How to integrate a differential equation which has the same variable on both sides of the equality For example, in a Physics-related question, we are given that $v=\sqrt{x}$. I simplified this to $\frac{dx} {dt}=\sqrt x,$ but I am unable to proceed further. 
This was also the case with the following question:
A particle moves with initial velocity $v_0$, and retardation $\beta v$, where v is the velocity at any time $t$, and $\beta$ is a positive constant.
I simplified this down to $\frac{d^2x}{dt^2}=-\beta\frac{dx}{dt}$. Again, I was unable to proceed further.I am aware that we can 'split' the differential term, and treat it as a fraction, but I don't know what to do after that.
Can someone kindly guide me on how to solve questions like this, where the differential term is on both sides of the equality?
Thanks in Advance.
 A: The method of separating variables is one of the fundamental methods of solving differential equations you learn in a first course (see also the AP Calculus syllabus). The method is clear: move variables of the same type to their own side, integrate, and do the rest. Here's a fundamental example:
$$ \frac{dx}{dt} = kx $$
Let us treat $dx$ and $dt$ like usual numbers. Rearranging gives
$$ \frac{dx}{x} = k \ dt $$
Okay, let's integrate.
$$ \int \frac{dx}{x} = \int k \ dt $$
Now we obtain
$$ \ln |x| = kt + C $$
to which we can grab
$$ x = C_0 e^{kt}. $$
Not every differential equation can be solved like this, but if you can find a way to get variables on their own side, you can employ this technique.
Let us now consider the second equation you posed: $x'' = -\beta x'$. Let $u = x'$, so $u' = x''$. This reduces to
$$ u' = -\beta u $$
Which we know how to solve! The solution?
$$ u = Ce^{-\beta t} $$
Now we need to solve
$$ \frac{dx}{dt} = Ce^{-\beta t} $$
This is another easy one. Let's separate variables and get
$$ dx = Ce^{-\beta t} \ dt $$
Now, integrate, and yield
$$ x = -\frac{C}{-\beta} e^{-\beta t} + D $$
as the most general form of the solution. Getting initial conditions will find particular values of $C$ and $D$.
