Question

Let $y=y\left(x\right)$ be the solution of the differential equation, $\frac{dy}{dx}+y\tan x=2x+x^2\tan x$, $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, such that $y\left(0\right)=1$. 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. $\left(\frac{\pi }{4}\right)+y(-\frac{\pi }{4})=\frac{{\pi }^2}{2}+2$

Answer 1

The correct option is B $y^{\prime }\left(\frac{\pi }{4}\right)-y^{\prime }\left(\frac{\pi }{4}\right)=\pi -\sqrt{2}$

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