numerical solution of differential equations

[Ether]

Originally Posted by Hitchhiker 42

Yes, sorry about that. That was what I meant.

OK.

x(t) = 1/4*t^3 + 1 is the analytical solution to the Initial Value Problem

a(t) = sqrt(3*v(t))

with initial values*

x(0.01) = 1.00000025

and

v(0.01) = 7.5e-5

As you can see from the spreadsheet, the Midpoint method gives much better results for that IVP than Forward Euler and Forward/Backward Euler.


* selected to avoid the stationary point at t=0

[GeeTwo]

Originally Posted by GeeTwo

No, as I read it you're setting x''=3x/2. This would be solved by a function of the form x=Ae^3x/2

Originally Posted by Hitchhiker 42

Could you explain what this means? I'm not sure I understand...

Sorry, that should have been x''=Ae^√3t/√2

if x=Ae^√3t/√2,
then x'=√3Ae^√3t/√2/√2 (chain rule)
and x''=3Ae^√3t/√2/2 (chain rule again) = 3x/2.

Edit: I knew there was more to it, and just figured out the other part:

x''=3x/2 ==> x= Ae^√3t/√2 + Be^-√3t/√2