# Inverse Operator Methods for Differential Equations

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

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I believe it should be $D-a$ in the denominator in the second paragraph? –  joriki Mar 31 '11 at 4:16
Yeah, that's right thanks. –  user38268 Mar 31 '11 at 13:04

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

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

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

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

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

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@RobertIsrael Hi Guys thanks for all your answers it really helps a lot. But one thing I don't understand. Say I wanted to solve the ODE $(D+2)[y]=g(x)$. Hence a particular solution would be $y_p = \frac{1}{D+2}[g(x)]$. Now from here if I were to expand $\frac{1}{D+2}$ using a series I would get one answer, but if I were to say that $\frac{1}{D+2}[g(x)] = e^{-2x}\int e^{2x} g(x) dx$, I would get a different answer, namely because of the $e^{-2x}$ out in front. Could anyone explain what's happening? Thanks. –  user38268 Apr 1 '11 at 7:25
@David: I don't see why you get a different answer. I believe you can get the same series that you get from expanding $(D+2)^{-1}$ by starting from the integral solution and repeatedly integrating by parts (integrating $\mathrm{e}^{2x}$ and differentiating $g$). –  joriki Apr 1 '11 at 8:39

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 $x \int \frac{g(x)}{x^2}\, dx$.

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Hi Guys thanks for all your answers it really helps a lot. But one thing I don't understand. Say I wanted to solve the ODE $(D+2)[y]=g(x)$. Hence a particular solution would be $y_p = \frac{g(x)}{D+2}$. Now from here if I were to expand $\frac{1}{D+2}$ using a series I would get one answer, but if I were to say that $\frac{1}{D+2} = e^{-2x}\int e^x g(x) dx$, I would get a different answer, namely because of the $e^(-2x)$ out in front. Could anyone explain what's happening? Thanks. –  user38268 Apr 1 '11 at 7:19

$(D+2)[y]=g(x)$

$y_p=g(x)/(D+2)$

$1/(D+2)=e−2x\int e2xg(x)dx$

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This symbol soup is not a useful answer to questions of the form "what happens when ...?", "what about ...?" "what does ... mean?" The question asks (asked! more than a year ago) for insight, so an appropriate answer should contain significantly more words than symbols. –  Henning Makholm Sep 13 '12 at 13:00