I am trying to solve a differential equation that has cumulative normal distributions on the non-homogenous part:

$$ay'' + by' + cy = \gamma e^{mx} \operatorname{erf}(\alpha x + \beta)$$

There are many more terms on the RHS, all of which are functions of erf. The Laplace Transform is easy, but the inverse is not. Matlab is not giving me an answer. Is there something I am missing to study? What can help me deal with these type of "hard" laplace/inverse laplace problems?

  • $\begingroup$ Maple 2019.2 can solve this equation and give general solution. $\endgroup$ – Mariusz Iwaniuk Dec 19 '19 at 15:07

In your other question I told you to look at the bilateral Laplace transform of $\text{erf}$ which has a nice closed-form :

for $s \in \mathbb{R}$ and by analytic continuation for $s \in \mathbb{C}$, $\ \ \int_{-\infty}^\infty e^{-x^2/2} e^{-sx}dx = e^{s^2/2}\int_{-\infty}^\infty e^{-(x+s)^2/2} dx= e^{s^2/2}\int_{-\infty}^\infty e^{-x^2/2} dx=e^{s^2/2} \sqrt{2\pi}$. Therefore for $\Re(s) > 0$, $\ \ \int_{-\infty}^\infty (\int_{-\infty}^x e^{-y^2/2}dy) e^{-sx}dx = -\int_{-\infty}^\infty e^{-x^2/2} \frac{e^{-sx}}{-s}dx = \frac{1}{s}e^{s^2/2} \sqrt{2\pi}$

For your ODE, applying partial fraction decomposition to $\frac{1}{s(as^2-bs+c)}$ you'll need to find the inverse Laplace transform of $\frac{1}{s-\alpha}e^{s^2/2}$, expressed in term of $\text{erf}$

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  • $\begingroup$ Thank you. What are a few advantages and disadvantages of the bilateral Laplace? I have not used it because there is very little material online about how to use it for solving odes $\endgroup$ – dleal Sep 11 '17 at 13:28
  • $\begingroup$ There is something about " not respecting causality" for the bilateral Laplace $\endgroup$ – dleal Sep 11 '17 at 13:31
  • $\begingroup$ math.stackexchange.com/questions/1719754/… $\endgroup$ – dleal Sep 11 '17 at 14:47

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