Question

If $y(x)$ satisfies the differential equation $y^{\prime }-ytanx=2xsecx$ and $y(0)=0$, then

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

Answer 1

Answer: (A), (D)

$\frac{dy}{dx}-y\tan x=2x\sec x$
$\cos x\frac{dy}{dx}+(-\sin x)y=2x$
$\frac{d}{dx}(y\cos x)=2x$
$y(x)\cos x=x^2+c,$ 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 $x=\frac{\pi }{4},y^{\prime }\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{8\sqrt{2}}+\frac{\pi }{\sqrt{2}}$
when $x=\frac{\pi }{3},y^{\prime }\left(\frac{\pi }{3}\right)=\frac{2{\pi }^2}{3\sqrt{3}}+\frac{4\pi }{3}$