We have learned in class how to use inverse operator methods to solve ODE's (i.e. with the symbolic $D$). E.g,
If I were asked to find a particular solution, $y_p$ to $(D-1)(D-2)[y] = x^{2}e^{x}$, then I would use the formula $\frac{g(x)}{D-a} = e^{ax} \int e^{-ax} g(x) dx$, where $g(x) = x^2 e^x$.
However, here's the difficulty I'm facing; all the ODE's that we have learned in class with operator methods have been dealing with constant coefficient ODEs, what happens when you have variable coefficients?
E.g. the ODE xy'' +(2x -1)y' - 2y = x^2 factorises to $(xD-1)(D+2)[y] = x^2$, so a particular solution $y_p$ should be $y_p$ = $\frac{x^2}{(xD-1)(D+2)}$. I can use the method outlined above for the $D+2$ bit, but what about $\frac{1}{xD-1}$?
What does the inverse operator $\frac{1}{xD-1}$ mean?
Thanks, Ben