Two Solutions for an ODE: $x' = x^{\frac45}$ Find two different solutions $x_1, x_2 : \mathbb{R} \to \mathbb{R}$ of
$$ \dot{x} = x^{\frac45}, \quad x(1) = 1. $$
This is a problem in a 60-minute exam, so it should be quite simple but still I'm failing. 
I can get $x(t) = \left(\frac{t+4}{5} \right)^5$ via separation of variables (might have miscomputed but that's not too important) but how can we find another solution?
The solutions on Wolfram-Alpha do not look too simple...

To get two different solutions, you need to have a violation of the Lipschitz condition, even the local one. This can be tested by looking for singularities of the derivative of the right side. You will find that there is such a singularity at $x=0$, and find further that the constant-zero function is a solution. 
You can check that your solution has derivative $0$ where it takes the value $0$, so that a continuation to the left with the zero solution is also a solution of the IVP. 
$$
x(t)=\begin{cases}0,&t<-4,\\\left(\frac{t+4}5\right)^5,&t\ge-4.\end{cases}
$$
There are other solutions in-between these two.
