Question

Let $y=y(x)$ be the solution of the differential equation $\cos x\frac{dy}{dx}+2y\sin x=\sin 2x$ $x\in \left(0,\frac{\pi }{2}\right).$ 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)

$\cos x\frac{dy}{dx}+2y\sin x=\sin 2x$
$\frac{dy}{dx}+\frac{2\sin x}{\cos x}y=2\sin x$
I.F. $=e^{\int 2\frac{\sin x}{\cos x}dx}$
$=e^{2\ln \sec x}={\sec }^{2}x$
$y\cdot {\sec }^{2}x=\int 2\sin x\cdot {\sec }^{2}xdx$
$y{\sec }^{2}x=2\int \tan x\sec xdx$
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 $=\sqrt{2}-2$