}y}=\int\frac{1}{1+y^{2}}\,dy+C$
$\displaystyle =\tan^{-1}y+C \qquad (2)$
$\displaystyle \text{Now, } y(0)=0$
$\displaystyle \Rightarrow 0\cdot e^{0}=\tan^{-1}0+C$
$\displaystyle \Rightarrow 0=0+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Putting } C=0 \text{ in (2),}$
$\displaystyle xe^{\tan^{-1}y}=\tan^{-1}y$

$\displaystyle \text{(vi)}~\frac{dy}{dx}+y\tan x=2x+x^{2}\tan x,\ y(0)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\tan x=2x+x^{2}\tan x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\tan x \text{ and } Q=2x+x^{2}\tan x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\tan x\,dx}$
$\displaystyle =e^{\log(\sec x)}=\sec x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sec x$
$\displaystyle \sec x\left(\frac{dy}{dx}+y\tan x\right)=\sec x\left(2x+x^{2}\tan x\right)$
$\displaystyle \Rightarrow \sec x\left(\frac{dy}{dx}+y\tan x\right)=2x\sec x+x^{2}\tan x\sec x$
$\displaystyle \Rightarrow \frac{d}{dx}(y\sec x)=2x\sec x+x^{2}\tan x\sec x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\sec x=\int 2x\sec x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2\int x\sec x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2\left(x\sec x-\int x\sec x\tan x\,dx\right)+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2\int x\sec x\tan x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2\left(x^{2}\tan x\sec x-\int 2x\tan x\sec x\,dx\right)+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2x^{2}\tan x\sec x+2\int 2x\tan x\sec x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2x^{2}\tan x\sec x+4\int x\tan x\sec x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2x^{2}\tan x\sec x+4\left(\frac{x^{2}}{2}\sec x-\int x^{2}\sec x\tan x\,dx\right)+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x-2x^{2}\tan x\sec x+2x^{2}\sec x-4\int x^{2}\tan x\sec x\,dx+\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =2x\sec x+2x^{2}\sec x-2x^{2}\tan x\sec x-3\int x^{2}\tan x\sec x\,dx+C$
$\displaystyle =x^{2}\sec x+C$
$\displaystyle \Rightarrow y=x^{2}+C\cos x \qquad (2)$
$\displaystyle \text{Now, } y(0)=1$
$\displaystyle \Rightarrow 1=0+C\cos 0$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Putting } C=1 \text{ in (2),}$
$\displaystyle y=x^{2}+\cos x$

$\displaystyle \text{(vii)}~x\frac{dy}{dx}+y=x\cos x+\sin x,\ y\!\left(\frac{\pi}{2}\right)=1. \hspace{2.0cm} \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+y=x\cos x+\sin x$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=\cos x+\frac{\sin x}{x} \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\frac{1}{x} \text{ and } Q=\cos x+\frac{\sin x}{x}$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac{1}{x}\,dx}$
$\displaystyle =e^{\log|x|}=|x|$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=|x|$
$\displaystyle |x|\left(\frac{dy}{dx}+\frac{1}{x}y\right)=|x|\left(\cos x+\frac{\sin x}{x}\right)$
$\displaystyle \Rightarrow \frac{d}{dx}\left(y|x|\right)=|x|\cos x+\sin x$
$\displaystyle \text{For } x>0,\ |x|=x$
$\displaystyle \Rightarrow \frac{d}{dx}(xy)=x\cos x+\sin x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle xy=\int x\cos x\,dx+\int \sin x\,dx+C$
$\displaystyle =x\sin x-\int \sin x\,dx-\cos x+C$
$\displaystyle =x\sin x+\cos x-\cos x+C$
$\displaystyle =x\sin x+C \qquad (2)$
$\displaystyle \text{Now, } y\left(\frac{\pi}{2}\right)=1$
$\displaystyle \Rightarrow \frac{\pi}{2}\cdot 1=\frac{\pi}{2}\sin\left(\frac{\pi}{2}\right)+C$
$\displaystyle \Rightarrow \frac{\pi}{2}=\frac{\pi}{2}+C$
$\displaystyle \Rightarrow C=0$
$\displaystyle \text{Putting } C=0 \text{ in (2),}$
$\displaystyle xy=x\sin x$
$\displaystyle \Rightarrow y=\sin x$

$\displaystyle \text{(viii)}~\frac{dy}{dx}+y\cot x=4x\,\mathrm{cosec}\,x,\ y\!\left(\frac{\pi}{2}\right)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\cot x=4x\,\mathrm{cosec}\,x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\cot x \text{ and } Q=4x\,\mathrm{cosec}\,x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}=\sin x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sin x$
$\displaystyle \sin x\left(\frac{dy}{dx}+y\cot x\right)=\sin x\left(4x\,\mathrm{cosec}\,x\right)$
$\displaystyle \Rightarrow \sin x\left(\frac{dy}{dx}+y\cot x\right)=4x$
$\displaystyle \Rightarrow \frac{d}{dx}(y\sin x)=4x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\sin x=\int 4x\,dx+C$
$\displaystyle =2x^{2}+C \qquad (2)$
$\displaystyle \text{Now, } y\left(\frac{\pi}{2}\right)=0$
$\displaystyle \Rightarrow 0\cdot \sin\left(\frac{\pi}{2}\right)=2\left(\frac{\pi}{2}\right)^{2}+C$
$\displaystyle \Rightarrow 0=\frac{\pi^{2}}{2}+C$
$\displaystyle \Rightarrow C=-\frac{\pi^{2}}{2}$
$\displaystyle \text{Putting } C=-\frac{\pi^{2}}{2} \text{ in (2),}$
$\displaystyle y\sin x=2x^{2}-\frac{\pi^{2}}{2}$

$\displaystyle \text{(ix)}~\frac{dy}{dx}+2y\tan x=\sin x;\ y=0\ \text{when}\ x=\frac{\pi}{3}. \hspace{2.0cm} \text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+2y\tan x=\sin x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=2\tan x \text{ and } Q=\sin x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int 2\tan x\,dx}$
$\displaystyle =e^{2\log(\sec x)}=\sec^{2}x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sec^{2}x$
$\displaystyle \sec^{2}x\left(\frac{dy}{dx}+2y\tan x\right)=\sec^{2}x\sin x$
$\displaystyle \Rightarrow \sec^{2}x\left(\frac{dy}{dx}+2y\tan x\right)=\tan x\sec x$
$\displaystyle \Rightarrow \frac{d}{dx}(y\sec^{2}x)=\tan x\sec x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\sec^{2}x=\int \tan x\sec x\,dx+C$
$\displaystyle =\sec x+C \qquad (2)$
$\displaystyle \text{Now, } y\left(\frac{\pi}{3}\right)=0$
$\displaystyle \Rightarrow 0\cdot \sec^{2}\left(\frac{\pi}{3}\right)=\sec\left(\frac{\pi}{3}\right)+C$
$\displaystyle \Rightarrow 0=2+C$
$\displaystyle \Rightarrow C=-2$
$\displaystyle \text{Putting } C=-2 \text{ in (2),}$
$\displaystyle y\sec^{2}x=\sec x-2$
$\displaystyle \Rightarrow y=\cos x-2\cos^{2}x$

$\displaystyle \text{(x)}~\frac{dy}{dx}-3y\cot x=\sin 2x;\ y=2\ \text{when}\ x=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}-3y\cot x=\sin 2x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-3\cot x \text{ and } Q=\sin 2x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-3\cot x)\,dx}$
$\displaystyle =e^{-3\log(\sin x)}=\sin^{-3}x=\mathrm{cosec}^{3}x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\mathrm{cosec}^{3}x$
$\displaystyle \mathrm{cosec}^{3}x\left(\frac{dy}{dx}-3y\cot x\right)=\mathrm{cosec}^{3}x\sin 2x$
$\displaystyle \Rightarrow \mathrm{cosec}^{3}x\left(\frac{dy}{dx}-3y\cot x\right)=2\cot x\,\mathrm{cosec}x$
$\displaystyle \Rightarrow \frac{d}{dx}\left(y\,\mathrm{cosec}^{3}x\right)=2\cot x\,\mathrm{cosec}x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\,\mathrm{cosec}^{3}x=2\int \cot x\,\mathrm{cosec}x\,dx+C$
$\displaystyle =-2\,\mathrm{cosec}x+C$
$\displaystyle \Rightarrow y=-2\sin^{2}x+C\sin^{3}x \qquad (2)$
$\displaystyle \text{Now, } y\left(\frac{\pi}{2}\right)=2$
$\displaystyle \Rightarrow 2=-2\sin^{2}\left(\frac{\pi}{2}\right)+C\sin^{3}\left(\frac{\pi}{2}\right)$
$\displaystyle \Rightarrow 2=-2+C$
$\displaystyle \Rightarrow C=4$
$\displaystyle \text{Putting } C=4 \text{ in (2),}$
$\displaystyle y=-2\sin^{2}x+4\sin^{3}x$
$\displaystyle \Rightarrow y=4\sin^{3}x-2\sin^{2}x$

$\displaystyle \text{(xi)}~\frac{dy}{dx}+y\cot x=2\cos x,\ y\!\left(\frac{\pi}{2}\right)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}+y\cot x=2\cos x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\cot x \text{ and } Q=2\cos x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}=\sin x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sin x$
$\displaystyle \sin x\left(\frac{dy}{dx}+y\cot x\right)=2\sin x\cos x$
$\displaystyle \Rightarrow \sin x\frac{dy}{dx}+y\cos x=2\sin x\cos x$
$\displaystyle \Rightarrow \frac{d}{dx}(y\sin x)=2\sin x\cos x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\sin x=\int 2\sin x\cos x\,dx+C$
$\displaystyle =\int \sin 2x\,dx+C$
$\displaystyle =-\frac{\cos 2x}{2}+C \qquad (2)$
$\displaystyle \text{Now, } y\left(\frac{\pi}{2}\right)=0$
$\displaystyle \Rightarrow 0\cdot \sin\left(\frac{\pi}{2}\right)=-\frac{\cos \pi}{2}+C$
$\displaystyle \Rightarrow 0=\frac{1}{2}+C$
$\displaystyle \Rightarrow C=-\frac{1}{2}$
$\displaystyle \text{Putting } C=-\frac{1}{2} \text{ in (2),}$
$\displaystyle y\sin x=-\frac{\cos 2x}{2}-\frac{1}{2}$
$\displaystyle \Rightarrow 2y\sin x=-(1+\cos 2x)$
$\displaystyle \Rightarrow 2y\sin x=-2\cos^{2}x$
$\displaystyle \Rightarrow y=-\cot x\cos x$

$\displaystyle \text{(xii)}~dy=\cos x\,(2-y\,\mathrm{cosec}\,x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } dy=\cos x\left(2-y\,\mathrm{cosec}\,x\right)dx$
$\displaystyle \Rightarrow \frac{dy}{dx}=2\cos x-y\cot x$
$\displaystyle \Rightarrow \frac{dy}{dx}+y\cot x=2\cos x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\cot x \text{ and } Q=2\cos x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}=\sin x$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sin x$
$\displaystyle \sin x\left(\frac{dy}{dx}+y\cot x\right)=2\sin x\cos x$
$\displaystyle \Rightarrow \sin x\left(\frac{dy}{dx}+y\cot x\right)=\sin 2x$
$\displaystyle \Rightarrow \frac{d}{dx}(y\sin x)=\sin 2x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle y\sin x=\int \sin 2x\,dx+C$
$\displaystyle =-\frac{\cos 2x}{2}+C$

$\displaystyle \text{(xiii)}~\tan x\,\frac{dy}{dx}=2x\tan x+x^{2}-y;\ \tan x\neq 0\ \text{given that}\ y=0\ \text{when}\ \\ x=\frac{\pi}{2}. \hspace{2.0cm} \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \tan x\frac{dy}{dx}=2x\tan x+x^{2}-y$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{1}{\tan x}y=\frac{2x\tan x+x^{2}}{\tan x}$
$\displaystyle \Rightarrow \frac{dy}{dx}+(\cot x)y=2x+x^{2}\cot x$
$\displaystyle \text{This is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\cot x \text{ and } Q=2x+x^{2}\cot x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\cot x\,dx}$
$\displaystyle =e^{\log(\sin x)}=\sin x$
$\displaystyle y\cdot \text{(I.F.)}=\int Q\cdot \text{(I.F.)}\,dx+C$
$\displaystyle \Rightarrow y\sin x=\int(2x+x^{2}\cot x)\sin x\,dx+C$
$\displaystyle \Rightarrow y\sin x=\int 2x\sin x\,dx+\int x^{2}\cos x\,dx+C$
$\displaystyle \Rightarrow y\sin x=\int 2x\sin x\,dx+\left[x^{2}\sin x-\int 2x\sin x\,dx\right]+C$
$\displaystyle \Rightarrow y\sin x=x^{2}\sin x+C$
$\displaystyle \Rightarrow y=x^{2}+C\,\mathrm{cosec}\,x \qquad (1)$
$\displaystyle \text{It is given that } y=0 \text{ when } x=\frac{\pi}{2}$
$\displaystyle \Rightarrow 0=\left(\frac{\pi}{2}\right)^{2}+C\,\mathrm{cosec}\left(\frac{\pi}{2}\right)$
$\displaystyle \Rightarrow C=-\frac{\pi^{2}}{4}$
$\displaystyle \text{Putting } C=-\frac{\pi^{2}}{4} \text{ in (1),}$
$\displaystyle y=x^{2}-\frac{\pi^{2}}{4}\,\mathrm{cosec}\,x$

$\displaystyle \textbf{Question 38: }~\text{Find the general solution of the differential equation } \\ x\frac{dy}{dx}+2y=x^{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } x\frac{dy}{dx}+2y=x^{2}$
$\displaystyle \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=\frac{2}{x} \text{ and } Q=x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac{2}{x}\,dx}$
$\displaystyle =e^{2\log|x|}=x^{2}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=x^{2}$
$\displaystyle x^{2}\left(\frac{dy}{dx}+\frac{2}{x}y\right)=x^{2}\cdot x$
$\displaystyle \Rightarrow \frac{d}{dx}(x^{2}y)=x^{3}$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle x^{2}y=\int x^{3}\,dx+C$
$\displaystyle =\frac{x^{4}}{4}+C$
$\displaystyle \Rightarrow y=\frac{x^{2}}{4}+\frac{C}{x^{2}}$

$\displaystyle \textbf{Question 39: }~\text{Find the general solution of the differential equation } \\ \frac{dy}{dx}-y=\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dy}{dx}-y=\cos x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-1 \text{ and } Q=\cos x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-1)\,dx}$
$\displaystyle =e^{-x}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=e^{-x}$
$\displaystyle e^{-x}\left(\frac{dy}{dx}-y\right)=e^{-x}\cos x$
$\displaystyle \Rightarrow \frac{d}{dx}(ye^{-x})=e^{-x}\cos x$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle ye^{-x}=\int e^{-x}\cos x\,dx+C$
$\displaystyle \text{Let } I=\int e^{-x}\cos x\,dx$
$\displaystyle I=e^{-x}\sin x-\int(-e^{-x}\sin x)\,dx$
$\displaystyle \Rightarrow I=e^{-x}\sin x+\int e^{-x}\sin x\,dx$
$\displaystyle I=e^{-x}\sin x-e^{-x}\cos x-\int e^{-x}\cos x\,dx$
$\displaystyle \Rightarrow I=e^{-x}(\sin x-\cos x)-I$
$\displaystyle \Rightarrow 2I=e^{-x}(\sin x-\cos x)$
$\displaystyle \Rightarrow I=\frac{e^{-x}}{2}(\sin x-\cos x)$
$\displaystyle \text{From above,}$
$\displaystyle ye^{-x}=\frac{e^{-x}}{2}(\sin x-\cos x)+C$
$\displaystyle \Rightarrow y=\frac{1}{2}(\sin x-\cos x)+Ce^{x}$

$\displaystyle \textbf{Question 40: }~\text{Solve the differential equation } (y+3x^{2})\frac{dx}{dy}=x. \hspace{2.0cm} \text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } (y+3x^{2})\frac{dx}{dy}=x$
$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y+3x^{2}}{x}$
$\displaystyle \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=3x \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dy}{dx}+Py=Q$
$\displaystyle \text{where } P=-\frac{1}{x} \text{ and } Q=3x$
$\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\left(-\frac{1}{x}\right)dx}$
$\displaystyle =e^{-\log|x|}=\frac{1}{|x|}$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\frac{1}{|x|}$
$\displaystyle \frac{1}{|x|}\frac{dy}{dx}-\frac{1}{x|x|}y=\frac{3x}{|x|}$
$\displaystyle \Rightarrow \frac{d}{dx}\left(\frac{y}{|x|}\right)=\frac{3x}{|x|}$
$\displaystyle \text{For } x>0,\ |x|=x$
$\displaystyle \Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right)=3$
$\displaystyle \text{Integrating both sides with respect to } x,$
$\displaystyle \frac{y}{x}=3\int dx+C$
$\displaystyle \Rightarrow \frac{y}{x}=3x+C$

$\displaystyle \textbf{Question 41: }~\text{Find the particular solution of the differential equation } \\ \frac{dx}{dy}+x\cot y=2y+y^{2}\cot y,\ y\neq 0,\ \text{given that } x=0\ \text{when}\ y=\frac{\pi}{2}. \hspace{2.0cm} \text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \frac{dx}{dy}+x\cot y=2y+y^{2}\cot y \qquad (1)$
$\displaystyle \text{Clearly, it is a linear differential equation of the form } \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where } P=\cot y \text{ and } Q=2y+y^{2}\cot y$
$\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int\cot y\,dy}$
$\displaystyle =e^{\log(\sin y)}=\sin y$
$\displaystyle \text{Multiplying both sides of (1) by I.F.}=\sin y$
$\displaystyle \sin y\left(\frac{dx}{dy}+x\cot y\right)=\sin y\left(2y+y^{2}\cot y\right)$
$\displaystyle \Rightarrow \sin y\frac{dx}{dy}+x\cos y=y^{2}\cos y+2y\sin y$
$\displaystyle \Rightarrow \frac{d}{dy}(x\sin y)=y^{2}\cos y+2y\sin y$
$\displaystyle \text{Integrating both sides with respect to } y,$
$\displaystyle x\sin y=\int y^{2}\cos y\,dy+\int 2y\sin y\,dy+C$
$\displaystyle =y^{2}\sin y-\int 2y\sin y\,dy+\int 2y\sin y\,dy+C$
$\displaystyle =y^{2}\sin y+C$
$\displaystyle \text{Now, } y=\frac{\pi}{2} \text{ at } x=0$
$\displaystyle \Rightarrow 0\cdot \sin\left(\frac{\pi}{2}\right)=\left(\frac{\pi}{2}\right)^{2}\sin\left(\frac{\pi}{2}\right)+C$
$\displaystyle \Rightarrow 0=\frac{\pi^{2}}{4}+C$
$\displaystyle \Rightarrow C=-\frac{\pi^{2}}{4}$
$\displaystyle \text{Putting } C=-\frac{\pi^{2}}{4},$
$\displaystyle x\sin y=y^{2}\sin y-\frac{\pi^{2}}{4}$

$\displaystyle \textbf{Question 42: }\text{Solve the following differential equation } \\ (\cot^{-1}y+x)\,dy=(1+y^{2})\,dx. \hspace{2.0cm} \text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given differential equation is } (\cot^{-1}y+x)\,dy=(1+y^{2})\,dx$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{\cot^{-1}y+x}{1+y^{2}}$
$\displaystyle \Rightarrow \frac{dx}{dy}-\frac{1}{1+y^{2}}x=\frac{\cot^{-1}y}{1+y^{2}}$
$\displaystyle \text{This is a linear differential equation of the form } \frac{dx}{dy}+Px=Q$
$\displaystyle \text{where } P=-\frac{1}{1+y^{2}} \text{ and } Q=\frac{\cot^{-1}y}{1+y^{2}}$
$\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int\left(-\frac{1}{1+y^{2}}\right)dy}$
$\displaystyle =e^{\cot^{-1}y}$
$\displaystyle \text{Multiplying the differential equation by I.F., we get}$
$\displaystyle e^{\cot^{-1}y}\frac{dx}{dy}-\frac{x}{1+y^{2}}e^{\cot^{-1}y}=\frac{\cot^{-1}y}{1+y^{2}}e^{\cot^{-1}y}$
$\displaystyle \Rightarrow \frac{d}{dy}\left(xe^{\cot^{-1}y}\right)=\frac{\cot^{-1}y}{1+y^{2}}e^{\cot^{-1}y}$
$\displaystyle \text{Integrating both sides with respect to } y,$
$\displaystyle xe^{\cot^{-1}y}=\int\frac{\cot^{-1}y}{1+y^{2}}e^{\cot^{-1}y}\,dy+C$
$\displaystyle \text{Put } t=\cot^{-1}y$
$\displaystyle \Rightarrow dt=-\frac{1}{1+y^{2}}\,dy$
$\displaystyle \Rightarrow xe^{\cot^{-1}y}=-\int te^{t}\,dt+C$
$\displaystyle \Rightarrow xe^{\cot^{-1}y}=-e^{t}(t-1)+C$
$\displaystyle \Rightarrow xe^{\cot^{-1}y}=e^{\cot^{-1}y}(1-\cot^{-1}y)+C$