QJEEAdv12Ma531a: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Ma531a Solution: If satisfies the differential equation and then

Given, $\frac{d y}{d x}-y \tan x=2 x \sec x$

Comparing with the liner differential form

$\begin{gathered}\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \\\Rightarrow \mathrm{P}=-\tan x \text { and } \mathrm{Q}=2 x \sec x\end{gathered}$

Now, Integrating factor (I.F) $=e^{\int P d x}$

$\begin{aligned}& =e^{\int-\tan x d x} \\& =e^{-\log \sec x} \\& =\cos x\end{aligned}$

$\begin{aligned}& \text { The solution will be } y \text { (I.F) }=\int Q \text { (I.F) } d x \\& \Rightarrow \quad y \cos x=\int 2 x \sec x \cdot \cos x d x \\& \Rightarrow \quad y \cos x=2 \frac{x^2}{2}+c \\& \Rightarrow \quad y=x^2 \sec x+c \sec x \\\end{aligned}$

Now, $y(0)=0$

$\begin{aligned}& \Rightarrow \quad 0=0^2 \sec 0+\mathrm{c} \mathrm{sec} 0 \\& \Rightarrow \quad c=0 \\& \therefore \quad y=x^2 \sec x \\& y^{\prime}=x^2 \sec x \cdot \tan x+2 x \sec x \quad \text{[using Product Rule in Differentiation]}\\\end{aligned}$

Now,

$\begin{aligned}y\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4}=\frac{\pi^2}{16} \sqrt{2}=\frac{\pi^2}{8 \sqrt{2}} \\y^{\prime}\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4} \tan \frac{\pi}{4}+2 \cdot \frac{\pi}{4} \sec \frac{\pi}{4} \\& =\frac{\pi^2}{16} \cdot \sqrt{2} \cdot 1+\frac{\pi}{2} \cdot \sqrt{2} \\& =\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}}\end{aligned}$

$\begin{aligned}y\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \\& =\frac{\pi^2}{9} \cdot 2=\frac{2 \pi^2}{9} \\y^{\prime}\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \cdot \tan \frac{\pi}{3}+2 \cdot \frac{\pi}{3} \cdot \sec \frac{\pi}{3} \\& =\frac{\pi^2}{9} \cdot 2 \cdot \sqrt{3}+\frac{2 \pi}{3} \cdot 2 \\& =\frac{2 \pi^2}{3 \sqrt{3}}+\frac{4 \pi}{3}\end{aligned}$