What happens when one squares an ODE in general? Suppose one has a ODE of the form $(y^{'})^{\frac{1}{2}}=f(y,x)$. What happens if we square it? Are there any pitfalls?
 A: Well one has
$$y' = f^2(y, x)\quad\Longleftrightarrow\quad (y')^{\frac{1}{2}} = |f(y, x)|$$
so the squared equation is equivalent to the first one in domains where $f(y, x)\ge 0$.
A: The only thing I can think of is that the Picard-Lindelof theorem still is true. This is the foundational tool for studying ordinary differential equations. It says that if you have an ODE of the form $$y' = f(x,y(x))$$ with some initial data, then this problem has a unique solution on a small time interval if $f$ is Lipschitz continuous in $y$.
An important catch is that even if $f(x,y)$ is Lipschitz in $y$, $f^2$ may not be, so your ODE may not have a unique solution even locally. 
A: It matters, as the solution may be dramatically different. Consider the ODE
$y'-y=0$. We know this is first order linear with solution $y(t)=ce^{t}$. Suppose we changed it to $y'-y=cos(t)$. This has solution $y(t)=ce^t+\frac{sin(t)-cos(t)}{2}$. However, suppose we square the $y'$ term: $(y')^2+y'=cos(t)$. This actually changes the ODE from linear to non-linear, this does not have a closed form solution. 
In general, squaring one or more terms of the ODE changes it from linear to non-linear, which are generally much harder to solve. Typically, we need numeric approximations to solve it. 
Note the ODE is still linear if we had $y'-y=cos^2(t)$, as it becomes non-linear when we square the function we are interested in or its unknown derivatives. 
