Question

Find the general solution of given differential equation.

$(1+x^2)\frac{dy}{dx}+y={\tan }^{-1}x$

• A. $y={\tan }^{-1}x+1+c{e}^{-{\tan }^{-1}x}.$
• B. $y={\tan }^{-1}x-1+c{e}^{-{\tan }^{-1}x}.$
• C. $y={\tan }^{-1}x+1+c{e}^{{\tan }^{-1}x}.$
• D. $y={\tan }^{-1}x-1+c{e}^{{\tan }^{-1}x}.$

Answer 1

Answer: (B)

$y={\tan }^{-1}x-1+c{e}^{-{\tan }^{-1}x}.$

$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{{\tan }^{-1}\left(x\right)}{1+x^2}$

$IF=e^{\int \frac{1}{1+x^2}.dx}$

$=e^{{\tan }^{-1}\left(x\right)}$

$y.e^{{\tan }^{-1}\left(x\right)}=\int e^{{\tan }^{-1}\left(x\right)}\frac{{\tan }^{-1}\left(x\right)}{1+x^2}.dx$

Let

$I=\int e^{ta{n}^{-1}\left(x\right)}\frac{{\tan }^{-1}\left(x\right)}{1+x^2}.dx$

Let

${\tan }^{-1}\left(x\right)=t$

Hence

$\frac{1}{1+x^2}.dx=dt$

Hence,

$I=\int e^t.t$

$=e^t(t-1)$

$=e^{{\tan }^{-1}\left(x\right)}\left({\tan }^{-1}\right(x)-1)$

Hence,

$\therefore y.e^{{\tan }^{-1}\left(x\right)}=\int e^{{\tan }^{-1}\left(x\right)}\frac{{\tan }^{-1}\left(x\right)}{1+x^2}.dx$

$\therefore y.e^{{\tan }^{-1}\left(x\right)}=e^{{\tan }^{-1}\left(x\right)}\left({\tan }^{-1}\right(x)-1)+C$

$\therefore y={\tan }^{-1}\left(x\right)-1+C{e}^{-{\tan }^{-1}\left(x\right)}$.