Question

If $y(x)$ satisfies the differential equation $y'-y \tan \, x=2 \, x \sec \, x $ and $ y(0) = 0 $ then

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

Answer 1

Answer: (A), (D)

PLAN Linear differential equation under one variable.
$ \frac {dy}{dx}+Py=Q; \, IF=e^{\int \limits Pdx}$
$ \therefore $ Solution is, $ y(IF)=\int \limits Q.(IF) dx+C $
$ y'-y \tan\, x=2x \sec \, x $ and $y(0)=0$
$\Rightarrow \frac {dy}{dx}-y\, \tan \, x =2x \sec \, x $
$\therefore IF=\int \limits e^{-\tan \, x}dx=e^{\log|\cos \, x|}=\cos \, x$
Solution is $y.\cos x= \int \limits 2 x \sec \, x.\cos\, x dx+C$
$\Rightarrow y.\cos \, x=x^2+ C $
As $ y(0)=0$
$\Rightarrow C=0 $
$\therefore y=x^2 \sec \, x $
Now, $ y\bigg (\frac {\pi}{4} \bigg )=\frac {\pi^2}{8 \sqrt 2}$
$\Rightarrow y'\bigg (\frac {\pi}{4} \bigg )= \frac {\pi}{\sqrt 2}+ \frac {\pi^2}{8 \sqrt 2}$
$ y\bigg (\frac {\pi}{3} \bigg )= \frac {2\pi^2}{9}$
$\Rightarrow y'\bigg ( \frac {\pi}{3}\bigg )= \frac {4 \pi}{3}+ \frac {2\pi^2}{3 \sqrt 3}$