Question

If $y(x)$ satisfies the differential equation $y' - y tan x = 2x\, secx$ and $y(0) = 0$, then

• A. $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$
• B. $y'\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}$
• C. $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}$
• D. $y'\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$

Answer 1

Answer: (A), (D)

$\frac{d y}{d x}-y \tan x=2 x \sec x$
$\cos x \frac{d y}{d x}+(-\sin x) y=2 x$
$\frac{d}{d x}(y \cos x)=2 x$
$y ( x ) \cos x = x ^{2}+ c ,$ where $c =0$ since $y (0)=0$
when $x =\frac{\pi}{4}, y \left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}},$ when $x =\frac{\pi}{3}, y \left(\frac{\pi}{3}\right)=\frac{2 \pi^{2}}{9}$
when $x =\frac{\pi}{4}, y'\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}}$
when $x =\frac{\pi}{3}, y '\left(\frac{\pi}{3}\right)=\frac{2 \pi^{2}}{3 \sqrt{3}}+\frac{4 \pi}{3}$