JEE Main 2019 — Differential Equations Question with Solution

Given $\frac{dy}{dx}+\left(3{\sec }^{2}x\right)y={\sec }^{2}x$

This is linear differential equation, of the type $\frac{dy}{dx}+Py=Q,$ where $P \& Q$ are functions of $x$ or constants.

Here, $P=3se{c}^{2}x \& Q=se{c}^{2}x$

Now, integrating factor $I.F.=e^{\int Pdx}$

$=e^{\int 3{\sec }^{2}x dx}=e^{3\tan x}$

Hence, the solution of the differential equation is $y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$

$\Rightarrow y·e^{3\tan x}=\int e^{3\tan x}·{\sec }^{2}xdx$

Let, $\tan x=t, \Rightarrow se{c}^{2}xdx=dt$

$\Rightarrow y·e^{3\tan x}=\int e^{3t}dt+c$

$\Rightarrow y·e^{3\tan x}=\frac{e^{3t}}{3}+c$

$\Rightarrow y·e^{3\tan x}=\frac{e^{3\tan x}}{3}+c$

$\Rightarrow y=c{e}^{-3\tan x}+\frac{1}{3}$

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

$\frac{4}{3}=c{e}^{-3}+\frac{1}{3}$

$\Rightarrow c=e^3$

Thus, $y=e^3·e^{-3\tan x}+\frac{1}{3}$

Hence, $y\left(-\frac{\pi }{4}\right)=e^3·e^{-3\left(\tan \left(-\frac{\mathrm{\pi }}{4}\right)\right)}+\frac{1}{3}$

$\Rightarrow y\left(-\frac{\pi }{4}\right)=e^3·e^{-3\left(-1\right)}+\frac{1}{3}$

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