Question

Let \(y=y(x)\) be the solution of the differential equation \( \cos x\,\frac{dy}{dx}+2y\sin x=\sin2x,\ x\in\left(0,\frac{\pi}{2}\right) \). If \( y\left(\frac{\pi}{3}\right)=0 \), then \( y\left(\frac{\pi}{4}\right) \) is equal to :

• A. \( \frac{1}{\sqrt2}-1 \)
• B. \( \sqrt2-2 \)
• C. \( 2-\sqrt2 \)
• D. \( 2+\sqrt2 \)

Hint:

Divide by \(\cos x\) to get a linear equation \(y'+2\tan x\,y=2\sin x\) and use the integrating factor \(\sec^2x\).

Answer 1

The Correct Option is B

Divide by \(\cos x\): \(\frac{dy}{dx}+(2\tan x)\,y=2\sin x\).
Integrating factor \(=e^{\int2\tan x\,dx}=e^{2\ln\sec x}=\sec^2x\).
\(y\sec^2x=\int2\sin x\sec^2x\,dx=2\int\sec x\tan x\,dx=2\sec x+C\), so \(y=2\cos x+C\cos^2x\).
\(y\left(\frac\pi3\right)=0\Rightarrow 2\cdot\frac12+C\cdot\frac14=0\Rightarrow C=-4\).
\(y\left(\frac\pi4\right)=2\cdot\frac{1}{\sqrt2}-4\cdot\frac12=\sqrt2-2\).
Answer: (2) \( \sqrt2-2 \)