Using integrating factor I have the following differential equation
$$\frac{dN(t)}{dt} = A - \mu N(t)$$
I understand that I will need to use an integrating factor but am not sure how to proceed. I think I should use $e^{-\mu t}$ as the integrating factor but am not entirely sure.
I believe I should end up with 
$$N(t)=N_{0}e^{-\mu t}+\frac{A}{\mu}(1-e^{-\mu t})$$ where $N(0)=N_{0}$ is the initial condition.
Any help would be much appreciated!
 A: You have $N'(t)+\mu N(t)=A$. Multiply by $e^{\mu t}$ to get $$(N(t)e^{\mu t})'=Ae^{\mu t}$$
A: Hint:
$$
\dfrac{dN}{dt} + \mu N = A \quad \Rightarrow \quad e^{\mu t}\dfrac{dN}{dt} + \mu e^{\mu t} N = A e^{\mu t} \quad (1)
$$
But, 
$$
\dfrac{d}{dt}(e^{\mu t}N(t)) =  e^{\mu t}\dfrac{dN}{dt} + \mu e^{\mu t} N \quad (2)
$$
Replacing, (2) in (1), 
$$
\dfrac{d}{dt}(e^{\mu t}N(t)) = A e^{\mu t} \quad \Rightarrow \quad d(e^{\mu t}N(t)) = A e^{\mu t}dt
$$
A: There are at least two ways you could solve this equation.  Looking at
$$
N'(t)+\mu N(t)
$$
you can multiply both sides my something:
$$
\begin{align}
& N'(t)M(t) + \mu M(t)N(t) \\[8pt]
= {} & N'(t)M(t) + M'(t) N(t) \\[8pt]
= {} & \Big(N(t) M(t)\Big)'
\end{align}
$$
and then since $\Big(N(t) M(t)\Big)'= A$, you have $N(t)M(t) = At+C$ and then $N(t)= \dfrac{At+C}{M(t)}$.
This works only if $\mu M(t)=M'(t)$, and that is a differential equation solved by $M(t) = e^{\mu t}$.  That is your integrating factor.
Another way is separation of variables:
$$
\frac{dN}{A-\mu N} = dt.
$$
Integrating both sides, you get
$$
\frac{-1}{\mu} \log|A-\mu N| = t+C.
$$
Hence
$$
\log|A-\mu N| = -\mu t-\mu C = -\mu t+ C_1
$$
$$
|A-\mu N| = e^{-\mu t} e^{C_1} = \left( e^{-\mu t}\cdot\text{positive constant} \right)
$$
$$
A-\mu N = e^{-\mu t}\cdot\text{consant},
$$
etc.
A: Why not just separate the variables?
We have $$\frac{dN}{dt}=A-\mu N \iff\int\frac{1}{A-\mu N}dN=\int dt$$
$$\iff \int_{N_0}^{N}\frac{1}{A-\mu \nu}d\nu=\int_0^td\tau$$
$$\iff-\frac{1}{\mu}\bigg[\ln|A-\mu\nu|\bigg]_{N_0}^{N}=\bigg[\tau\bigg]_0^t$$
$$\iff-\frac{1}{\mu} \ln\left| \frac{A-\mu N}{A-\mu N_0}\right|=t \iff \frac{A-\mu N}{A-\mu N_0}=e^{-\mu t}$$
$$\iff \mu N=A-(A-\mu N_0)e^{-\mu t}$$
$$\iff \boxed{N(t)=\frac{A-(A-\mu N_0 ) e^{-\mu t}}{\mu}}$$
