Is this function Lipschitz and therefore has a unique solution?

Question

I'm struggling to understand if the differential equation $y'=ye^x$ with initial value $y(0)=1$ or rather the $f(x,y)= ye^x$ part is Lipschitz, as from that it would follow that it has a unique solution. I have already solved the IVP, which gave me $y(x)=e^{e^x-1}$, but I don't think this answers whether it has a unique solution. Thanks for any answer.

Answer 1

$y(x)=e^{e^x}$ is a solution of the differential equation, but not a solution of the IVP, since $y(0)=e \ne 1.$

The correct solution of the IVP is given by $ y(x)=\frac{1}{e}e^{e^x}$.

In order to show that this is the unique solution of the IVP, you need no "Lipschitz":

Asumme that $z$ is also a solution of the IVP and consider the function $h:= y/z.$

Now show that $h'=0$ on $ \mathbb R.$ Hence $h$ is constant. Since $h(0)=1$, we get $h=1,$ thus $z=y.$