Question

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

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