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, $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$, you give are solutions to the "characteristic equation". If a second order equation has repeated eigenvalues, $\displaystyle \lambda_1= \lambda_2= \lambda$, that means that the characteristic equation is of the form $\displaystyle (x- \lambda_1^2= x^2- 2\lambda x+ \lambda^2= 0$ and so the differential equation is $\displaystyle \frac{d^2x}{dt^2}- 2\lambda \frac{dy}{dx}+ \lambda^2= 0$.

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It is easy to show that $\displaystyle x(t)= e^{\lambda t}$ satisfies that equation. What about $\displaystyle x(t)= te^{\lambda t}$? What do you get when you put that into the equation? Can you show that those two functions are independent?