Differential equation with 3d root of x

Question

How can I solve the following differential equation?

$x'(t)=-\sqrt[3]{x(t)}$

How can I solve the following initial value problem?

$x'(t)=-\sqrt[3]{x(t)}\\ x(0)=-1$

I thought x must be non negative...because of $\sqrt[3]{x(t)}\\$?

Answer 1

$$\frac{dx}{dt}=-x^{1/3} \implies \int x^{-1/3} dx=-\int dt + C \implies \frac{3 x^{2/3}}{2}=-t+C \implies C=3/2.$$ Finally we choose $-$ branch of square root to write $$x(t)=-\sqrt{(1-2t/3)^3}.$$

Note to OP: $(1)^{1/3}=1,\omega, \omega^2; (-1)^{1/3}=-1, -\omega, -\omega^2$, so here $x(0)=-1.$