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QJEEM25Ma4dcd: JEE Main 2025 Mathematics - Class 12 JEEM25Ma4dcd Solution: Let be the solution of the differential equation . If , then is …

Let y=y(x)y=y(x) be the solution of the differential equation

2cos⁡x dy dx=sin⁡2x−4ysin⁡x,x∈(0,π2)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(π3)=0y\left(\frac{\pi}{3}\right)=0 , then y′(π4)+y(π4)y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right) is equal to _________.

Step-by-Step Verified Solution

amp;dydx+2ytan⁡x=sin⁡xamp; I.F. =e2∫tan⁡xdx=sec⁡2xamp;ysec⁡2x=∫sin⁡xcos⁡2xdxamp;=∫tan⁡xsec⁡xdxamp;=sec⁡x+Camp;C=−2amp;y=cos⁡x−2cos⁡2xamp;y(π4)=12−1amp;y′=−sin⁡x+4cos⁡2xsin⁡xamp;y′(π4)=−12+2amp;y′(π4)+y(π4)=1\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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