$5y''+6y'=x^2+5x +3$ What is the $ y_p$ here? How to derive $y_p$? $5y''+6y'=x^2+5x +3$ What is the $ y_p$ here?
So I tried the usual. when the right hand side is polynomial degree of 2, my $ y_p = Ax^2 + Bx + C$. I then differentiate once and twice and sub them back in. But this time it doesn't work. I think it is because the left hand side is missing a constant term.
So for example in general,
$ay'' + by'= \sum _{i=0}^n\:\beta _i\:x^i$
Can someone help out? I don't want the answer. I just want to know how do you get the functional form of $y_p$? How do you derive it?
Can someone prove it (ie what y_p) is  or link a proof to me? Also, how do you mathematically express "some polynomial"?
 A: This is how we know that the particular solution is a polynomial of degree $3$. Since the right hand side is polynomial of degree $2$, differentiate thrice to obtain the following equation:
$$
\begin{align}
0&=5\frac{d^{5}y}{dx^{5}}+6\frac{d^{4}y}{dx^{4}}\\
\\
&=5\left(\frac{d^{5}y}{dx^{5}}+\frac{6}{5}\frac{d^{4}y}{dx^{4}}\right)\\
\\
&= 5\left(e^{\frac{6}{5}x}\frac{d^{5}y}{dx^{5}}+\frac{6}{5}e^{\frac{6}{5}x}\frac{d^{4}y}{dx^{4}}\right)\\
\\
&=5\frac{d}{dx}\left(e^{\frac{6}{5}x}\frac{d^{4}y}{dx^{4}}\right)
\end{align}
$$
This is equivalent to:
$$
\begin{align}
\frac{d^{4}y}{dx^{4}}&=Ce^{-\frac{6}{5}x}\\
\\
y&=C_{1}e^{-\frac{6}{5}x}+C_{2}+C_{3}x+C_{4}x^{2}+C_{5}x^{3}
\end{align}
$$
If we substitute this back to the original equation, notice that any constant $C_{1},C_{2}$ will result in the equation evaluating to zero, that is why we call the first two terms homogenous solution. However, there are only a particular $C_{3},C_{4},C_{5}$ that will result in the right hand side, that is why we call the last three term particular solution.
A: I hope this post develops an intuition for you for how things are working here. The procedure is not quite rigorous, but I think having a heuristic proof is always better before moving to a rigourous one.
In your case, if the degree of $y$ is $n$, then the RHS is a degree-$2$ polynomial, and the LHS is degree $n-1$ polynomial. Clearly, $n=3$ is the only way this can happen.
Consider a general differential equation
$$ay^{(m)}+by^{(m-1)}+\cdots=P(x)$$
where $y^{(k)}$ denotes the $k$th derivative of $y$, and $P(x)$ is some polynomial of degree $p$.
Note that $y$ has to be a polynomial. (Think why!) Now, if the degree of $y$ is $n$, the LHS is a degree $n-m$ polynomial and the RHS is a degree $p$ polynomial. You can write it mathematically using the Big-O Notation as
$$\mathcal O(x^{n-m})=\mathcal O(x^p)$$
And for these two polynomials to be equal, we must have $n-m=p$ i.e. the function $y$ is a degree $(m+p)$ polynomial. Then you can simply proceed by doing the usual - taking the general form of a polynomial and proceeding. There might be simpler ways depending upon the question though.
Hope this helps. Ask anything if not clear :)
A: It is easy to see that the equation in its original form is integrable once to give
$$
5y'+6y=\frac13x^3+\frac52x+3x,
$$
leaving the integration constant at zero. Now the method of undetermined coefficients tells us that a particular solution can be found that is a cubic polynomial.
