Solution differential equation |QuadraticEquationJEEMAIN2019

JEE MAIN 2019

10-04-2019 S2

Question

Let $y=y(x)$ be the solution of the differential equation, $\frac{d y}{d x}+y \tan 2 x=x^2+x^2 \tan x$, $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $y(0)=1$.Then :

Select the correct option:

$y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^2}{2}+2$

$y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=\sqrt{2}$

$y^{\prime}\left(\frac{\pi}{4}\right)+y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt{2}$

$y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Hard

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