How to solve $\textbf{x}''(t)=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\textbf{x}(t)$

Question

How would I solve this second order matrix equation:

$\textbf{x}''(t)=\begin{bmatrix}-1&0 \\ 0&-1\end{bmatrix}\textbf{x}(t)$

What I'm thinking is to consider the first order version

$\textbf{x}'(t)=\begin{bmatrix}-1&0 \\ 0&-1\end{bmatrix}\textbf{x}(t)$

which I know how to solve, the solution is

$\textbf{x}(t)=\textbf{c}\begin{bmatrix}e^{-t} \\ e^{-t}\end{bmatrix}$

How do I use this to solve the second-order equation?

Answer 1

Let $y(t) = (x(t), x'(t))$. Then we have

$$y'(t) = \begin{pmatrix} 0 & I_2 \\ -I_2 & 0 \end{pmatrix} y(t).$$

Then once you have the solution $y(t) = e^{At}y(0)$, you can get a solution for $x$.

This is the general strategy. However, in your case, we have a simpler problem because the matrix is diagonal. Here,

$$\begin{pmatrix} x_1''(t) \\ x_2''(t) \end{pmatrix} = \begin{pmatrix} -x_1(t) \\ -x_2(t) \end{pmatrix}$$

and hence $x_1$ and $x_2$ are solutions to the one-dimensional equation $u'' = -u$. So $x_i = a_i \cos(t) + b_i \sin(t)$ for $i = 1, 2$.

Answer 2

Notice that the matrix you have at the beginning is the unit matrix of dimention 2, $\mathbb{I}_2$. Then you have $(x"(t),y"(t))=(-x(t),-y(t))$ so $x"(t)+x(t)=0$. The characteristic polynomial for this ODE is $r²=-1$ which means $r_1=i$ and $r_2=-i$ hence $x(t)=c_1\cdot e^{it}+c_2\cdot e^{-it}$ and hence your solution (by doing the same thing for $y(t)$) is $$(x(t),y(t))=(c_1 e^{it}+c_2e^{-it},c_3 e^{it}+c_4 e^{-it}).$$ I didnt wrote it with the use of matrices because i dont remember how to make a matrix in latex. Sorry about that.