By considering expressions of the form $Y(t) = e^{tw}V$, find the general solution to $\frac{d^2Y}{dt^2} = AY$.

Question

By considering expressions of the form $Y(t) = e^{tw}V$, find the general solution to $\frac{d^2Y}{dt^2} = AY$.

We were given a matrix: $$A = \begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 & \end{bmatrix}$$

The eigenvalues are 3 and 0. The corresponding eigenvectors I found are:

For $\lambda = 3$: $$\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}. $$ For $\lambda = 0$: $$\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}$$

How do I find the general solution?

Thank you!

P.S. $V$ is the eigenvectors.

Answer 1

Plug $Y(t)$ into the differential equation. You'll get $w^2e^{tw}V=Ae^{tw}V$, which can be rewritten as

$$e^{tw}(A-w^2I)V=0.$$

It follows that $w^2=\lambda$ and $V$ is the corresponding eigenvector of $A$.