If y=y(x) is the solution of the differential equation
Question:

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 :

  1. (1) $\mathrm{e}-2$

     

     

  2. (2) $\frac{1}{2}-e$

  3. (3) $2+\frac{1}{e}$

  4. (4) $\frac{1}{e}-2$


Correct Option: 1

Solution:

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

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,

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

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