# Differential equation [closed]

I need to solve the following equation

$$\left(v\frac{\partial L}{\partial v}-L\right)^{2}=\left(\frac{\partial L}{\partial v}\right)^{2}c^{2}+\frac{1}{M^{2}}\left(\frac{\partial L}{\partial v}\right)^{4}+m^{2}c^{4}$$

for $L=L(q,v)$, where $c$, $M$, and $m$ are constants, $M>>m$.

I tried to use a Fourier transform: (let $F_{v}[L](\omega)=F$):

$$F_{v}\left[\frac{\partial L}{\partial v}\right](\omega)=-i\omega F_{v}[L]=-i\omega F$$

$$F_{v}[v](\omega)=-i\sqrt{2\pi}\delta'(\omega)$$

Here $\delta'(\omega)$ is the derivative of the delta function. Substituting these expressions in the given equation, we set $t=F^{2}$, to obtain the quadratic equation

$$c^4 m^2-t \left(c^2 \omega ^2-\omega\sqrt{2\pi} \delta (\omega )'-1\right)+\frac{t^2\omega^4}{M^2}=0$$

At this point it seemed that all is well, but Mathematica provides a cumbersome expression for the roots and can't do the inverse Fourier transform.

Can you help me to solve this equation? Maybe there are more artful ways?

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Could you add the Mathematica expression that you tried? – Sjoerd C. de Vries Feb 25 at 19:30
what does $\delta(\omega)'$ mean? – chris Feb 25 at 20:27
It is a derivative of the Dirac delta function, I think. That is $\delta(\omega)^\prime = \delta^\prime(\omega)$. – Sasha Feb 25 at 21:55

## closed as not a real question by Jens, m_goldberg, whuber, Sjoerd C. de Vries, ArtesMar 13 at 22:13

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