Cauchy Euler ODE $2x^2 y''-6xy'+6y=x^3$ with resonance Question:
$$2x^{2}y''-6xy'+6y=x^{3} $$
I have tried solving it, and obtained the roots 3 and 1. Apparently this is a case of resonance. Even so, I wasn't able to get the particular solution through variation of parameters.
Perhaps there are other methods to solve this?
 A: The general solution is the sum of the complementary solution and a particular solution.
First, we calculate the complementary solution.
Assume a solution to this Euler-Cauchy equation will be proportional to $x^{\lambda}$ for some constant $\lambda$.
Substitute $y(x)=x^{\lambda}$ and its derivatives into:
$$2x^{2}y^{''}-6xy^{'}+6y=0 $$
You get, for $x\neq 0$:
$$
\begin{align}
(2\lambda^2-8\lambda+6)x^{\lambda} &= 0 \\
2\lambda^2-8\lambda+6 &= 0 \\
2(\lambda-3)(\lambda-1)&=0
\end{align}
$$
So, $$\lambda=\{1;3\}$$
Then, you have the complementary solution with $c_1$ and $c_2$ are constants:
$$\begin{align}
y_c(x) &= c_1y_1(x)+c_2y_2(x)\\
&=c_1x+c_2x^3
\end{align}$$
Now, let's compute the particular solution that will be given by:
$$y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$$
where $W(x)$ is the Wronskian of $y_1(x)$ and $y_2(x)$, and:$$u_1(x)=-\int\dfrac{f(x)y_1(x)}{W(x)}dx, \\u_2(x)=\int\dfrac{f(x)y_2(x)}{W(x)}dx$$ 
Compute the Wronskian:
$$\begin{align}
W(x)&=\begin{vmatrix}x & x^3\\1 & 3x^2\end{vmatrix}\\
&=2x^3
\end{align}$$
From:
$$y^{''}+b(x)y^{'}+c(x)y=f(x),$$
let: $$f(x)=\dfrac{x}{2}.$$ 
Thus, 
$$u_1(x)=-\int\dfrac{x}{4}dx=-\dfrac{x^2}{8}$$
$$u_2(x)=\int\dfrac{1}{4x}dx=\dfrac{ln(x)}{4}$$
Therefore, 
$$y_p(x)=\dfrac{1}{8}x^3(2ln(x)-1)$$
We conclude, with the general solution:
$$\begin{align}
y(x)&=y_c(x)+y_p(x)\\
y(x)&=c_1x+c_2x^3+\dfrac{1}{8}x^3(2ln(x)-1)\\
&=\dfrac{1}{4}x^3ln(x)+c_1x+c_2x^3
\end{align}
$$
A: Variation of parameters is a valid method to solve this. Undetermined coefficients will also work here.
Let $x = e^t$, then
$$ y''(t) - 4y'(t) + 3 = \frac12 e^{3t} $$
The characteristic roots are still $1$ and $3$, so we can make an ansatz of $y_p(t) = Ate^{3t}$. Converting this back to $x$ gives
$$ y_p(x) = Ax^3\ln x $$
You can plug in this solution form to find the constant
Conclusion: When resonant occurs in a non-homogeneous Cauchy-Euler, add in a power of $\ln x$ to your ansatz.
