# Solve this following

Question:

If $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{3}{\cos ^{2} \mathrm{x}} \mathrm{y}=\frac{1}{\cos ^{2} \mathrm{x}}, \mathrm{x} \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$, and

$\mathrm{y}\left(\frac{\pi}{4}\right)=\frac{4}{3}$, then $\mathrm{y}\left(-\frac{\pi}{4}\right)$ equals :

1. $\frac{1}{3}+e^{6}$

2. $\frac{1}{3}$

3. $-\frac{4}{3}$

4. $\frac{1}{3}+e^{3}$

Correct Option: 1

Solution:

$\frac{d y}{d x}+3 \sec ^{2} x \cdot y=\sec ^{2} x$

I.F. $=e^{3 \int \sec ^{2} x d x}=e^{3 \tan x}$

or $y \cdot e^{3 \tan x}=\int \sec ^{2} x \cdot e^{3 \tan x} d x$

or $y \cdot \mathrm{e}^{3 \tan x}=\frac{1}{3} e^{3 \tan x}+C$     .................(1)

Given

$y\left(\frac{\pi}{4}\right)=\frac{4}{3}$

$\therefore \quad \frac{4}{3} \cdot \mathrm{e}^{3}=\frac{1}{3} \mathrm{e}^{3}+\mathrm{C}$

$\therefore \mathrm{C}=\mathrm{e}^{3}$

Now put $x=-\frac{\pi}{4}$ in equation (1)

$\therefore \quad y \cdot e^{-3}=\frac{1}{3} e^{-3}+e^{3}$

$\therefore \quad y=\frac{1}{3}+\mathrm{e}^{6}$

$\therefore \quad y\left(-\frac{\pi}{4}\right)=\frac{1}{3}+e^{6}$