JEE Main 2019 — Differential Equations Question with Solution

The given differential equation can be written as
$\frac{dy}{dx}+y{\mathrm{s}\mathrm{e}\mathrm{c}}^{2}x=\mathrm{t}\mathrm{a}\mathrm{n}x.{\mathrm{s}\mathrm{e}\mathrm{c}}^{2}x$
Integrating factor $=e^{\int {\mathrm{s}\mathrm{e}\mathrm{c}}^{2}xdx}=e^{\mathrm{t}\mathrm{a}\mathrm{n}x}$
Hence, solution of given differential equation,
$y.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}=\int \mathrm{t}\mathrm{a}\mathrm{n}x.{\mathrm{s}\mathrm{e}\mathrm{c}}^{2}x.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}dx$
$\Rightarrow y.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}=\mathrm{t}\mathrm{a}\mathrm{n}x.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}-\int \mathrm{s}\mathrm{e}{\mathrm{c}}^{2}x.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}dx$   (using integration by parts)

$\Rightarrow y.e^{\tan x}=\tan x.e^{\tan x}-e^{\tan x}+c$

Given, $y\left(0\right)=0\Rightarrow \mathrm{c}=1$

Solution of given differential is

$y.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}=\mathrm{t}\mathrm{a}\mathrm{n}x.e^{\mathrm{t}\mathrm{a}\mathrm{n}x}-e^{\mathrm{t}\mathrm{a}\mathrm{n}x}+1$

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

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