Find the particular solution that satisfies the differential equation and the initial condition.

Question

f ''(x) = x^2, f '(0) = 8, f(0) = 8
f (x) = ?

Answer 1

d^2y/dx^2 = x^2

dy/dx = (1/3)x^3 + c
8 = 0 + c so c = 8
or
dy/dx = (1/3)x^3 + 8

y = (1/12)x^4 + 8 x + k
8 = 0 + k
so k = 8
y = (1/12)x^4 + 8 x + 8