Using integration by parts, we can show that: $$\frac{1}{D + a} f = \frac{1}{a}\left(f - \frac{1}{D + a} f'\right)$$

where $a$ is a real number, $D$ is differentiation, and $D+a$ is the corresponding linear differential operator. This formula is very helpful for computing $\frac{1}{D+a}$ applied to a polynomial, exponential, trigonometric or hyperbolic function.

For example, to solve

$$y' - 5y = x^7$$

rewrite it as $$(D-5)y = x^7$$ which is equivalent to $$y \in \frac{1}{D-5} x^7$$ which an be solved relatively painlessly using the above formula, which essentially manages your integration by parts for you and allows you to just focus on the algebra.

Question. Is there a name for this formula?

  • $\begingroup$ Do you have a link with that formula Goblin ? $\endgroup$ – Isham Oct 30 '18 at 0:37
  • 2
    $\begingroup$ I'm not sure one even needs integration to obtain it: $$ af = af+f'-f' $$ $$ af = (D+a)f-f' $$ $$ \frac{a}{D+a} f = f - \frac{1}{D+a}f' $$ $$ \frac{1}{D+a} f = \frac{1}{a} \left(f - \frac{1}{D+a}f'\right) $$ $\endgroup$ – rafa11111 Oct 30 '18 at 0:40
  • 1
    $\begingroup$ Oh that's nice @rafa11111 $\endgroup$ – Isham Oct 30 '18 at 0:41
  • 1
    $\begingroup$ @rafa11111, nice observation. $\endgroup$ – goblin Oct 30 '18 at 0:41
  • 1
    $\begingroup$ @Isham sounds interesting. Link? $\endgroup$ – goblin Oct 30 '18 at 0:41

I think this sort of thinking is contained in conventional wisdom about $D$-operator methods.

You can basically write an infinite series for $1/(D+a)$, i.e. $$ \frac{1}{D+a} = \tfrac{1}{a}(1+a^{-1}D)^{-1} = \frac{1}{a}\bigl(1 - a^{-1}D+ a^{-2}D^2 - a^{-3}D^3 + \dots \bigr). $$ Then when you hit this with a polynomial it obviously collapses to a finite sum. And you can imagine when you hit with an exponential, you can basically re-sum the resulting series to get another exponential. With this in mind, what you've written corresponds to the fact that $$ \frac{1}{1+x} = 1 - \frac{1}{1+x}x $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.