Differential Equations 2 Question 17

17.

The function y=f(x) is the solution of the differential equation dydx+xyx21=x4+2x1x2 in (1,1) satisfying f(0)=0. Then, 323/2f(x)dx is

(2014 Adv.)

(a) π332

(b) π334

(c) π634

(d) π632

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Answer:

Correct Answer: 17. (b)

Solution:

  1. PLAN (i) Solution of the differential equation dydx+Py=Q is

y(IF)=Q(IF)dx+c where, F=ePdx

(ii) aaf(x)dx=20af(x)dx, if f(x)=f(x)

Given differential equation

dydx+xx21y=x4+2x1x2

This is a linear differential equation.

IF=exx21dx=e12ln|x21|=1x2 Solution is y1x2=x(x3+2)1x21x2dx or y1x2=(x4+2x)dx=x55+x2+cf(0)=0c=0f(x)1x2=x55+x2 Now, 3/23/2f(x)dx=3/23/2x21x2dx=203/2x21x2dx=20π/3sin2θcosθcosθdθ [taking x=sinθ]=20π/3sin2θdθ=0π/3(1cos2θ)dθ=θsin2θ20π/3=π3sin2π/32=π334



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