Reduction of Order being done in two different ways? I was working through some questions in my math textbook and the following stuff really confused me. We have to find a second solution and a particular solution for each. Please read all the details that I'm providing.
The first question, $y'' - 4y' = 2$ with a given solution $y_1 = e^{-2x}$ when solved through reduction of order, my textbook eventually gets to a point where it's $u'' - 4u' = 0$ I understand how to get here, but what I don't get is why it's set to 0? 
Because for the other question, $y'' - 3y' + 2y = 5e^{3x}$ with a given solution of $y_1 = e^x$ this one eventually gets to a point where it's labelled as $w' - w = 5e^{2x}$
Why is the second one set to $5e^{2x}$ but the first one is set to $0$? How do I know when to set these to $0$ and when not to? 
 A: Note: There is a typo somewhere in the first problem. This is very clear given that the solution they give is not a solution to the homogeneous DEQ.
I am going to write out the details for the second question. There is no guesswork in these problems.
We are asked to use Reduction of Order to solve:
$$y'' - 3y' + 2y = 5e^{3x}$$
We can find a solution to the homogeneous equation $y'' - 3y' + 2y = 0$, as:
$$y_h(x) = c_1 e^x + c_2 e^{2x}$$
We are free to choose either of those as our solution, so lets choose as the author did, $y_1 = e^x$.
Using Reduction of Order, we have:
$$y = y_1 u = e^x u \rightarrow y' = e^x u + e^x u' \rightarrow y'' = e^x u + 2 e^x u' + e^x u''$$
Substituting these into the original equation yields:
$$y'' - 3y'+2y = e^x u'' -e^x u' = 5e^{3x}$$
Dividing by $e^x$ yields:
$$u''-u' = 5 e^{2x}$$
Now, we make the substitution $w = u'$, yielding (exactly what the author has):
$$w' - w = 5e^{2x}$$
Solving for $w(x)$ yields:
$$w(x) = c_1 e^x+5 e^{2x}$$
We know $u' = w$, hence
$$u' = c_1e^x+5 e^{2x} \rightarrow u(x) = c_1 + c_2 e^x + \dfrac{5}{2}e^{2x}$$
Lastly, we know that $y = y_1 u$, hence
$$y(x) = c_1e^x + c_2e^{2x} + \dfrac{5}{2}e^{3x}$$
If you use something like Undetermined Coefficients, you will see it matches this result exactly.
Because of the error in the first problem, I cannot comment on it, but there is definitely a typo and that is throwing results off (I may be able to reverse engineer something, but it is only guesswork). Maybe you can state which book this is from and we can see if there are multiple versions.
