# Systems of Linear Homogenous Differential equations with Constant Coefficients

1. Mar 6, 2010

### raisin_raisin

Hello,
I am looking at different ways to solve Systems of Linear Homogenous Differential equations with Constant Coefficients that is $$\acute{x}=Ax$$ (x and x' are vectors A is a matix) then the solutions are $$x= \xi e^{\lambda t}$$ where $$\xi$$ are the eigenvectors and $$\lambda$$ the eigenvalues of A and the general solution is the sum of all the eigenvectors with constants inserted.
i.e $$x= c_{1}\xi^{(1)} e^{\lambda t}+c_{2}\xi^{(2)} e^{\lambda t}$$

The problem I have with this is that I can't figure out how to get mixing since surely if n=2, say, then x=(x1,x2) but to get x1 you are just adding weighted amounts of x1 doesn't ever couple to x2 i.e $$x_{1} = c_{1}\xi^{(1)}_1 e^{\lambda t}+c_{2}\xi^{(2)}_{1} e^{\lambda t}$$I assume I am being idiot but if someone could point out where I am going wrong that would be brilliant.

Thanks very much,

P.S I normally solve it like this http://physics.ucsc.edu/~peter/114A/coupled_fol.pdf [Broken] if you know any links to other ways to solve them I would be grateful.

Last edited by a moderator: May 4, 2017
2. Mar 7, 2010

### kikushiyo

If you have :
d X/dt = AX where A is a diagonalisable matrix and X a column vector.
U^-1AU=D where D is the diagonal matrix : diag(lambda_1...lambda_n)

d X/dt = AX <=>
d X/dt = UDU^-1X <=>
U^-1dX/dt = DU^-1X <=>
dY/dt = D Y with Y =U^-1X provided that A is time independant. The mixing comes from the fact the eigendirections are Y and not X : you need the transfer matrix.