Let g be a differentiable function such that 0x g(t) dt = x - 0x t g(t) dt, x 0 and let y = y(x) satisfy the differential equation dydx - y x = 2(x+1) x g(x), x [0, 2 ). If y(0) = 0, then y ( 3 ) is equal to
- 4 3
- 2 3
- 2 3 3
- 4 3 3
Answer
(1) 4 3
Solution
Differentiating 0x g(t) dt = x - 0x t g(t) dt w.r.t. x:
g(x) = 1 - x g(x)
g(x) = 11+x
the differential equation becomes
dydx - y x = 2(x+1) x 11+x = 2 x
It is linear in y; I.F. = e- x dx = e x = x
ddx(y x) = 2 x x = 2
y x = 2x + c
y(0) = 0 c = 0
y = 2x x
y ( 3 ) = 2 3 2 = 4 3
Answer: (1)