JEE Main 2019 — Differential Equations Question with Solution

$\frac{dy}{dx}-y\tan x=6x\sec x ...\left(i\right),$ which is a liner differential equation of the form $\frac{dy}{dx}+Py=Q,$ where $P\hspace{0.17em}\&\hspace{0.17em}Q$ are the function of $x$ or constants.

Here, $P=-\tan x \& Q=6xsecx$

Integrating factor $=e^{\int Pdx}=e^{-\int \tan x dx}$

$=e^{-\left(-\ln \left(\cos x\right)\right)}=e^{\ln \left(\cos x\right)}=\cos x$

And, the solution of the linear differential equation is,

$y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$

 $\Rightarrow y\cdot \cos x=\int \left(6x\sec x\right)\cdot \cos xdx$

$\Rightarrow y\cdot \cos x=\int 6xdx$

$\Rightarrow y·\cos x=3x^2+c ...\left(ii\right)$

Given $y\left(\frac{\pi }{3}\right)=0,$ hence

$0=3{\left(\frac{\mathrm{\pi }}{3}\right)}^2+c$

$\Rightarrow c=-\frac{{\pi }^2}{3}$

Thus, the curve is, $y·\cos x=3x^2-\frac{{\pi }^2}{3}$

Now, put $x=\frac{\pi }{6},$ to get

$y\left(\frac{\pi }{6}\right)·\cos \left(\frac{\mathrm{\pi }}{6}\right)=3{\left(\frac{\mathrm{\pi }}{6}\right)}^2-\frac{{\pi }^2}{3}$

$\Rightarrow y\left(\frac{\pi }{6}\right)\cdot \frac{\sqrt{3}}{2}=3\cdot \frac{{\pi }^2}{36}-\frac{{\pi }^2}{3}$

$\Rightarrow y\left(\frac{\pi }{6}\right)\cdot \frac{\sqrt{3}}{2}=\frac{{\pi }^2}{12}-\frac{{\pi }^2}{3}$

$\Rightarrow y\left(\frac{\pi }{6}\right)\cdot \frac{\sqrt{3}}{2}=-\frac{{\pi }^2}{4}$

$\Rightarrow y\left(\frac{\pi }{6}\right)=-\frac{{\pi }^2}{2\sqrt{3}}.$