If $y=y(x)$ is the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=(\tan x-y) \sec ^{2} x, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $\mathrm{y}(0)=0$
then $y\left(-\frac{\pi}{4}\right)$ is equal to :
Correct Option: 1
$\frac{d y}{d x}+y \sec ^{2} x=\sec ^{2} \times \tan x$
Given equation is linear differential equation.
IF $=e^{\int \sec ^{2} x d x}=e^{\tan x}$
$\Rightarrow y \cdot e^{\tan x}=\int e^{\tan x} \sec ^{2} \times \tan x d x$
Put $\tan x=u=\sec ^{2} x d x=d u$
$y e^{\tan x}=\int e^{u} u d u \Rightarrow y e^{\tan x}=u e^{u}-e^{u}+c$
$\Rightarrow y e^{\tan x}=(\tan x-1) e^{\tan x}+c$
$\Rightarrow y=(\tan x-1)+c \cdot e^{-\tan x}$
$\therefore y(0)=0($ given $) \Rightarrow 0=-1+c \Rightarrow c=1$
Hence, solution of differential equation,
$y\left(-\frac{\pi}{4}\right)=-1-1+e=-2+e$