TITLE: How to solve the differential equation: $y'=\sqrt{|y|}$
QUESTION [5 upvotes]: While trying to solve the differential equation: $y'=\sqrt{|y|}$. I got confused how to deal with the absolute value. I want to draw a sketch for the direction field for that equation, and see for what initial values does this equation fulfill the conditions of the existence and uniqueness Theorem.
I tried to do the integral according to the sign, depends if $(y>0)$ or $(y<0)$. but I'm still not sure if my result is right. 
result that I got:

while $y>0$ : $y=({\frac{x}{2}+c})^{2}$
while $y<0$ : $y=({-\frac{x}{2}+c})^{2}$

if that's the case, Its hard for me to imagine the Direction field by these two equations, because they are always above axis: $x$. How does it look? any hints?

REPLY [2 votes]: For the $y<0$ case your DE is evaluated incorrectly:
$$\frac{dy}{dx}=\sqrt{-y}$$
$$(-y)^{-\frac{1}{2}}\frac{dy}{dx}=1$$
$$-2\sqrt{-y}=x+c$$
$$-4y=(x+c)^2$$
$$y=-\frac{(x+c)^2}{4}$$
This should hopefully help you to visualize the direction field now. Something like this: