# Let y=y(x) be the solution of the differential equation,

Question:

Let $y=y(x)$ be the solution of the differential equation,

$\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $\mathrm{y}(0)=1$

Then :

1. (1) $y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{2}+2$

2. (2) $y,\left(\frac{\pi}{4}\right)+y \cdot\left(\frac{\pi}{4}\right)=-\sqrt{2}$

3. (3) $y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=\sqrt{2}$

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

Correct Option: , 4

Solution:

Given differential equation is,

$\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$

Here, $\mathrm{P}=\tan x, \mathrm{Q}=2 x+x^{2} \tan x$

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

$\therefore y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x$

$=\int x^{2} \tan x \sec x d x+\int 2 x \sec x d x=x^{2} \sec x+c$

Given $y_{2}(0)=1 \Rightarrow c=1$

$\therefore y=x^{2}+\cos x$..........(i)

Now put $x=\frac{\pi}{4}$ and $x=\frac{-\pi}{4}$ in equation (i),

$y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16}+\frac{1}{\sqrt{2}}$ and $y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{16}+\frac{1}{\sqrt{2}}$

$\Rightarrow y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=0$

$\frac{d y}{d x}=2 x-\sin x$

$\therefore y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi}{2}-\frac{1}{\sqrt{2}}$ and $y^{\prime}\left(-\frac{\pi}{4}\right)=-\frac{\pi}{2}+\frac{1}{\sqrt{2}}$

$\Rightarrow y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}$