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# 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 $y(0)=1$. Then :

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

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

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

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

Correct Option: , 2

Solution:

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

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

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

$=\int 2 x \sec x d x+\int x^{2}(\sec x \cdot \tan x) d x$

$y \sec x=x^{2} \sec x+\lambda$

$\Rightarrow y=x^{2}+\lambda \cos x$

$y(0)=0+\lambda=1 \quad \Rightarrow \lambda=1$

$y=x^{2}+\cos x$

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

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

$y^{\prime}(x)=2 x-\sin x$

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

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

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