Question

The solution of the differential equation $\left[\right(\cos x)dx–dy](1+x^2)+e^x\left[\right(1+x^2){\tan }^{–1}x+1]dx=0$ is
(where $C$ is arbitrary constant)

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

Answer 1

The correct option is A $y=e^x{\tan }^{-1}x+\sin x+C$

$\left[\right(\cos x)dx–dy](1+x^2)+e^x\left[\right(1+x^2){\tan }^{–1}x+1]dx=0$
Dividing the equation by $(1+x^2)dx$, we get
$\Rightarrow \cos x-\frac{dy}{dx}+e^x({\tan }^{-1}x+\frac{1}{1+x^2})=0\Rightarrow \frac{dy}{dx}=\cos x+e^x({\tan }^{-1}x+\frac{1}{1+x^2})\Rightarrow \int dy=\int \cos x+e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx+C\Rightarrow y=\sin x+\int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx+C\therefore y=\sin x+e^x{\tan }^{-1}x+C$