Given that $y' = \sqrt{y}+1$, $y(0)=0$, $x\in [0,1] =: I$, how does one show that this ODE has a unique solution on I? I was thinking that one might be able to show that this ODE satisfies a Lipschitz condition on some $J \subset I$. We might consider $\lvert \sqrt{y_1} - \sqrt{y_2} \lvert^2 = \lvert y_1 - 2\sqrt{y_1 y_2}+y_2\lvert\le2\lvert y_1-\sqrt{y_1y_2}+y_2 \lvert = 2\lvert -y_1+\sqrt{y_1^2+y_2^2}-y_2 \lvert$ $\le2\lvert -y_1+\sqrt{y_1^2+2y_1y_2+y_2^2}-y_2 \lvert $ $=2\lvert -y_1+y_1+y_2-y_2 \lvert \le 2\lvert -y_1+y_2\lvert + 2\lvert y_1-y_2 \lvert = 4\lvert y_1 - y_2\lvert$.
I'm concerned, however, that one would need to arrive at something like $K\lvert y_1-y_2 \lvert^2$ in this case.
Would appreciate some clarification.
Write the differential equation as $$ \dfrac{1}{1+\sqrt{y}} \dfrac{dy}{dx} = 1 $$ We can write this as $\dfrac{d}{dx} F(y(x)) = 1$, where $$F(y) = \int_0^y \dfrac{ds}{1+\sqrt{s}}$$ Thus we must have $F(y(x)) = x + c$ for some constant $c$. Since $F(0)=0$, the initial condition $y(0)=0$ tells us $c = 0$. Since $F$ is an increasing function, $F(y(x)) = x$ specifies $y(x)$ uniquely for any $x$.