Let $y=y(x)$ be the solution of the differential equation $2\cos x\frac{dy}{dx}=\sin2x-4y\sin x,\ x\in\left(0,\frac\pi2\right)$. If $y\left(\frac\pi3\right)=0$, then $y'\left(\frac\pi4\right)+y\left(\frac\pi4\right)$ is equal to ________.

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $y'+2y	an x=\sin x$. Integrating factor $\sec^2x$: $y\sec^2x=\int\sin x\sec^2x\,dx=\sec x+C$, so $y=\cos x+C\cos^2x$. $y\left(\frac\pi3ight)=0\Rightarrow\frac12+\frac C4=0\Rightarrow C=-2$. $y=\cos x-2\cos^2x$, $y\left(\frac\pi4ight)=\frac{\sqrt2}2-1$, $y'=-\sin x+4\sin x\cos x$ gives $y'\left(\frac\pi4ight)=2-\frac{\sqrt2}2$. Sum $=1$. Answer: 1.

Q. Let $y=y(x)$ be the solution of the differential equation $2\cos x\frac{dy}{dx}=\sin2x-4y\sin x,\ x\in\left(0,\frac\pi2\right)$. If $y\left(\frac\pi3\right)=0$, then $y'\left(\frac\pi4\right)+y\left(\frac\pi4\right)$ is equal to ________.

Explanation

Reviewed by Dr. Abubacker

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