Solving for the integrating factor in a Linear Equation with Variable Coefficients So I am studying Diff Eq and I'm looking through the following example.
Solve the following equation:
$(dy/dt)+2y=3 \rightarrow μ(t)*(dy/dt)+2*μ(t)*y=3*μ(t) \rightarrow (dμ(t)/dt)=2*μ(t) \rightarrow (dμ(t)/dt)/μ(t)=2 \rightarrow 
(d/dt)\ln|μ(t)|=2 \rightarrow \ln|μ(t)|=2*t+C \rightarrow μ(t)=c*e^{2*t} \rightarrow μ(t)=e^{2*t}$
So I have two questions regarding this solved problem. It appears that the absolute value sign is just tossed out of the problem without saying that as a result $c \ge 0$, is this not correct and if not why?
Secondly and more importantly, I was confused by the assumption that $c=1$. Why should it be $1$ and would the answer differ if another number were selected is it just an arbitrary selection that doesn't influence the end result and just cancels out anyways?

Method 1: Calculus
We have: $y' + 2y = 3$.
Lets use calculus to solve this and see why these statements are okay. We have:
$$\displaystyle \frac{\dfrac{dy}{dt}}{y - \dfrac{3}{2}} = -2$$
Integrating both sides yields:
$$\displaystyle \int \frac{dy}{\left(y - \dfrac{3}{2}\right)} = -2 \int dt$$
We get: $\ln\left|y - \dfrac{3}{2}\right| = -2t + c$.
Lets take the exponential of both sides, we get:
$$\left|y - \dfrac{3}{2}\right| = e^{-2t + c} = e^{c}e^{-2t} = c e^{-2t}$$
Do you see what happened to the constant now? 
Now, lets use the definition of absolute value and see why it does not matter.
For $y \ge \dfrac{3}{2}$, we have:
$$\left(y - \dfrac{3}{2}\right) = c e^{-2t} \rightarrow y = c e^{-2t} +\dfrac{3}{2}$$
For $y \lt \dfrac{3}{2}$, we have:
$$-\left(y - \dfrac{3}{2}\right) = c e^{-2t} \rightarrow y = -c e^{-2t} + \dfrac{3}{2}$$
However, we know that $c$ is an arbitrary constant, so we can rewrite this as:
$$y = c e^{-2t} + \dfrac{3}{2}$$
We could also leave it as $-c$ if we choose, but it is dangerous to keep those pesky negatives around.
Note: look at this graph of $\left|y - \dfrac{3}{2}\right|$:

Now, can you use this approach and see why it is identical to the integrating factor (it is exactly the same reasoning)?
For your second question:
You could make $c$ be anything you want. Let it be $y = ke^{-2t} + \dfrac{3}{2}$. Take the derivative and substitute back into ODE and see if you get $3 = 3$ (you do). If they gave you initial conditions, then it would be a specific value, so the authors are being a little sloppy. They should have said something like $y(0) = \dfrac{5}{2}$, which would lead to $c = 1$.
Lets work this statement:
$y = ke^{-2t} + \dfrac{3}{2}$
$y' = -2 k e^{-2t}$
Substituting back into the original DEQ, yields:
$y' + 2y = -2 k e^{-2t} + 2(ke^{-2t} + \dfrac{3}{2}) = 3$, and $3 = 3$.
What if we let $c = 1$, we have:
$y' + 2y = -2 e^{-2t} + 2(e^{-2t} + \dfrac{3}{2}) = 3$, and $3 = 3$.
So, you see that we can let $c$ be anything, unless given an IC.
Method 2: Integrating Factor
Here is a step-by-step solution using integrating factor.


*

*$y' + 2 y = 3$

*$\mu y' + 2 \mu y = 3 \mu$

*$\dfrac{d}{dt}(\mu y) = uy' + u' y$

*Choose $\mu$ so that $\mu' = 2 \mu \rightarrow \mu = e^{2t}$

*We have: $y'+2y = 3$, so:

*$e^{2t}y' + 2e^{2t}y = 3e^{2t}$

*$\dfrac{d}{dt}(e^{2t}y) = 3 e^{2t}$

*$e^{2t} y = \dfrac{3}{2}e^{2t} + c$, thus

*$y(t) = \dfrac{3}{2} + c e^{-2t} = c e^{-2t}+ \dfrac{3}{2}$

