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