# General Question about 2nd ODE's

1. Sep 11, 2009

### string_656

how do you determine whether a second order differential equation can be solved or not?

2. Sep 11, 2009

### HallsofIvy

Staff Emeritus
What do you mean by "solved"?

The fundamental existence and uniqueness theorem for first order differential equations says that there exist a unique solution to y'= f(x,y), y(x0)= y0 as long as f(x,y) is continuous and "Lipgarbagez" in y in some neighborhood of (x0,y0). If f is continuous but not Lipschitz, a solution will exist but may not be unique.

A general second order equation is of the form y''= f(x, y, y') and, by setting z= y' so that y''= z', we have the two first order equations y'= z, z'= f(x, y, z) which we can then cast as a single first order vector equation, Y'= f(x, Y) where Y is now <y, z>. Basically, that says, then, that the "initial value" problem y"= f(x, y, y'), y(x0)= y0, y'(x0)= y1 has a unique solution as long as f is continuous and Lipschitz in both y and y' in some neighborhood of (x0,y0,y1).

Those are the conditions for which an initial value problem HAS a solution. If by "can be solved" you mean that you can find a reasonable formula for the solution, that's much too vague a question.