Recent content by Jwill

Okay, thanks. I know what you mean now... I actually feel kinda stupid lol. I've actually found exact error from numerical solutions of different orders.

What do you mean by error functions? Could you give me an example of such a situation? I am not exact familiar with that.

I mean if u(t) is completely unknown, is there no way to just generally integrate it even using terms like u'(t)? \frac{-1}{3u'(t)}cos(3u(t)) seems to be sort of close if you differentiate it... But this is an incorrect usage of U substitution.

What method of integration would I use to find the anti-derivative of \int sin(3u(t)) dt when u(t) Is and unknown function of time?

When you said that the two first order functions were incorrect... I am confused because you said that was correct earlier. I am confused and not sure what to do. I don't see how one could use the runge kutta method with those functions. I am under the impression that in many cases when you...

Actually, I'm sure that u' = \frac{-1}{3v} cos(3u) is not correct. When I differentiate it, it does not work out to be the original equation. What method of integration would be used to solve it? I'm sure it's integrated over time and v and u are functions of time.

Okay. I might be on to something with this: u = y v = y' v' = sin(3u) u' = \frac{-1}{3v} cos(3u) This way it's in the form: \frac{dv}{dt} = f[u] \frac{du}{dt} = F[v, u] So... during runge kutta iterations, I'd have to solve for v to solve for u for each iteration...

Well... anyway my attempt at this was: k1 = z_{i} l1 = sin(3.*(y_{i})); k2 = z_{i}+\frac{h}{2}*l1 l2 = sin(3*(y_{i}+\frac{h}{2}*k1)) k3 = z_{i}+\frac{h}{2}*l2 l3 = sin(3*(y_{i}+\frac{h}{2}*k2)) k4 = z_{i}+h*l2 l4 = sin(3*(y_{i}+h*k2)) z_{i+1} = z_{i} +...

Well... anyway my attempt at this was: k1 = z_{i} l1 = sin(3.*(y_{i})); k2 = z_{i}+\frac{h}{2}*l1 l2 = sin(3*(y_{i}+\frac{h}{2}*k1)) k3 = z_{i}+\frac{h}{2}*l2 l3 = sin(3*(y_{i}+\frac{h}{2}*k2)) k4 = z_{i}+h*l2 l4 = sin(3*(y_{i}+h*k2)) z_{i+1} = z_{i} +...

If these equations are correct then I must not understand how to do the stage constants for the iteration. If \frac{dz}{dt} = sin(3y(t)) \frac{dy}{dt} = z then \frac{dz}{dt} = f[t, y(t)] \frac{dy}{dt} = f[t, z(t)] (I think) How are they "coupled" in that I don't see...

If that is correct then I must not understand how to do the stage constants for the iteration. If \frac{dz}{dt} = sin(3y(t)) \frac{dy}{dt} = z then \frac{dz}{dt} = f[t, y(t)] \frac{dy}{dt} = f[t, z(t)] (I think) How are they "coupled" in that I don't see that they...

Can someone tell me how to solve this differential equation in MATLAB so that I can check my numerical solution? (I'm not asking for an answer, just a way to check my answer)

Homework Statement I have a large project involving Runge Kutta numerical solutions of differential equations. I understand the Runge Kutta method and I've done it before, but my problem involves taking the differential equation y''=sin(3y(t)), t>=0 and reexpressing this IVP into coupled...

I have a large project involving Runge Kutta numerical solutions of differential equations. I understand the Runge Kutta method and I've done it before, but my problem involves taking the differential equation y''=sin(3y(t)), t>=0 and reexpressing this IVP into coupled first order...

I might not quite get what you're saying, but if the cylinder and the hole have the same diameter why would the hole have any stress on it's inner diameter? Are you talking about a press fit of some sort?