Question

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

• A. $\sqrt{2}-2$
• B. $\frac{1}{\sqrt{2}}-1$
• C. $2-\sqrt{2}$
• D. $2+\sqrt{2}$

Answer 1

Answer: (A)

$\cos x \frac{d y}{d x}+2 y \sin x=\sin 2 x$
$\frac{d y}{d x}+\frac{2 \sin x}{\cos x} y=2 \sin x$
I.F. $=e^{\int 2 \frac{\sin x}{\cos x} d x}$
$=e^{2 \ln \sec x}=\sec ^{2} x$
$y \cdot \sec ^{2} x=\int 2 \sin x \cdot \sec ^{2} x d x$
$y \sec ^{2} x=2 \int \tan x \sec x d x$
y $\sec ^{2} x=2 \sec x+c$
At $x=\frac{\pi}{3}, y=0$
$\Rightarrow 0=2 \sec \frac{\pi}{3}+C $
$\Rightarrow C=-4$
$y \sec ^{2} x=2 \sec x-4$
Put $x=\frac{\pi}{4}$
y $\cdot 2=2 \sqrt{2}-4$
y $=\sqrt{2}-2$