f''(x) = x^2 f'(0) = 8, f(o)=4
if f''(x) = x^2, then
f'(x) = (1/3)x^3 + c
for the given:
8 = (1/3)(0) + c ---> c = 8 , and
f'(x) = (1/3)x^3 + 8
f(x) = (1/12)x^4 + 8x + k
f(0) = 4
4 = (1/12)(0) + 8(0) + k ----> k = 4
f(x) = (1/12)x^4 + 8x + 4
do the 2nd question in the same way
Find the particular solution that satisfies the differential equation and the initial conditions.
f''(x) = x^2 f'(0) = 8, f(o)=4
f'(t) 10t - 12t^3, f(3) =2
1 answer
1 answer