Question

Solution of differential equation $\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{ta{n}^{-1}x}}{1+x^2}$ is

• A. $y=\frac{e^{ta{n}^{-1}x}}{2}+c$
• B. $y=\frac{e^{2ta{n}^{-1}x}}{2}+c$
• C. $y=\frac{e^{ta{n}^{-1}x}}{2}+c{e}^{-ta{n}^{-1}x}$
• D. $y=e^{ta{n}^{-1}x}+c$

Answer 1

Answer: (C)

$y=\frac{e^{ta{n}^{-1}x}}{2}+c{e}^{-ta{n}^{-1}x}$

Integrating factor =IF =$e^{\int 1/(1+x^2)dx}=e^{{\tan }^{-1}x}$

Solution :

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

RHS :let ${\tan }^{-1}x=p\Rightarrow \frac{1}{1+x^2}dx=dp$

RHS =$\int e^{2p}dp=\frac{e^{2p}}{2}=\frac{e^{2{\tan }^{-1}x}}{2}$

Solution:

$\Rightarrow y{e}^{{\tan }^{-1}x}=\frac{e^{2{\tan }^{-1}x}}{2}+c$

$\Rightarrow y=\frac{e^{{\tan }^{-1}x}}{2}+c{e}^{-{\tan }^{-1}x}$