Q.

The function y=f(x) is the solution of the differential equation dydx+xyx2-1=x4+21-x2 in (-1,1) satisfying f(0)=0. Then 2∫-3/23/2f(x) dx is

1 2π3+32  
2 2π3-32  
3 π6-34  
4 π6-32  

Ans.

(2)

dydx+xx2-1y=x4+2x1-x2

The I.F. of this differential equation is     e∫xx2-1 dx=e-∫x1-x2 dx=e12log(1-x2)=1-x2

The solution is given by   y1-x2=∫x3+21-x21-x2 dx+λ

=∫(x4+2x) dx+λ=x55+x2+λ

At y(0)=0⇒λ=0⇒y1-x2=x55+x2

∫-3/23/2x55+x21-x2 dx=∫-3/23/2x21-x2 dx

(The other part is odd) =2∫03/2x21-x2 dx

Let x=sinθ, we get

I=2∫0π/3sin2θcosθcosθ dθ

  =2∫0π/3sin2θ dθ

   =∫0π/3(1-cos2θ) dθ

    =[θ-sin2θ2]0π/3

     =π3-34

∴  2I=2π3-32