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

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