Question

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