numerical solution of differential equations

[Ether]

Originally Posted by Hitchhiker 42

I've attached your spreadsheets modified for 1/4*x^3 + 1

Originally Posted by Ether

I am assuming what you actually meant by that is

x(t) = 1/4*t^3 + 1

Is that correct?

Originally Posted by Hitchhiker 42

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

Originally Posted by Hitchhiker 42

I took the derivative of the actual velocity function. Is this not correct?

Yes, but you got x and t confused.

given:

x(t) = 1/4*t^3 + 1

... take time derivative of x(t) to get:

x'(t) = (3/4)*t^2

...take time derivative of x'(t) to get:

x''(t) = (3/2)*t


Now solve for the differential equation:

solve x'(t) for t:

x'(t) = (3/4)*t^2 => t=sqrt(4x'(t)/3)

... and then substitute for t in x''(t):

x''(t) = (3/2)*t = (3/2)*sqrt(4x'(t)/3) = sqrt(3*x'(t))