Solve y²d²y/dx² + ay = b(cx-d) for y(x)

[adamjacobs173]

Homework Statement

$$y^{2}\frac{d^{2}y}{dx^2}$$ + ay = b(cx-d)

Find y as a function of x,a,b,c & d (a,b,c & d are all constant(!))

Homework Equations

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The Attempt at a Solution

Not a clue, this is actually how far I got with my own take on an orbital mechanics problem I made up. Not my homework (I swear!). My guess would be some kind of substitution, but I don't know how to go about it.

P.S. Sorry about the alignment, but my mastery of LaTeX is minimal . As is probably obvious, the unformatted bit should be on a level with the y^2 at the start.

 

[hunt_mat]

Nothing immediately screams out at me, the think I might be tempted to try is write $u=dy/dx$ and then treat it as a dynamical system. Other than that, solve it numerically?

 

[HallsofIvy]

That is a very non-linear equation and, like all non-linear equations, will be very difficult to solve, even for specified initial conditions. In fact, there might not be a single formula that will give all solutions and, even if there is, I would not expect it to be in terms of elementary functions. You say this came from "an orbital mechanics problem". Those tend to give elliptic function solutions.

 

[adamjacobs173]

Those tend to give elliptic function solutions.

Yes, I think that's probably the case, although this was to do with radius/time, rather than cartesian. I was just wondering if there was a standard method for solving 2nd order differentials involving powers of y as a coefficient?

 

[HallsofIvy]

No, there isn't. As I said before, non-linear equations tend to be very difficult. There is no general way of solving even the simplest non-linear equations.

Sometimes it helps, not so much to "solve" the equation, but to get information about the solution, to write the single equation as a system of equations. If you let v= dy/dx, then $d^2y/dx^2= dv/dx$ so your equation becomes $y^2 dv/dt= -y+ bcx- bd$ so you have the system of equations
$$\begin{pmatrix} \frac{dy}{dx}= v \\ \frac{dv}{dx}= -\frac{1}{y}+ \frac{bcx- bd}{y^2}\end{pmatrix}$$