To show that this IVP does/does not have a unique solution

Question

Given the IVP $y'=2√y, y(0)=0$. I find $y=x^2$ as the only solution to the problem. But I am asked to show that it does not have a unique solution. Is there a mistake or am I just wrong? Is there a sufficient condition for the existence of more than one solution to an IVP?

Answer 1

This is a standard type of example problem to illustrate the difference of the Cauchy-Peano and Picard-Lindelöf theorems. You get any of $$ y(x)=\begin{cases}0,&x<c,\\(x-c)^2,&x\ge c,\end{cases} $$ as valid solutions to this IVP.

Answer 2

When you separate variables, recall that either $y=0$, which is easy to see is a solution, or $$ \int\frac{\mathrm dy}{\sqrt{y}}=2\int \mathrm dt\implies \sqrt{y}=t+c\\ \implies y=t^2+2ct+c $$ and $c=0$ thanks to BC, you have, as you found $$ y=t^2 $$