Question

If $y(x)$ satisfies the differential equation $y^{\prime }-y\tan x=2x\sec x$ 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)

$\frac{dy}{dx}-y\tan x=2x\sec x$
$y(0)=0$
I.F. $=e^{-\int \tan xdx}=e^{-\ell n\sec x}$
I.F. $=\cos x$
$\cos x\cdot y=\int 2x$ sec $x\cdot \cos dx$
$\cos x\cdot y=x^2+c$
$c=0$
$y=x^2\sec x$
$y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}\cdot \sqrt{2}=\frac{{\pi }^2}{8\sqrt{2}}$
$y^{\prime }\left(\frac{\pi }{4}\right)=\frac{\pi }{2}\cdot \sqrt{2}+\frac{{\pi }^2}{16}\cdot \sqrt{2}$
$y\left(\frac{\pi }{3}\right)=\frac{{\pi }^2}{9}\cdot 2=\frac{2{\pi }^2}{9}$
$y^{\prime }\left(\frac{\pi }{3}\right)=2\frac{\pi }{2}\cdot 2+\frac{{\pi }^2}{9}\cdot 2\cdot \sqrt{3}$
$\frac{4\pi }{3}+\frac{2{\pi }^2\sqrt{3}}{9}$