ExamBro
ExamBro
JEE Advanced · Mathematics · 29. Differential Eqns

The function y=fx is the solution of the differential equation dydx+xyx2-1=x4+2x1-x2 in (- 1 , 1) satisfying f0=0. Then -3232fx dx is

  1. A π3-32
  2. B π3- 34
  3. C π6-34
  4. D π6-32
Verified Solution

Answer & Solution

Correct Answer

(B) π3- 34

Step-by-step Solution

Detailed explanation

dydx+xx2-1y=x4+2x1-x2
This is a linear differential equation
I.F. =exx2-1 dx=e12 In x2-1= 1-x2
solution is
y 1-x2=xx3+21-x2 . 1-x2 dx
Or y1-x2=x4+2x dx=x55+x2+c
f0=0 c=0
fx 1-x2=x55+x2
Now, -3232fxdx=-3232x21-x2 dx (Using property)
=2032x21-x2 dx=2 0π3sin2θcosθcosθdθ (Taking x=sinθ )
=20π3sin2θdθ=2θ2-sin2θ40π3=2π6-238=π3-34
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app