Lipschitz continuity and differential equations 
Does anyone have any ideas for this one? could use some help.
 A: Your guess $f(t)=\sqrt t$ works out fine.
So now you want to find a solution $y$ to $y'(x)=\sqrt {y(x)}$. This is a separable equation. If $y'(x)\neq 0$, $$y'(x)=\sqrt {y(x)}\iff y'(x)(y(x))^{-1/2}=1\iff (y(x))^{1/2}=\dfrac x 2+\dfrac C 2,$$
which implies $y(x)=\left(\dfrac x2+\dfrac C 2\right)^2$.
But you want $y(0)=y_0$. Let us choose $y_0=0$ for simplicity. "So $C$ must be zero", not quite because the solution we got was for whenever $y(x)\neq 0$. It could be that $y(0)=0$, so this suggests looking at $$y_1(x)=\dfrac {x^2}4 \text{ and } y_2(x)=\begin{cases} 0, &\text{if }x\leq -C\\\left(\dfrac x2+\dfrac C2\right)^2, &\text{if }x\ge -C \end{cases},$$ where $C$ is a negative number, (you want to shift to the right, so the solution is worth $0$ at $0$, hence picking a  negative $C$). This $y_1$ is only good on $[0, +\infty[$.So make another adjustement:
Take $$\varphi _1(x)=\begin{cases} 0, &\text{if }x\leq 0\\ \dfrac {x^2}4, &\text{if }x\ge 0\end{cases}\text{ and }\varphi_2(x)=\begin{cases} 0, &\text{if }x\leq 1\\\left(\dfrac {x-1}2\right)^2, &\text{if }x\ge 1 \end{cases}$$
You should check that $\varphi _1$ and $\varphi _2$ are differentiable and are solutions to the I.V.P..
