How do I find a function from a differential equation? Hey, I'm looking for a guide on how I find Q given the following, where $a$ and $b$ are constants:
\begin{equation}
\frac{dQ}{dt} = \frac{a + Q}{b}
\end{equation}
I have the answer and working for specific case I'm trying to solve but do not understand the steps involved. A guide on how I can solve this, with an explanation of each step would be much appreciated.

Yet, another method for solving this differential equation is to look at it as a linear differential equation, whose general form is:
$$
y'(x) = a(x) y(x) + b(x) \ , \qquad\qquad\qquad [1]
$$
where $a(x), b(x)$ are arbitrary functions depending on the variable $x$. In your case:
$$
x = t \ , y(x) = Q(t) \ , b(x) = \frac{a}{b} \quad \text{and}\quad a(x) = \frac{1}{b} \ .
$$
A general procedure for solving [1] is the following:
1. First, try to solve the associated homogeneous linear differential equation
$$
y' = a(x)y \ .  \qquad\qquad\qquad    [2]
$$
This is easy: the general solution is
$$
y = K e^{A(x)} \ ,   \qquad\qquad\qquad   [3]
$$
where $K\in \mathbb{R}$ is an arbitrary constant and $A(x) = \int a(x)dx$ is a primitive function of $a(x)$.
2. Once you have the general solution [3] of [2], you apply variation of constants; that is, you look for solutions of [1] of the following kind:
$$
y = K(x) e^{A(x)} \ .   \qquad \qquad \qquad  [4]
$$
Here, we have replaced the arbitrary constant $K$ by an arbitrary unkown function $K(x)$ (hence the name "variation of constants") to be determined. How? Imposing that we want [4] to be a solution of our first differential equation [1]. It goes like this: if you replace $y$ in [4] into [1], you get
$$
K'(x) e^{A(x)} + K(x) A'(x) e^{A(x)} = a(x) K(x) e^{A(x)} + b(x) \ .
$$
Since $A'(x) = a(x) $, this is the same as
$$
K'(x) e^{A(x)} = b(x) \ .
$$
So
$$
K(x) = \int b(x)e^{-A(x)}dx + C \ ,
$$
where $C \in \mathbb{R}$ is an arbitrary constant. Now you put this $K(x)$ into [4] and get the general solution of your differential equation:
$$
y(x) = Ce^{A(x)} + e^{A(x)}\int b(x) e^{-A(x)}dx \ . \qquad\qquad\qquad   [5]
$$
Since I've never could remember formula [5], I use to repeat the whole process for each particular linear differential equation, which is not hard and you can do it for yours.
