# How do you solve the differential equation given g'(x)=6x^2, g(0)=-1?

Nov 1, 2016

$g \left(x\right) = 2 {x}^{3} - 1$

#### Explanation:

$g ' \left(x\right) = 6 {x}^{2}$

So we can integrate to get;

$g \left(x\right) = \int 6 {x}^{2} \mathrm{dx}$
$\therefore g \left(x\right) = 2 {x}^{3} + C$

Using the initial condition $g \left(0\right) = - 1 \implies - 1 = 0 + C$
$\therefore C = - 1$

Hence, the solution is:
$g \left(x\right) = 2 {x}^{3} - 1$