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.

To deal well with substitution of variables in differential equations you should know when there is a differential operator and when there is a derivative. In the first case a variable is linked with differential operator and changing it you should also change a differential operator, in the second a derivative is just a function so you can simultaneously change the same argument in all equation's functions.
Let's see on your equation.
$$\frac{dQ}{dt}=\frac{a+Q}{b}$$
And rewrite it using a differential operator notation:
$$\frac{d}{dt}[Q(t)]=\frac{a+Q(t)}{b}$$
Let's define a functional map $F_1(f)(x)=f(x)-a$, obvious this map is bijection and let $g$ be a solution of $F_1(g)=Q$. So $Q(t)=F_1(g)(t)=g(t)-a$ and we rewrite the equation:
$$\frac{d}{dt}[g(t)-a]=\frac{a+g(t)-a}{b}$$
Reduce it and calculate a derivative:
$$g'(t)=\frac{1}{b}g(t)$$
And back to a differential operator notation:
$$\frac{d}{dt}[g(t)]=\frac{1}{b}g(t)$$
Repeat the trick with a map $F_2(f)(x)=f(\frac{1}{b}x)$, let $w$ be a solution of $F_2(w)=g$, so $g(t)=F_2(w)(t)=w(\frac{1}{b}t)$ and substitute it into the equation:
$$\frac{d}{dt}[w(\frac{1}{b}t)]=\frac{1}{b}w(\frac{1}{b}t)$$
Calculate a derivative and reduce:
$$w'(\frac{1}{b}t)=w(\frac{1}{b}t)$$
But there is no a differential operator, so we change an argument $\frac{1}{b}t$ to $t$ in whole equation:
$$w'(t)=w(t)$$
It's known that a solution is $w(t)=C_1e^t$ and $Q=F_1(F_2(w))$ so $Q(t)=C_1e^{\frac{1}{b}t}-a$
This is a very verbose solution, but then understand it you can just use rules of substitution of variables and solve an equation in a few lines.
