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Short (well...) version of the question

I am having a bit of a problem understanding one thing about the solution to non-homgeneous second-degree linear differential equations using the constant coefficients method.

Given a DE of the form: $$ a_2y''+a_1y'+a_0y=f(x) $$ where $f(x)$ is some polynomial of degree $n$ in $x$, the solution that can be tried is $$ y_p=x^m A_n(x) $$ where $y_p$ is the particular solution, $A_n$ an $n$-degree polynomial in $x$ and $m$ is 0, 1 or 2. It's the choice of $m$ I don't get. My book says that

$m$ is the smallest of the integers 0, 1 and 2, that ensures that no term of $y_p$ is a solution of the corresponding homogeneous equation $$ a_2y''+a_1y'+a_0y=0.$$

What does this mean? How do I choose $m$? How do I check this? What is the intuition?

I'm guessing there is a (somewhat) easy and (hopefully) intuitive explanation, but I haven't found one.

An example

As an example, consider the DE $$ y''+y'=2x $$ where the homogeneous solution is $$ y_h=c_1e^{-x}+c_2. $$

For the particular solution, I need to use $m=1$ it seems. That is, $$ y_p=x(b_0+b_1x). $$ which yields $$ y_p=x^2-2x. $$ However, my way of thinking is that I would rewrite the DE as $$ y'+y=x^2+C $$ giving $$ \begin{align} y_p&=b_0+b_1x+b_2x^2\\ y_p'&=b_1+2b_2x^2\\ &\begin{cases}b_0&=2\\b_1&=-2\\b_2&=1 \end{cases} \end{align} $$ and for this to be the same as the solution above, I'd like $b_0$ to be 0. But it isn't. Is my thinking here wrong? Is this constant sort of accounted for in the general solution because of the $c_2$ constant? Or what is the motivation for multiplying the polynomial in the solution by $x$ here (i.e. why is $m=1$)?

Sorry for the wall of text... Thanks for reading.

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up vote 1 down vote accepted

Multiplication by $x$ or $x^2$ is necessary when your trial solution is already a solution of the homogenous system.

If '$y_P(x)$' is a solution of the homogenous system then it can't output $f(x)$ when you substitute it in because it will produce zero.

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Are there any restrictions such that the coefficients in the trial solution cannot all be 0? In my example, $y_p=b_0+b_1x$ is inadmissible it seems, and is this then because if $b_0$ is anything and $b_1$ is 0, this is a solution of the homogeneous equation. So then we need to use $y_p=x(b_0+b_1x)$, which gives $y''+y'=b_1+2b_1x+b_0$ which is a solution to the homogeneous equation only if $b_0=b_1=0$. Is this correct? – hejseb Nov 28 '13 at 9:27
The problem with $y_p=b_0+b_1x$ is that when operate on it with differential operator $D^2+D$ you just get zero; i.e. it is a solution of the homogenous system. Now try $x\cdot y(x)=b_1x^2+b_0x$ and you get $y''+y'=2b_1+2b_1x+b_0$ and you want this to equal $2x$... – Jp McCarthy Nov 28 '13 at 9:32

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