How to solve DE: $x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$ Find particular solution for:
$x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$
$\forall x > 0$
I believe that this problem can be solved by means of the method of undetermined coefficients. Something like this:
$x^3y_1''' + 6x^2y_1'' + 4y_1' -4y_1 = 1 - x \rightarrow y_1 = C_1 - C_2x$
$x^3y_2''' + 6x^2y_2'' + 4y_2' -4y_2 = -2x^2 \rightarrow y_2 = - \frac{2}{3}C_3x^3$
Unfortunately, I don't really know all of the steps. Can someone provide further explanation on this method of solving differential equations?
 A: Its been a while. I believe this is called an Euler equation.
Try assuming a homogeneous solution of the form $y=x^n$ and work from there.  You should end up having to solve a cubic equation for $n$ that will provide you with three homogeneous solutions.  Some linear combination will be in your final answer.
$x^3 y''' + 6x^2 y'' + 4xy' - 4y = 0$
$x^3 (x^n)''' + 6x^2 (x^n)'' + 4x(x^n)' - 4(x^n) = 0$
$x^3 n(n-1)(n-2)x^{n-3} + 6x^2 n(n-1)x^{n-2} + 4xnx^{n-1} - 4x^n = 0$
$n(n-1)(n-2)x^{n} + 6n(n-1)x^{n} + 4nx^{n} - 4x^n = 0$
$n(n-1)(n-2) + 6n(n-1) + 4n - 4 = 0$
$n^3 + 3n^2 - 4 = 0$
$(n-1)(n+2)^2 = 0$
This implies that your homogeneous solutions are $y = x^1$, $y=x^{-2}$ and, because a double root, the third solution is of the form $y=\ln(x)x^{-2}$. 
Your homogeneous solution is therefore $y(x)=Ax + Bx^{-2}+ C\ln(x)x^{-2}$.
Now you can do your magic for the particular solutions.
Then evaluate the constants with the IVP, if you have one.
A: After change $x=e^t$ we get equation with constant coefficients:
$$\frac{{{d}^{3}}}{d {{t}^{3}}} y+3 \left( \frac{{{d}^{2}}}{d {{t}^{2}}} y\right) -4 y=1-e^t-2e^{2t}$$
Solution this equation is
$$y=c_1e^t+c_2e^{-2t}+c_3te^{-2t}-\frac14-\frac{te^t}{9}-\frac{e^{2t}}{8}$$
After inverse change $t=\ln x$ we have final answer
$$y=c_1x+c_2x^{-2}+c_3x^{-2}\ln x-\frac14-\frac{x\ln x}{9}-\frac{x^2}{8}$$
