$y'' + y' = \frac 43x - 1$ I was able to get the solution to $y'' + y' = \frac 43x - 1$ by using variation of parameters. I got $y = \frac 23 x^2 - \frac 73 x+c_1+c_2e^{-x}.$ But I wanted to know how to solve this using undetermined coefficients since that method is easier.
The characteristic polynomial is $r^2+r=r(r+1)=0$ which has roots $r=0,-1$ so $y_c = c_1+c_2 e^{-x}.$ What I learned is that if the "forcing function" ($\frac 43x -1$ in this problem) shares a term that is $x^k$ times a term in $y_c,$ where k is a nonnegative integer, then $y_p$ will contain a term that is $x^{k+1}$ times the shared term. Since $y_c$ and the forcing function have the term "1" in common, I thought this meant that $y_p$ would contain a term $x^{0+1}*1=x.$ So $y_p$ would have the form $Ax+B.$ Since B is accounted for in $y_c$, I tried to use $y_p=Ax.$
This gave me the wrong answer, which made me think I need to try a solution of the form $y_p=Ax^2+Bx.$ However, I don't know where the $x^2$ term comes from! If this is the right form, can someone explain where the quadratic term comes from? If I'm wrong, please correct me. Also, sorry if there are any typos. Thank you. :)
 A: I like to think of it like this:
If you assume that $y$ is a solution to your differential equation, and differentiate both sides of your equation, you will get
$$
y''''+y'''=((4/3)x-1)''=0.
$$
Thus, $y$ has to be a solution to the homogeneous differential equation $y''''-y'''=0$. This differential equation has characteristic equation $\lambda^4-\lambda^3=0$. You see that you have a root $\lambda=0$ of multiplicity $3$ and a root $\lambda=1$ of multiplicity $1$. Hence
$$
y(x)=(a+bx+cx^2)e^{0x}+de^{x}=a+bx+cx^2+de^x.
$$
I think you recognize the terms $a$ and $de^x$ from your $y_c$. The other parts must be your particular solution. Hence, the correct ansatz is
$$
y_p(x)=bx+cx^2.
$$
Now just insert and differentiate, to find $b$ and $c$.
A: There is no $y$-term on the LHS. Hence the term with the smallest degree of $x-$ polynomial on the LHS is $y'$. In this case you should let $y' = Ax+B$ instead. This is the same as claiming that $y$ is a quadratic polynomial in $x$.  
Then as you claimed the constant term is absorbed by the solution to the homogeneous counterpart. So we let $y=Dx^2 + Ex $
