1. Aug 5, 2012

### diffEQult

Hit a wall on this one...can anyone help with the next step?

A(d2y/dt2)+B*(dy/dt)^2*(C*y^2+D*y+E)+F*y^-2=0

Thanks!

Last edited: Aug 5, 2012
2. Aug 5, 2012

### HallsofIvy

Staff Emeritus
Because t does not expicitely appear in that equation, you can simplify a little by "quadrature". Let v= dy/dt. The $d^2y/dt^2= dv/dt= (dv/dx)(dx/dt)= v(dv/dt)$ so your equation becomes a first order equation for v in terms of y:
$$VA\frac{dv}{dy}+ Bv^2(Cy^2+ Dy+ E)= -Fy^{-2}$$
I don't know that that equation is solvable but if you can find v then you solve v= dy/dt.

3. Aug 5, 2012

### diffEQult

Yea that's how far I could get before getting stuck. The equation isn't really separable.

Is there a way to do this analytically or numerically?

4. Aug 6, 2012

### JJacquelin

Hi !
The equation derived by HallsofIvy is a Riccati ODE which can be transformed to a second order linear ODE (attachment).
Since this linear ODE is much too complicated to be analytically solved in terms of a finite combination of standard functions, we can conclude that the problem raised by DiffEQlt cannot be analytically solved (except in case of some particular values of the coefficients). So, in the general case, numerical process is required.

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5. Aug 6, 2012

### diffEQult

I dont see how you made the jump to your second line in the attached jpg? Where does the w come from and what does it represent.

In the end, what would you solve the equation for if you have w, y, and v's throughout it?

Also, V should be v in Halls equation.

Last edited: Aug 6, 2012
6. Aug 7, 2012

### JJacquelin

v is the unknown function in the previous ODE.
w is the unknown function in the new ODE after the change of function.
Of course, one have to define a relationship between the previous unknown function and the new unknown function. The second line in the attached jpg is the chosen relationship.
This relationship was chosen so that the quadratic term in the previous ODE be eliminated and that the new ODE be linear.
It is not always possible to do that. A convenient relationship can be found only for few kind of non-linear ODEs in order turn them to linear ODEs. In is possible in the case of the Riccati equations.

If it is possible to analytically solve the linear ODE, one obtains the literal expression for the function w, which becomes known. Then, one have to express w' and bring back w and w' into the relationship which leads to the function v(y). Since v=dy/dt, the integration of dt=dy/v(y) leads to t as a function of y. Then, if possible, the inverse of the function t(y) gives y as a function of t.

Ha! ha! ha! I didn't check the Halls equation. So, I thought that V was a constant.
You are right: V is not a constant, but V=v. As a consequence all that was done is of no use. There is NO Riccati equation at all !!!
It's much simpler : You just have to let z=v² and the EDO directly becomes a first order linear ODE with the unknown function z.

Last edited: Aug 7, 2012