Question

Let $y = y(x)$ be the solution of the differential equation, $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, x \in \left( - \frac{\pi}{2} , \frac{\pi}{2} \right) $ , such that $y(0) = 1$. Then :

• A. $y'\left(\frac{\pi}{4}\right)+y'\left(\frac{-\pi}{4}\right)=- \sqrt{2} $
• B. $y'\left(\frac{\pi}{4}\right) - y'\left(\frac{-\pi}{4}\right)= \pi - \sqrt{2} $
• C. $y\left(\frac{\pi}{4}\right) - y\left( - \frac{\pi}{4}\right)= \sqrt{2} $
• D. $y\left(\frac{\pi}{4}\right)+y\left(\frac{-\pi}{2}\right)= \frac{\pi^2}{2} + 2 $

Answer 1

Answer: (B)

$\frac{dy}{dx} +y\left(\tan x\right)=2x+x^{2} \tan x $
$ I.F = e^{\int\tan xdx} = e^{ln.\sec x} =\sec x $
$ \therefore y .\sec x =\int\left(2x+x^{2} \tan x\right)\sec x. dx $
$ = \int 2x \sec x dx + \int x^{2} \left(\sec x. \tan x\right) dx $
$ y \sec x =x^{2} \sec x+\lambda $
$ \Rightarrow y = x^{2} +\lambda\cos x $
$ y\left(0\right) =0+\lambda=1 \Rightarrow \lambda=1 $
$ y=x^{2} +\cos x$
$ y\left(\frac{\pi}{4}\right) = \frac{\pi^{2}}{16} + \frac{1}{\sqrt{2}} $
$ y\left(- \frac{\pi }{4}\right) = \frac{\pi ^{2}}{16} + \frac{1}{\sqrt{2}} $
$ y'\left(x\right)=2x -\sin x$
$ y'\left(\frac{\pi}{4}\right)=\frac{\pi}{2} - \frac{1}{\sqrt{2}} $
$ y'\left(\frac{-\pi}{4}\right) = \frac{-\pi}{2} + \frac{1}{\sqrt{2}} $
$ y'\left(\frac{\pi}{4}\right) -y'\left(\frac{-\pi}{4}\right) = \pi - \sqrt{2} $