Is There an Analytic Solution for This Crazy ODE?

[Brin]

Can anyone tell me if this ODE has an analytic solution? And if it does, how the heck might I go about it?

$$\left(\frac{1}{y^{2}}\frac{dy}{dx}\right)^{2}-\frac{A}{y^{3}}-\frac{B}{y^{2}}=D$$

 

[Ben Niehoff]

It looks like it is separable. Just isolate dy/dx. It looks like the solution will be some kind of elliptic function.

(Mathematica can find an analytic solution for x as a function of y. It involves elliptic functions and finding roots of a cubic polynomial.)

 

[JJacquelin]

The integral of :
dx = Sqrt[Ay+By²+Dy^4] dy
involves elliptic integrals.
In the general case, the function x(y) which can be obtained would be rather complicated. Then, inverting it in order to express y(x) would be a big chore. Better use numerical integration.

 

[jackmell]

. . . ohhhhhh . . . you guys give up too easy. Assume we are given that the solution to:

$$\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2z^2)=\Delta^2(z)$$

is:

$$y=sn(z,k)$$

where sn is the Jacobi elliptic function. Then we seek a transformation z=z(y) that transforms:

$$\begin{align*}<br /> \left(\frac{dy}{dx}\right)^2&=a+by+cy^2+dy^3+ey^4\\<br /> &=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta) \\<br /> &=h^2 \Delta_2^2(y)<br /> \end{align*}$$

into this standard form. To this end we let:

$$z^2=\frac{(\beta-\gamma)}{(\alpha-\delta)}\frac{(y-\alpha)}{(y-\beta)}=s\frac{(y-\alpha)}{(y-\beta)},\quad k^2=\frac{(\beta-\delta)}{(\alpha-\gamma)}\frac{(\alpha-\delta)}{(\beta-\delta)},\quad M^2=\frac{(\beta-\delta)(\alpha-\delta)}{4}$$

for which we obtain:

$$\frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2(y)}\frac{dy}{dx}=Mh$$

so that:

$$\frac{dz}{dx}=Mh\Delta(z)=Mh\sqrt{(1-z^2)(1-k^2z^2)}$$

and therefore:

$$z=sn(hMv,k),\quad v=x-x_0$$

or:

$$y=\frac{z^2\beta-s\alpha}{z^2-s}$$

I believe though the actual implementation of this would be difficult as I have never worked a real problem using this method but I think would be a nice project for someone taking non-linear DEs next semester. :)

 

[JJacquelin]

Hi !
Very nice job jackmell, but . . .

. . . ohhhhhh . . . you guys give up too easy.

. . . ohhhhhh . . . even easier . . . just a few seconds to have the explicit formula :
I let you try WolframAlpha and see the result