Differential Equation - $y'=|y|+1, y(0)=0$ The equation is $y'=|y|+1, y(0)=0$.
Suppose $y$ is a solution on an interval $I$. Let $x\in I$.
If $y(x)\ge 0$ then $$y'(x)=|y(x)|+1\iff y'(x)=y(x)+1\iff \frac{y'(x)}{y(x)+1}=1\\ \iff \ln (y(x)+1)=x+C\iff y(x)+1=e^{x+C}\\ \iff y(x)=e^{x+C}-1$$
Then $y(0)=0\implies C=0$. So $y(x)=e^x-1$ if $y(x)\ge 0$.
If $y(x)\leq 0$ then $y(x)=1-e^x$.
Now I want to say $y(x)=\begin{cases} e^x-1, \text{if } x\ge 0\\1-e^x, \text{if } x\leq 0\end{cases}$
Is this correct? Is there only one solution?

I think my solution above is correct.
there are a few details missing:
it is necessary to show that $y(x)\leq 0\iff x\leq0$ and $y(x)\ge 0\iff x\ge 0$ which allows me to define $y$ the way I do.
also it is necessary to check that $y$ is differentiable at $x=0$ and it is because:
$$\lim _{x\to 0^+}\frac{e^x-1}{x-0}=1=\lim _{x\to 0^-}\frac{1-e^{-x}}{x-0}$$
