General Solution of x'(t)=Ax(t)

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- July 26th 2009, 02:13 PMdiddledabbleGeneral Solution of x'(t)=Ax(t)
General Solution of x'(t)=Ax(t)

- July 26th 2009, 03:13 PMpickslides
You need to solve these DEs

- July 27th 2009, 12:52 AMMoo
Hello,

Treat A as a constant (which it is).

We know that the general solution of this type of equation is

And when A is a matrix,

See here : http://www.mathhelpforum.com/math-he...al-matrix.html for an example of computing this exponential.

If you don't understand, I can explain. - July 27th 2009, 02:13 AMCaptainBlack
You can also convert this into a pair of uncoupled second order homogeneous constant coefficient ODE's.

To do this write them out in full as two first order ODE's differnetiate one of them again and substitute the derivative from the other to eliminate one of the variables ...

CB - July 29th 2009, 03:53 PMdiddledabbleStill don't get it
I still am not sure what to do with the matrix.

- July 31st 2009, 05:56 PMHavenEigenValue Method
Okay here's an easy way to solve it:

We assume the form of the solution that

It turns out the r, corresponds to the eigenvalue of the matrix A and are the eigenvectors

for an aribitrary constant

so the roots of this polynomial are:

Now we plug our values of r back in

So we see that , which means our eigenvector is and our first solution is

Similarly for

we compute

eigenvector:

second solution

Our final solution is a linear combination to the two solutions

Where are arbitrary constants