Help with $y'' + ay'+ by= h(x)$ and finding solutions. I need help with the equation $y'' + ay'+ by= h(x)$ and how to find certain solutions to it.
My mathbook state:
That if $h(x) = constant$
we can see that $y'' + ay'+ by = c$
and that:
(a) if $b \neq 0$ then $y = \frac{c}{b}$ is a solution.
(b) $b = 0, a \neq 0$ then $y = \frac{c}{a}x$ is a solution.
(c) $b = a = 0 $then $y = \frac{c}{2}x^2$ is a solution.
Can somebody please explain to me how they arrive at these solutions and what their line of thinking is? That would make my day :)
Thank you!
 A: For your first equation, $y''+ay'+by=c$ with $b\ne 0$, we know that if $y$ is a polynomial, that it must be constant. If it had any higher degree, then there would be a non-constant term on the LHS, which does not make sense. Does it make more sense now?
A: You have $y'' + ay'+ by = c$ and the conditions are:


*

*if $b \neq 0$ then $y = \frac{c}{b}$ is a solution: Lets plug this in into our second order differential equation. We get:


$$(\frac{c}{b})'' + a(\frac{c}{b})'+ b(\frac{c}{b}) = c$$ 


*

*Since $\frac{c}{b}$ is just a constant you get $c = c$ which is a solution. 


Now do that same method with your other two solutions. 
Let me know if you still do not understand it. 
A: I think the study of differential equations is notorious (at least at an undergraduate level) for being filled with 'divine inspiration' - ie. there are many things involved that were discovered hundreds of years ago by mathematical greats that are given and which you are not expected to derive yourself but simply to take for granted.  I too found this frustrating.
Really, what it amounts to is solution by inspection (it's unfortunate schools don't try and have students do this themselves...).  Given the equation, you think to yourself, "what could I possibly put in here to make this true?"  For some forms of differential equations (particularly linear ones), this type of reasoning can get you very far.  Consider how the derivative operator affects different functions, and try to piece together what kind of function would work.
In your example, say for b not=0, we know that, if the RHS is constant, then so must be the left.  We know that the derrivative of a constant is 0, so the y'' and y' terms vanish, so its feasible that we can find a constant k that will work as a solution, ie. y = k.  Plugging in y = k and rearranging, we find k = c/b.
For b=a=0, we think "okay, so now the second derrivative of my function y must give me a constant.'  Well, what kind of functions have second derrivatives that are constant? clearly, quadratics.  So we assume something like y = kx^2, plug in, and again solve for k.
Starting to make sense?  Try thinking about y' = ky .  What kind of function y satisfies this - ie. what kind of function has a derivative that is proportional to itself? 
This is really the type of reasoning that is used to build up much of the framework for studying the solutions to ODEs.
A: This seems to concern the particular solution for such problems.  If you assume $y$ equals some constant $k$, the first, second, and any higher derivatives are all $0$.  The solution then becomes algebra: $bk=c,k=\frac cb$.
For the second case, make the substitution $z=y'$, leaving you with $z'+az=c$.  The same logic yields that $z=y'=\frac ca$ is a solution, or $y=\frac{cx}a$.
If $a$ and $b$ are both $0$, you can just integrate twice to find the general solution immediately, the particular solution of which is $\frac{cx^2}2$.
