Find the particular solution of the differential equation that satisfies the initial condition(s).

Question

(a) f' (x) = 5x², f (2) = 7

(b) f'' (x) = 4, f'(-1) = 4, f (-1) = -6

(c) f'' (x) = cos x, f ' (π) = 3, f (π) = -1

(d) f'' (x) = ex, f '(0) = -3, f (0) = 6

Answer 1

Dusan

Here we integrate (once or twice), using the initial conditions to set the integration constants

(a) f(x) = ∫ 5x² dx = 5 (x³/3) + C f(2) = 5 (8/3) + C = 7, so C = 7 - 5 (8/3)

So f(x) = (5/3) x³ + 7 - (40/3)

(b) f'(x) = ∫ 4 dx = 4x + C. f'(-1) = -4+C = 4 so C = 8

f'(x) = 4x + 8

f(x) = ∫ (4x+8) dx = 2x²+8x+C

f(-1) = 2 - 8 + C = -6 so C = 0

f(x) = 2x² + 8x

(c) (d) Integrate twice

Mike