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

Question

f ''(x) = 3x²,    f '(−1) = −4    f(2) = 7

Answer 1

f'(x) = ∫f"(x)dx = x³ + C₁

Since f'(-1) = -4, we have -1 + C₁ = -4. So, C₁ = -3

f'(x) = x³ - 3

f(x) = ∫f'(x)dx = (1/4)x⁴ - 3x + C₂

Since f(2) = 7, we get 4 - 6 + C₂ = 7. So, C₂ = 9

Therefore, f(x) = (1/4)x⁴ - 3x + 9