Let y = y(x) be the solution of the differential equation (x2 + 1)y' - 2xy = (x4 + 2x2 + 1) x, y(0) = 1. Then -33 y(x) dx is :
- 36
- 24
- 18
- 30
Answer
(2) 24
Solution
Divide by (x2 + 1); note x4 + 2x2 + 1 = (x2 + 1)2:
dydx - 2xx2 + 1y = (x2 + 1) x
I.F. = e- 2xx2 + 1dx = e- (x2 + 1) = 1x2 + 1
yx2 + 1 = x dx = x + C
y(0) = 1 C = 1
y = (x2 + 1)( x + 1) = (x2 + 1) x + (x2 + 1)
(x2 + 1) x is odd, so its integral over [-3, 3] is zero:
-33 y dx = -33 (x2 + 1) dx = 2 [ x33 + x ]03
= 2(9 + 3) = 24
Answer: (2)