Question

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

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

$\frac{dy}{dx}-y\text{tan}x=2x\text{sec}x$
$y\left(0\right)=0$

$I.F.=e^{-\int \text{tan}xdx}=e^{-\log \text{sec}x}$

$I.F.=\text{cos}x$

$\text{cos}x.y=x^2+c$
As $y\left(0\right)=0$
Put $x=0$
$c=0$
$y=x^2\text{sec}x$
$y^{\prime }=2x\sec x+x^2\sec x\tan x$

$y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}.\sqrt{2}=\frac{{\pi }^2}{8\sqrt{2}}$

$y^{\prime }\left(\frac{\pi }{4}\right)=\frac{\pi }{2}.\sqrt{2}+\frac{{\pi }^2}{16}.\sqrt{2}$

$y\left(\frac{\pi }{3}\right)=\frac{{\pi }^2}{9}.2=\frac{2{\pi }^2}{9}$

$y^{\prime }\left(\frac{\pi }{3}\right)=2\frac{\pi }{3}.2+\frac{{\pi }^2}{9}.2.\sqrt{3}$

$=\frac{4\pi }{3}+\frac{2{\pi }^2\sqrt{3}}{9}$