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I am trying to find the solutions to this differential equation:

\begin{align} \frac{d^2y}{dx^2}+a e^{-x^2}y=0\ , \end{align} where $a\in\Re$. I know that equations of the form \begin{align} \frac{d^2y}{dx^2}-\left(f(x)^2+\frac{df}{dx}\right)y=0 \end{align} have the solution \begin{align} y(x)=\exp\left(\int f(x)dx\right)\ . \end{align}

Thus, to solve my first equation, I need to solve \begin{align} f(x)^2+\frac{df}{dx}=-ae^{-x^2}\ , \end{align} which is a type of Riccati equation. I know that the homogenous part of this equation gives me a Bernoulli equation with solution

\begin{align} y(x)=\frac{1}{x+c_1} \end{align}

where $c_1$ is a constant. However, I am now stuck with finding the particular solution. I am not sure of the best method to solve for it.

I also tried solving it with both Maple and Mathematica, but they were unable to do so. I found the book, Handbook of Exact Solutions for Ordinary Differential Equations, but their equations contain exponential functions (Sections 1.2 and 2.1.3) include $e^{-x}$, rather than a Gaussian.

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  • $\begingroup$ It seems unlikely that this differential equation can be solved in closed form. $\endgroup$ – Robert Israel Jun 26 '18 at 19:54
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Hint:

Let $t=x^2$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=2x\dfrac{dy}{dt}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{dt}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{dt}\right)+2\dfrac{dy}{dt}=2x\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)\dfrac{dt}{dx}+2\dfrac{dy}{dt}=2x\dfrac{d^2y}{dt^2}2x+2\dfrac{dy}{dt}=4x^2\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}=4t\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}$

$\therefore4t\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}+ae^{-t}y=0$

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