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

Answer 1

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

Given equation is linear differential of the form $\frac{dy}{dx}+Py=Q$
Here, $P=\tan x$ and $Q=2x+x^2\tan x$

$I.F.=e^{\int \tan xdx}=e^{\ln \sec x}=\sec x$

$\Rightarrow y\cdot \sec x=\int (2x+x^2\tan x)\sec xdx$
$\Rightarrow y\sec x=x^2\sec x+\lambda$
$\Rightarrow y=x^2+\lambda \cos x$
$\Rightarrow y\left(0\right)=0+\lambda =1$
$\Rightarrow \lambda =1$
$\Rightarrow y=x^2+\cos x$
$\Rightarrow y^{\prime }=2x-\sin x$
$\Rightarrow y^{\prime }\left(\frac{\pi }{4}\right)=\frac{\pi }{2}-\frac{1}{\sqrt{2}}$
$\Rightarrow y^{\prime }(-\frac{\pi }{4})=-\frac{\pi }{2}+\frac{1}{\sqrt{2}}$