Let be the solution of differential|Differential Equations

JEE MAIN 2020

04-09-2020 S1

Question

Let $y=y(x)$ be the solution of the differential equation, $x y^{\prime}-y=x^2(x \cos x+\sin x), x>0$. If $y(\pi)=\pi$, then $y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right)$ is equal to :

Select the correct option:

$1+\frac{\pi}{2}$

$1+\frac{\pi}{2}+\frac{x^2}{4}$

$2+\frac{\pi}{2}$

$2+\frac{\pi}{2}+\frac{\pi^2}{4}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Medium

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