Confused about system of linear differential equations?

theBEAST
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Homework Statement


So in my notes it says that the general solution to a system of linear equations:
x1'(t) = a11x1(t) + a12x2(t)
x2'(t) = a21x1(t) + a22x2(t)

is:
x(t) = c1eλ1tv1 + c2eλ2tv2

What I don't understand is how you go from the system of equations to the general solution, the notes did not explain how. This was for my linear algebra class and I have never taken a differential equation class. However I am familiar with calculus so could someone please explain this?

I know how to solve the general solution, just not sure how it was derived.

Thanks!
 
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From what I remember, it's basically setting up a matrix from the system of differential equations and finding the eigenvectors and eigenvalues. I'd have to to dig out my diff eq book to see how close I am to the right process, but that should be a good starting point, and you can do an internet search to find out more.
 
The solutions of the system of equations are two-dimensional vectors of the form veλt, with components x1=v1eλt, x2=v2eλt. Substitute it back and you will find the values of λ. Determine v1 and v2 for each λ: these are the independent solutions of the system of equation, and the general solution is a linear combination of them.

ehild
 
You can write the system of equations as the "matrix" equation
\frac{d\begin{pmatrix}x_1(t) \\ x_2(t)\end{pmatrix}}{dt}= \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}x_1(t) \\ x_2(t)\end{pmatrix}

Which is effectively dx/dt= Ax so that dx/x= Adt and, integrating, ln(x)= At+ C from which x= e^{At+ C}= e^{At}e^C= C'e^{At}.

The simplest way to find the exponential of a matrix is to find its eigenvalues and eigenvectors so that it can be changed to diagonal or Jordan Normal form.

The form you give is correct if and only if the matrix can be diagonalized which is true if the matrix has two independent eigenvectors which, in turn, is true if (but not "only if") the two eigenvalues, \lambda_1 and \lambda_1, are distinct.
 
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