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wofsy
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Can someone tell me how to solve the ODE,
d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?
d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?
Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.
If d^{2}x/dt^{2}= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":
Let v= dx/dt so that d^{2}x/dt^{2}= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v^{2}= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or
[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]
That left side is an "elliptical integral".
Of course, y will be exactly the same, though possibly with different constants of integration.