Question

If $y\left(x\right)$ satisfies the differential equation $y^{\prime }-y\tan x=2x\sec x$ and $y\left(0\right)=0$, then:

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

Answer 1

Answer: (A), (D)

The correct options are
A $y\left(\pi /4\right)=\frac{{\pi }^2}{8\sqrt{2}}$
D $y^{\prime }\left(\pi /3\right)=\frac{4\pi }{3}+\frac{2{\pi }^2}{3\sqrt{3}}$

$\frac{dy}{dx}-ytanx=2xsecx$

$cosx\frac{dy}{dx}+(-sinx)y=2x$

$\frac{d}{dx}\left(ycosx\right)=2x$

$y\left(x\right)cosx=x^2+e$, Where $e=0$

Since $y\left(0\right)=0$.

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 $x=\frac{\pi }{3},y^{\prime }\left(\frac{\pi }{3}\right),=\frac{2{\pi }^2}{3\sqrt{3}}+\frac{4\pi }{3}$.