JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 23)
Let $y=y(x)$ be the solution of the differential equation
$2 \cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}=\sin 2 x-4 y \sin x, x \in\left(0, \frac{\pi}{2}\right)$. 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 _________.
$2 \cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}=\sin 2 x-4 y \sin x, x \in\left(0, \frac{\pi}{2}\right)$. 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 _________.
Answer
1
Explanation
$$\begin{aligned}
& \frac{d y}{d x}+2 y \tan x=\sin x \\
& \text { I.F. }=e^{2 \int \tan x d x}=\sec ^2 x \\
& y \sec ^2 x=\int \frac{\sin x}{\cos ^2 x} d x \\
& =\int \tan x \sec x d x \\
& =\sec x+C \\
& C=-2 \\
& y=\cos x-2 \cos ^2 x \\
& y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}-1 \\
& y^{\prime}=-\sin x+4 \cos ^2 x \sin x \\
& y^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}+2 \\
& y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)=1
\end{aligned}$$
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