Question

Let \( y=y(x) \) be the solution of differential equation \( \sin x \frac{dy}{dx}+y\cos x=4x \). If \( y\left(\frac{\pi}{2}\right)=0 \) and \( y\left(-\frac{\pi}{6}\right)=\frac{\alpha}{\beta}\pi^{2} \), where \( \alpha \) and \( \beta \) are co-prime numbers, then value of \( (\alpha+\beta) \) is ________

Hint:

The left side is the derivative of \( y\sin x \); integrate both sides.

Answer 1

Correct Answer: 17

The left side is \( \frac{d}{dx}(y\sin x) \), so \( \frac{d}{dx}(y\sin x)=4x \).
Integrating: \( y\sin x = 2x^2 + C \).
Using \( y(\pi/2)=0 \): \( 0 = 2\cdot\frac{\pi^2}{4} + C \Rightarrow C = -\frac{\pi^2}{2} \).
At \( x=-\pi/6 \): \( y\cdot\left(-\frac12\right) = 2\cdot\frac{\pi^2}{36} - \frac{\pi^2}{2} = \frac{\pi^2}{18}-\frac{9\pi^2}{18} = -\frac{4\pi^2}{9} \).
So \( y = \frac{8}{9}\pi^2 \), giving \( \alpha=8, \beta=9 \) (co-prime).
\( \alpha+\beta = 17 \).
Answer: 17