Let y=y(x) be the solution of the differential equation 2 cos x d y/d x=sin 2 x-4 y sin x, x in (0, π/2 ) . If y (π/3 )=0 , then y prime (π/4 )+y (π/4 ) is equal to \ \ \ \ \ .

Mathematics · Differential Equations

JEE Main 2025 — 24 January, Evening Shift — Question 21

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x, x \in\left(0, \frac{\pi}{2}\right)$ .

If $y\left(\frac{\pi}{3}\right)=0$ , then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to $\_\_\_\_\_$ .

Numerical answer — enter this value.

$\frac{d y}{d x}+2 y \tan x=\sin x$

I.F. $=e^{2 \int \tan x d x}=\sec ^{2} x$

$y \sec ^{2} x=\int \frac{\sin x}{\cos ^{2} x} d x$

$=\int \tan x \sec x d x$

$=\sec x+C$ ,, $\mathrm{C}=-2$

$y=\cos x-2 \cos ^{2} x$

$y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}-1$

$y^{\prime}=-\sin x+4 \cos x \sin x$

$y^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}+2$

$y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)=1$

Answer key and solution verified before publishing.

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