How do you solve the differential equation given f''(x)=x^2, f'(0)=6, f(0)=3?

Jul 23, 2017

$f \left(x\right) = \frac{1}{12} {x}^{4} + 6 x + 3$

Explanation:

We just integrate twice.

$\int \left(f ' ' \left(x\right)\right) = \int {x}^{2} \mathrm{dx}$

$f ' \left(x\right) = \frac{1}{3} {x}^{3} + C$

We now solve for $C$.

$6 = \frac{1}{3} {\left(0\right)}^{3} + C$

$C = 6$

Therefore,

$f ' \left(x\right) = \frac{1}{3} {x}^{3} + 6$

Accordingly,

$\int f ' \left(x\right) = \int \frac{1}{3} {x}^{3} + 6 \mathrm{dx}$

$f \left(x\right) = \frac{1}{12} {x}^{4} + 6 x + C$

We solve for $C$ one more time.

$3 = \frac{1}{12} {\left(0\right)}^{4} + 6 \left(0\right) + C$

$3 = C$

The differential equation therefore has solution $f \left(x\right) = \frac{1}{12} {x}^{4} + 6 x + 3$.

Hopefully this helps!