Inverse Operator Methods for Differential Equations

Solution 1:

The reason you can write

$$\frac{g(x)}{D-a}=\mathrm{e}^{ax}\int\mathrm{e}^{-ax}g(x)\mathrm{d}x$$

is that

$$(D-a)\mathrm{e}^{ax}=0$$

and therefore

$$(D-a)\left(\mathrm{e}^{ax}\int\mathrm{e}^{-ax}g(x)\mathrm{d}x\right)=g(x)\;.$$

You can generalize this for any operator of the form $s(x)D-t(x)$ (where in the above case $s(x)=1$ and $t(x)=a$): if you can find $h(x)$ with $(sD-t)h=0$, then the ansatz

$$ \begin{eqnarray} (sD-t)\left(h\int gn\mathrm{d}x\right)=shD\int gn\mathrm{d}x=shgn\stackrel{!}{=}g \end{eqnarray}$$

leads to

$$n=\frac{1}{sh}$$

and thus to

$$\frac{g}{sD-t}=h\int \frac{g}{sh}\mathrm{d}x\;.$$

For instance, if we generalize your case slightly to inverting $xD-a$, i.e. $s=x$ and $t=a$, then $h$ must satisfy $(xD-a)h=0$. That leads to $h=x^a$, and thus to

$$\frac{g}{xD-a}=x^a\int x^{-a-1} g\mathrm{d}x\;.$$

Solution 2:

First of all, these operators don't commute when there are nonconstant coefficients, so be careful about the orders you put them in. If $(xD - 1)(D + 2) y = g$, that says $ (D+2) y = (xD - 1)^{-1} g$, and then $y = (D+2)^{-1} (xD - 1)^{-1} g$. Now $(x D - 1)^{-1} g$ would be the solution of $(x D - 1) v = g$, namely $x \int \frac{g(x)}{x^2}\, dx$.