find the particular solution that satisfies the differential equation and the initial condition
f’’(x)=x^2 , f’(0)=4, f(0)=8
f(x)=
2 answers
idk sorry
f" = x^2
f' = 1/3 x^3 + C
f'(0) = 4 ==> C=4, so
f' = 1/3 x^3 + 4
f = 1/12 x^4 + 4x + C
f(0)=8 ==> C=8, so
f(x) = 1/12 x^4 + 4x + 8
f' = 1/3 x^3 + C
f'(0) = 4 ==> C=4, so
f' = 1/3 x^3 + 4
f = 1/12 x^4 + 4x + C
f(0)=8 ==> C=8, so
f(x) = 1/12 x^4 + 4x + 8
6 answers