General Questions | JEE 2012 | Discussion Page

If y(x) satisfies the differential equation $y'-y \tan x=2 x \sec x$ and y(0) , then

$y\left(\frac{\pi}{4}\right)= \frac{\pi^{2}}{8 \sqrt{2}}$

$y' \left(\frac{\pi}{4}\right)= \frac{\pi^{2}}{18}$

$y\left(\frac{\pi}{3}\right)= \frac{\pi^{2}}{9}$

$y'\left(\frac{\pi}{3}\right)= \frac{4 \pi^{}}{3}+\frac{2\pi^{2}}{3 \sqrt{3}}$

Option A,D

Concept lnvolved. Linear differential equation under one variable.

$\frac{dy}{dx}+Py=Q$ ; $IF=e^{\int P dx}$

$\therefore$ Solution is , $y(IF)= \int Q. (IF) dx+C$

Sol. $y'- y \tan x=2x \sec x$ and y(0)=0$

$\Rightarrow$ $\frac{dy}{dx}-y \tan x=2 x \sec x$

$\therefore$ $IF= \int e^{-\tan x} dx=e^{\log |\cos x|}=\cos x$

Solution is

$y. \cos x=\int 2x.\sec x.\cos x dx+C$

$\Rightarrow$ $y. \cos x=x^{2}+C$

As y(0)=0 $\Rightarrow$ C=0

$\therefore$ $y=x^{2}\sec x$

Now, $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8\sqrt{2}}$

$\Rightarrow$ $y'\left(\frac{\pi}{4}\right)=\frac{\pi^{}}{\sqrt{2}}+\frac{\pi^{2}}{8\sqrt{2}}$

$\Rightarrow$ $y\left(\frac{\pi}{3}\right)=\frac{2\pi^{2}}{9}$

$\Rightarrow$ $y'\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3\sqrt{3}}$

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