# Thread: 2D linear system help

1. ## 2D linear system help

I have that given we have 2 distinct eigenvalues for a matrix describing the 2D linear system, then the general solution to the system is:

A(e^lambda1*t)U + B(e^lambda2*t)V, where A,B constants and U,V eigenvectors

How do we derive, or what is the general solution where we have repeated eigenvalues?

Thanks

2. ## Re: 2D linear system help

What you state is only true of equations with constant coefficients. It is easy to show that the set of all solutions to an nth order linear, homogeneous, differential equation is a vector space of dimension n. That means that if we have two independent solutions to the equation, any solution can be written as a linear combination of those two solutions. In particular, if the coefficients are constant, the eigenvalues, $\lambda_1$ and $\lambda_2$, you give are solutions to the "characteristic equation". If a second order equation has repeated eigenvalues, $\lambda_1= \lambda_2= \lambda$, that means that the characteristic equation is of the form $(x- \lambda_1^2= x^2- 2\lambda x+ \lambda^2= 0$ and so the differential equation is $\frac{d^2x}{dt^2}- 2\lambda \frac{dy}{dx}+ \lambda^2= 0$.
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It is easy to show that $x(t)= e^{\lambda t}$ satisfies that equation. What about $x(t)= te^{\lambda t}$? What do you get when you put that into the equation? Can you show that those two functions are independent?