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 :
Correct Option: 1
$\frac{d y}{d x}+3 \sec ^{2} x \cdot y=\sec ^{2} x$
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)
$\therefore \quad y \cdot e^{-3}=\frac{1}{3} e^{-3}+e^{3}$
$\therefore \quad y=\frac{1}{3}+\mathrm{e}^{6}$
$\therefore \quad y\left(-\frac{\pi}{4}\right)=\frac{1}{3}+e^{6}$
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