Answer:
Option A,D
Explanation:
Concept lnvolved. Linear differential equation under one variable.
$\frac{dy}{dx}+Py=Q$ ; $IF=e^{\int P dx}$
$\therefore$ Solution is , $y(IF)= \int Q. (IF) dx+C$
Sol. $ y'- y \tan x=2x \sec x $ and y(0)=0$
$\Rightarrow$ $ \frac{dy}{dx}-y \tan x=2 x \sec x$
$\therefore$ $IF= \int e^{-\tan x} dx=e^{\log |\cos x|}=\cos x$
Solution is
$y. \cos x=\int 2x.\sec x.\cos x dx+C$
$\Rightarrow$ $y. \cos x=x^{2}+C$
As y(0)=0$\Rightarrow$ C=0
$\therefore$ $y=x^{2}\sec x$
Now, $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8\sqrt{2}}$
$\Rightarrow$ $y'\left(\frac{\pi}{4}\right)=\frac{\pi^{}}{\sqrt{2}}+\frac{\pi^{2}}{8\sqrt{2}}$
$\Rightarrow$ $y\left(\frac{\pi}{3}\right)=\frac{2\pi^{2}}{9}$
$\Rightarrow$ $y'\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3\sqrt{3}}$
