Question

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

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

Answer 1

The correct option is B $\sqrt{2}-2$

$\frac{dy}{dx}+\left(2\tan x\right)y=2\sin x$
I.F. $=e^{2\ln \left(\sec x\right)}={\sec }^{2}x$
$y\left({\sec }^{2}x\right)=\int \frac{2\sin x}{{\cos }^{2}x}dx$
$\Rightarrow y\left({\sec }^{2}x\right)=2\int \sec x\tan xdx$
$\Rightarrow y\left({\sec }^{2}x\right)=2\sec x+c$

$\because y\left(\frac{\pi }{3}\right)=0$
$0=2\times 2+c$
$\Rightarrow c=-4$
So, $y\left({\sec }^{2}x\right)=2\sec x-4$
At $x=\pi /4$
${\left[y\right]}_{x=\pi /4}\times (\sqrt{2}{)}^2=2\sqrt{2}-4$
$\therefore {\left[y\right]}_{x=\pi /4}=\sqrt{2}-2$