Question:

Let $y=y(x)$ be the solution of the differential equation $(1+x^2)\dfrac{dy}{dx}+y=e^{\tan^{-1}x}$, $y(1)=0$. Then $y(0)$ is:

$\dfrac{1}{2}(e^{\pi/2}-1)$

$\dfrac{1}{2}(1-e^{\pi/2})$

$\dfrac{1}{4}(1-e^{\pi/2})$

$\dfrac{1}{4}(e^{\pi/2}-1)$

Step-by-Step Solution

Key Concept: IF $=e^{\tan^{-1}x}$. $ye^{\tan^{-1}x}=\int\frac{e^{2\tan^{-1}x}}{1+x^2}dx=\frac{e^{2\tan^{-1}x}}{2}+C$.

$y(0)=\frac{1-e^{\pi/2}}{2}$.

Correct Answer: 2