- #1

- 6

- 0

for example:

y

^{(4)}+5y

^{(3)}-2y

^{''}+3y

^{'}-y=0

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- Thread starter Ojo Saheed
- Start date

If you're trying to solve this numerically, one way it is often done is to convert to a set of first order ODEs. In this particular case, there would be 12 first order ODEs to integrate. Define,zj=yj (j = 1,3)zj+3=y'j (j = 1,3)zj+6=y''j (j=1,3)zj+9=y'''j (j=1,3)Thenz1' = z4z4' = z7z7'f

- #1

- 6

- 0

for example:

y

- #2

- 22,822

- 5,470

This doesn't look like a system of 4th order ODEs. It looks like a single 4th order ODE.

Chet

Chet

- #3

- 6

- 0

Thanks for the reply Chet. Actually, I had thought that given an example of one the system of equations might suffice for a demonstration. In any case, a better representation of the system goes thus:This doesn't look like a system of 4th order ODEs. It looks like a single 4th order ODE.

Chet

y

y

y

where a, b and c terms are constant coefficients

- #4

- 22,822

- 5,470

If you're trying to solve this numerically, one way it is often done is to convert to a set of first order ODEs. In this particular case, there would be 12 first order ODEs to integrate.Thanks for the reply Chet. Actually, I had thought that given an example of one the system of equations might suffice for a demonstration. In any case, a better representation of the system goes thus:

y_{1}^{(4)}=a_{1}y_{1}^{''}+a_{2}y_{2}^{''}+a_{3}y_{3}^{''}+a_{4}y_{1}+a_{5}y_{2}+a_{6}y_{3}+a_{7}

y_{2}^{(4)}=b_{1}y_{1}^{''}+b_{2}y_{2}^{''}+b_{3}y_{3}^{''}+b_{4}y_{1}+b_{5}y_{2}+b_{6}y_{3}+b_{7}

y_{3}^{(4)}=c_{1}y_{1}^{''}+c_{2}y_{2}^{''}+c_{3}y_{3}^{''}+c_{4}y_{1}+c_{5}y_{2}+c_{6}y_{3}+c_{7}

where a, b and c terms are constant coefficients

Define,

z

z

z

z

Then

z

z

z

z

etc.

You end up with 12 coupled first order linear ODEs explicit in the derivatives. You can use an automatic integrator to solve them or use your own coding of Runge Kutta or forward Euler, or backward Euler, or whatever.

Chet

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