Solving $y'''-4y''+13y'=0$ using matrix method

Question

I have an exam in a couple of days and I am trying to do some past papers. One of the questions is about solving $$y'''-4y''+13y'=0$$ by the matrix method, i.e converting the 3rd order ode into the following form:

$$\vec{Y}'=A\vec{Y}$$

I converted the ODE into matrix form: $$\implies \begin{pmatrix}y'''\\y''\\y' \end{pmatrix}=\begin{pmatrix}4&-13&0\\1&0&0\\0&1&0 \end{pmatrix}\begin{pmatrix}y''\\y'\\y \end{pmatrix}$$

The question then asks:

Calculate the solution matrix $\vec{Y}$ that satisfies the matrix differential equation $\vec{Y}'=A\vec{Y}$. Hint: Express $\vec{Y}$ by using the eigenvalues of the matrix as well as the matrix $C$, where $C^{-1}AC=diag(\lambda_1,\lambda_2,\lambda_3)$. You do not have to state $C$ explicitly.

I am not sure what this matrix $C$ is.Don't I just have to find the eigenvalues $\lambda_n $ and eigenvectors $\vec{v}_n$ and then my solution will be $$y(x)=c_1 \vec{v}_1 e^{\lambda_1 t}+c_2 \vec{v}_2 e^{\lambda_2 t}+c_3 \vec{v}_3 e^{\lambda_3 t} \space ?$$

Can anyone explain what this matrix C does and what $\vec{Y}$ should look like? Thanks in advance

Answer 1

You found $\vec Y$ already, and it is a vector, not a matrix. $\vec Y = \begin{pmatrix}y''\\y'\\y \end{pmatrix}$. And $C$ is simply the eigenvectors written as colums of a matrix, $C=\begin{pmatrix}\vec v_1 \vec v_2 \vec v_3\end{pmatrix}$

Answer 2

You want your $Y$ vectors to be same entries.

$\begin{pmatrix}y'''\\y''\\y' \end{pmatrix}'=\begin{pmatrix}0&4&-13\\1&0&0\\0&1&0 \end{pmatrix}\begin{pmatrix}y'''\\y''\\y' \end{pmatrix}$

$Y' = A Y\\ Y(t)= e^{At} Y(0)\\ A = P^{-1}DP\\ Y = P^{-1}e^{Dt} PY(0)$