Solution to the following differential equation

Question

I am trying to find the solution for the differential equation $\frac{dz}{dt}$ = $z^{\alpha}$ for some $0<\alpha<1$. Can anyone help me out here!!

Thanks in advance

Answer 1

This is a separable differential equation. Divide both sides by $z^\alpha$: $$ z^{-\alpha} \frac{dz}{dt} = 1 $$

Thus: $$ \int z^{-\alpha} \, {dz} = \int \, dt $$

Integrate to get: $$ \frac{z^{1-\alpha}}{1-\alpha} = t + c $$

One thing to note here, when we divided by $z^\alpha$, we assumed that $z$ was not the constant function $0$. This function is also a solution to the differential equation. The constant of integration often takes care of "incorporating" this extra solution into the general one, but it's not the case here.

Answer 2

Hint: $z^{-\alpha}\,dz=dt$. Integrate. On the left-hand side you have a power. The right-hand side is even simpler. Don't forget the "$C$" (on one side only is fine).