solve the differential equation $\dot{x} + a\cdot x - b\cdot \sqrt{x} = 0$ Problem
I want to know how to solve the differential equation
$$ \dot{x} + a\cdot x - b\cdot \sqrt{x} = 0 $$ for $a>0$ and both situations: for $b > 0$ and $b < 0$. 
My work
One can separate the variables to obtain:
$$ \frac{dx}{b\cdot \sqrt{x} - a\cdot x} = dt$$ but I do not know how to proceed ...
https://www.wolframalpha.com/input/?i=solve+x%27(t)%2Bax(t)-bsqrt(x(t))+%3D+0 it seems to have an explicit solution ... 
Context
This problem occurs in the following context:
$$ \ddot{X} + a \cdot \dot{X} = f(X)$$ then multiplying both sides with $2\dot{X}^T$ one obtains:
$$ (\dot{X}^T\dot{X})' + 2a\cdot \dot{X}^T\dot{X} = 2\dot{X}^T f(X)$$ Let $v= \dot{X}^T \dot{X}$ and the above differential equation arises ... 
 A: The next step is to integrate,
$$t+c=\int\frac{dx}{b\sqrt x-ax}=\int\frac{d\sqrt x^2}{b\sqrt x-a\sqrt x^2}=\int\frac{2\,d\sqrt x}{b-a\sqrt x}=-\frac2a\log\left(\frac ba-\sqrt x\right).$$
From this you can draw $x$,
$$x=\left(\frac ba-e^{-a(t+c)/2}\right)^2.$$
A: Some caution has to be taken here because of the presence of the $\sqrt{}$ function--we need to keep track of signs carefully. We know that $x(t) \ge 0$ everywhere, so there is a function $u$ such that $u \ge 0$ and $x = u^2$. Additionally, we can scale $u$ and $t$ to eliminate the constants. The proper choice for this is $x(t) = (b/a)^2 u(a t/2)^2$, giving
$$
u\, [u'+u - \mathrm{sgn}(b)] = 0
$$
which has two solutions: $u(s)= 0$ and $u(s) = \mathrm{sgn}(b)+Ce^{-s}$. For this second solution, $x(t) = (b/a)^2 [\mathrm{sgn}(b)+Ce^{-s}]^2$. Factoring $|b|/a$ into the brackets gives
$$
x(t) = \left(\frac{b}{a} + Ce^{-at/2}\right)^2 \;\;\;\;\mathrm{or}\;\;\; x(t) = 0
$$
Except, there's one little problem here. We required $u\ge 0$, but that term in parentheses, which has the same sign as $u$, could be negative. The key is to note that for both solutions, when $x(t) = 0$, $x'(t)$ is also $0$. This allows them to be spliced together at that point and still be continuously differentiable. Thus, the general solution is
$$
x(t) = \left[\max \left(\frac{b}{a} + Ce^{-at/2},0\right)\right]^2
$$
for an arbitrary real constant $C$.
A: One way of working this out is to make the substitution $y = \sqrt{x}$.  Then,
$$\frac{dx}{b\sqrt{x}-ax} \rightarrow \frac{2ydy}{by-ay^2} = \frac{2ydy}{y(b-ay)}.$$
You can treat the integral in $y$ with partial fractions.
