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Can someone tell me what the point of numerical methods for ODE's is?

Like for instance my prof gave me this question:

Reconstruct the Milne Simpson model: y[n+1]=a[1]*y[n-1]+h(b[-1]*y'[n+1]+b[0]*y'[n]+b[1]*y'[n-1]), (where [ ] denote subscripts)

My prof. taught us the method to solve it and i can solve for the parameters but i dont get whats going on. I never took a ODE course before so that is why i guess i am even more lost. Can someone help me here?

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Because when you get right down to it, ODEs cannot be solved, and analytical methods (e.g. perturbations, asymptotics, phase plane, linearization, solution methods, estimates, etc.) only go so "far" in the sense of obtaining a solution. The same remarks go for PDE, but the situation for analytical methods is even worse in that realm. Since mathematicians don't really have any need for solutions to ODE (only their properties), numerical methods are often sidelined or looked down upon. But in applied/industrial mathematics, numerically solving DEs is a huge part of what you do. –  Taylor Martin Oct 14 '12 at 22:30
but why do have y[n+1]? –  CamNewton Oct 14 '12 at 22:32
And...if you want to numerically solve DEs, you must know the properties of available approximation methods, as they can be very equation-dependent (for example, if you have a "stiff" ODE, you will often get garbage using the explicit methods you learn first, as your time step will often be outside the region of stability; if you had no idea about the derivations of implicit methods, or their properties, or even what a "stiff" equation was, you would probably have no idea how to deal with your bad results, or worse yet, you would accept them as being accurate). –  Taylor Martin Oct 14 '12 at 22:33
Well for one, you don't even have the full "model." This method belongs to a class of schemes called "predictor-corrector" methods. That is, you use an explicit scheme first, and then use an implicit scheme using the "predicted" value from the explicit method; this is called the "corrector," and is named so since implicit methods are generally more accurate, and more importantly more stable. –  Taylor Martin Oct 14 '12 at 22:40
If your professor asked you to derive the method, then you first take four initial time steps at $x_{i-3},x_{i-2},x_{i-1},x_{i}$ where you have values (or approximations) $y_{i-3},\ldots,y_{i}$ (To obtain these values at the beginning, one begins with a one-step method like RKF). Then at each step you integrate over $[x_{i-3},x_{i+1}]$ the Lagrange interpolant of the known interior points. After evaluating this integral, you obtain an expression for $y_{i+1}$. You then compute this time step. –  Taylor Martin Oct 14 '12 at 22:48

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