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I would like to find an analytic solution (if possible) of the differential equation:

$g = c_1 f^2 + c_2 (f')^2$

Where $c_1$ and $c_2$ are constants, $g$ is a known function of $x$, $f$ is another function of $x$ which I'm trying to find, and $f'$ is the derivative of $f$ with respect to $x$. This equation came up when trying to find a sound pressure field related to $f$ such that one gets a required sound radiation force which is related to $g$.

If a general solution is not possible, what about the case of $g = c_3 x^2$?

EDIT:

As suggested by @MichaelEngelhardt, for $g = c_3 x^2$ being a potential function, one can add a constant to it and change the function to: $g = c_3 x^2 + c_3 c_2 / c_1$. In that case one solution to the ODE is $f = \sqrt{c_3/c_1} x$

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  • $\begingroup$ note that you can change $c_2$ using $f(\lambda x)$ and make it equal to $c_1$; so you can absorb both into $g$ and assume w.l.o.g. $c_1=c_2=1$ $\endgroup$
    – Pietro Majer
    Commented Feb 1, 2023 at 17:55
  • $\begingroup$ Look for weak solutions with convex integration. $\endgroup$
    – Funktorality
    Commented Feb 2, 2023 at 2:13
  • $\begingroup$ For $g=\lambda x^n$ this can be transformed to an Abel equation. Taking $f'=\lambda^{1/2}x^{n/2}\cos(\xi)$, $f=\lambda^{1/2}x^{n/2}\sin(\xi)$, and $x-n\tan(\xi)/2=u$ we have the equation \begin{align} uu'_\xi+\left(\frac n2\sec(\xi)^2-1\right)u=\frac{n}{2}\tan(\xi). \end{align} $\endgroup$
    – Eli Bartlett
    Commented Feb 5, 2023 at 18:43

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This ODE is extremely unlikely to have an explicit solution. Mathematica cannot do anything with this ODE even when $g(x)\equiv x$.

Here is an image of the corresponding Mathematica notebook:

enter image description here

And here is a link to the Mathematica notebook itself.

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  • $\begingroup$ Thank you losif. Yes it does not look like it can be solved, except maybe changing it to a Riccati equation? en.wikipedia.org/wiki/Riccati_equation $\endgroup$
    – Alex
    Commented Jan 31, 2023 at 21:46
  • $\begingroup$ @Alex : Riccati is a very different kind of ODE. $\endgroup$
    – Iosif Pinelis
    Commented Jan 31, 2023 at 22:10
  • $\begingroup$ Another thing to try is taking a derivative to get something like: $c_3 f f' + c_4 f' f'' - g' = 0$ but I'm not sure how to solve this either $\endgroup$
    – Alex
    Commented Jan 31, 2023 at 22:14
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    $\begingroup$ If you allow for a certain constant, $g=c_3 x^2 + c_3 c_2 /c_1 $, there's a simple solution. Maybe one can arrange the physics application to allow for this constant? $\endgroup$
    – Michael Engelhardt
    Commented Feb 1, 2023 at 17:28
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    $\begingroup$ Well, then $f=\sqrt{c_3 / c_1} x$ is a solution. $\endgroup$
    – Michael Engelhardt
    Commented Feb 1, 2023 at 19:13

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