Question

Let \(y=y(x)\) be the solution of the differential equation \(\sin x\frac{dy}{dx}+y\cos x=4x\), \(x\in(0,\pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to :

• A. \(-\frac{4}{9}\pi^2\)
• B. \(\frac{4}{9\sqrt3}\pi^2\)
• C. \(-\frac{8}{9\sqrt3}\pi^2\)
• D. \(-\frac{8}{9}\pi^2\)

Hint:

The left side is \(\frac{d}{dx}(y\sin x)\). Integrate \(4x\), use \(y(\pi/2)=0\) to find the constant.

Answer 1

The Correct Option is D

\(\sin x\frac{dy}{dx}+y\cos x=\frac{d}{dx}(y\sin x)=4x\).
Integrating: \(y\sin x=2x^2+C\).
At \(x=\frac\pi2\), \(y=0\): \(0=2\cdot\frac{\pi^2}{4}+C\Rightarrow C=-\frac{\pi^2}{2}\).
At \(x=\frac\pi6\): \(y\cdot\frac12=2\cdot\frac{\pi^2}{36}-\frac{\pi^2}{2}=\frac{\pi^2}{18}-\frac{9\pi^2}{18}=-\frac{8\pi^2}{18}=-\frac{4\pi^2}{9}\).
So \(y\left(\frac\pi6\right)=-\frac{8\pi^2}{9}\).
Answer: (4) \(-\frac{8}{9}\pi^2\)