# Runge Kutta Input

Hi,

Hi. I am trying to enter some differential equations into a Runge-Kutta 4th order spreadsheet which requires input in the form y' = ...? My differential equations are in the form :
dvx/dt = -FDvx/mv (i.e relate velocity derivative with respect to time to acceleration).
Can anyone please help or point me in the right direction as to how to convert this to to y' = ....? format?

Many Thanks

I like Serena
Homework Helper
Hi andyfive, welcome to PF!

Perhaps if you define y = vx you have the required form?

Or possibly y = (vx, vy)?

Many Thanks for your reply I like Serena. If I understand correctly, by defining y = ax I get :

dy/dt = -FDy/mv

If y' is equal to dy/dt then it is as simple as saying y' = -FDy/mv. Is this correct?

Thanks

I like Serena
Homework Helper
Many Thanks for your reply I like Serena. If I understand correctly, by defining y = ax I get :

dy/dt = -FDy/mv

If y' is equal to dy/dt then it is as simple as saying y' = -FDy/mv. Is this correct?

Thanks
Basically, yes!

However, you also have a "v" in your expression.
I suspect this should actually be sqrt(vx2+vy2), but I'm guessing here, since I think you did not list all your variables and equations.
If it is, "y" would come back into the expression for "v".

Basically, yes!

However, you also have a "v" in your expression.
I suspect this should actually be sqrt(vx2+vy2), but I'm guessing here, since I think you did not list all your variables and equations.
If it is, "y" would come back into the expression for "v".
Thanks again I like Serena. Yes, you are absolutely correct, the v will need to be defined actually as sqrt(vx2+vy2+vz2) - this is a model for a 3 dimensional ball flight. The initial velocity(v) at time(t) = 0 is known and hence the three initial velocity vectors are known. Not sure what happens though when other derivatives are part of the expression (ie vy & vz)?

Regards

I like Serena
Homework Helper
Thanks again I like Serena. Yes, you are absolutely correct, the v will need to be defined actually as sqrt(vx2+vy2+vz2) - this is a model for a 3 dimensional ball flight. The initial velocity(v) at time(t) = 0 is known and hence the three initial velocity vectors are known. Not sure what happens though when other derivatives are part of the expression (ie vy & vz)?

Regards
In this case you would have y = (y1, y2, y3) as a vector.

And you would effectively have 3 equations, the first being:

$$\frac {dy_1} {dt} = \frac {-F_D y_1} {m \sqrt{y_1^2+y_2^2+y_3^2}}$$

In this case you would have y = (y1, y2, y3) as a vector.

And you would effectively have 3 equations, the first being:

$$\frac {dy_1} {dt} = \frac {-F_D y_1} {m \sqrt{y_1^2+y_2^2+y_3^2}}$$
Brilliant. Many Thanks I like Serena. You really have helped clarify things for me. I'll get back to the spreadsheet and see what results I now obtain.

I like Serena
Homework Helper
Brilliant. Many Thanks I like Serena. You really have helped clarify things for me. I'll get back to the spreadsheet and see what results I now obtain.
Cheers!