Solve the differential equation: $dy/dx=\sqrt y,\ y(0)=0$ 
I am trying to solve this. But I see that this equation can not satisfy the LIPSCHITZ
  
  condition in the interval containing 0.  Can this be solved by separation of variables?

 A: By Peano existence theorem the ODE $y^\prime=\sqrt y,\ y(0)=0$ has at least one solution defined in an interval $[0,\alpha)$ with $\alpha >0$. But...that solution might not be unique because as you noticed, Picard-Lindelöf theorem hypothesis are not satisfied.
Hence unicity might not be satisfied...
And indeed it is the case. While you can solve the equation by separation of variables, as mentioned in the other solutions doing so, you're supposing that the solution is not equal to zero in a neighborhood of $0$. A pretty strong hypothesis as the always vanishing function is also a solution. A good example of an ODE having several solutions!
$y_1(x)=0$ is a first solution defined on $\mathbb R$. A second one is equal to $y_2(x)=\frac{1}{4}x^2$ for $x \ge 0$.
And as mentioned by mickep, for $c >0$ the function defined by $\begin{cases}
0 & \text { for } 0 \le x \le c \\
\frac{1}{4} (x-c)^2 & \text{ for } x \ge 0
\end{cases}$ is also a solution.
Conclusion: the ODE has an infinite number of solutions.
A: If an equation does not satisfy Lipschitz it still can have a solution. Picard-Lindelöf is not a necessary condition, it is only sufficient.
$$\frac{dy}{dx}=\sqrt{y}$$
$$\frac{dy}{\sqrt{y}}=dx$$
$$y^{-\frac{1}{2}}dy=dx$$
$$\frac{1}{-\frac{1}{2}+1}y^{-\frac{1}{2}+1}=x+c$$
$$\frac{1}{\frac{1}{2}}y^{\frac{1}{2}}=x+c$$
$$2y^{\frac{1}{2}}=x+c$$
$$y^{\frac{1}{2}}=\frac{1}{2}(x+c)$$
$$y=\frac{1}{4}(x+c)^2$$
Now apply IV: $y(x=0)=0=\frac{1}{4}c^2$ or $c=0$.
$$y=\frac{1}{4}x^2$$
A: If
$\frac{dy}{dx}
=y^a
$,
$y^{-a}dy = dx
$
or
$\frac{(y^{-a+1})'}{-a+1}
= dx
$.
Integrating,
$\frac{(y^{-a+1})}{-a+1}
= x+c
$,
so
$y 
=((-a+1)(x+c))^{1/(-a+1)}
$.
If
$a = \frac12$,
$y 
= (\frac12(x+c))^{1/(1/2)}
= (\frac12(x+c))^2
= \frac14(x+c)^2
$.
As a check,
$y'
=\frac12(x+c)
=\sqrt{y}
$.
