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

If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x$ $=2 x \sec x$ and $y(0)=0$, then
  1. $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$
  2. $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}$
  3. $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}$
  4. $y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$

Solution

$\frac{d y}{d x}-y \tan x=2 x \sec x$ I.F. $=e^{-\int \tan x d x}=\cos x$ $\therefore \quad y \cdot \cos x=\int 2 x d x=x^{2}+c$ Now, $y(0)=0 \Rightarrow c=0, \therefore y=x^{2} \sec x$ $\Rightarrow y^{\prime}=2 x \sec x+x^{2} \sec x \tan x$ Now, $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16} \times \sqrt{2}=\frac{\pi^{2}}{8 \sqrt{2}}$ $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9} \times 2=\frac{2 \pi^{2}}{9}$ $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{2 \pi}{4} \times \sqrt{2}+\frac{\pi^{2}}{8 \sqrt{2}} \times 1=\frac{\pi^{2}}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}}$ $y^{\prime}\left(\frac{\pi}{3}\right)=\frac{2 \pi}{3} \times 2+\frac{2 \pi^{2}}{9} \times \sqrt{3}=\frac{2 \pi^{2}}{3 \sqrt{3}}+\frac{4 \pi}{3}$

Asked in: JEE Advanced 2012 (Paper 1)

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