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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Mathematics
Chapter
Differential Equations

JEE Advanced 2012 · Paper 1 · Mathematics · Differential Equations

If y(x) satisfies the differential equation y^ -y x =2 x x and...

If y(x)y(x) satisfies the differential equation y′−ytan⁡xy^{\prime}-y \tan x =2xsec⁡x=2 x \sec x and y(0)=0y(0)=0, then

Options

  1. Ay(π4)=π282y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}
  2. By′(π4)=π218y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}
  3. Cy(π3)=π29y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}
  4. Dy′(π3)=4π3+2π233y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}

Correct answer

A. y(π4)=π282y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}

Step-by-step solution

dydx−ytan⁡x=2xsec⁡x\frac{d y}{d x}-y \tan x=2 x \sec x I.F. =e−∫tan⁡xdx=cos⁡x=e^{-\int \tan x d x}=\cos x ∴y⋅cos⁡x=∫2xdx=x2+c\therefore \quad y \cdot \cos x=\int 2 x d x=x^{2}+c Now, y(0)=0⇒c=0,∴y=x2sec⁡xy(0)=0 \Rightarrow c=0, \therefore y=x^{2} \sec x ⇒y′=2xsec⁡x+x2sec⁡xtan⁡x\Rightarrow y^{\prime}=2 x \sec x+x^{2} \sec x \tan x Now, y(π4)=π216×2=π282y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16} \times \sqrt{2}=\frac{\pi^{2}}{8 \sqrt{2}} y(π3)=π29×2=2π29y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9} \times 2=\frac{2 \pi^{2}}{9} y′(π4)=2π4×2+π282×1=π282+π2y^{\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′(π3)=2π3×2+2π29×3=2π233+4π3y^{\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}
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