Question

If $\frac{dy}{dx}+2y\tan x=\sin x,0<x<\frac{\pi }{2}$ and $y\left(\frac{\pi }{3}\right)=0$, then the maximum value of $y(x)$ is

• A. $\frac{1}{8}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. $\frac{3}{8}$

Answer 1

Answer: (A)

$\frac{dy}{dx}+2y\tan x=\sin x$
I.F $=e^{\int 2\tan xdx}=e^{\ln (\sec x{)}^2}={\sec }^{2}x$
$y\left({\sec }^{2}x\right)=\int \sin x{\sec }^{2}xdx+C$
$y\cdot {\sec }^{2}x=\sec x+C$
Put $x=\frac{\pi }{3},y=0$
$y=\cos x-2{\cos }^{2}x$
$=\frac{1}{8}-2{\left(\cos x-\frac{1}{4}\right)}^2$
$\therefore y_{\max }=\frac{1}{8}$