Eileen Garza

2022-04-25

How to solve $\mathbf{\text{x}}\text{'}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)$

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

$\mathbf{\text{x}}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)$

which I know how to solve, the solution is

$\mathbf{\text{x}}\left(t\right)=\mathbf{\text{c}}\left[\begin{array}{c}{e}^{-t}\\ {e}^{-t}\end{array}\right]$

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

ritmesysv

Beginner2022-04-26Added 12 answers

Step 1

Let $y\left(t\right)=(x\left(t\right),{x}^{\prime}\left(t\right))$. Then we have

$y\text{'}\left(t\right)=\left(\begin{array}{cc}0& {I}_{2}\\ -{I}_{2}& 0\end{array}\right)y\left(t\right)$.

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

Step 2

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

$\left(\begin{array}{c}{x}_{1}\text{'}\text{'}\left(t\right)\\ {x}_{2}\text{'}\text{'}\left(t\right)\end{array}\right)=\left(\begin{array}{c}-{x}_{1}\left(t\right)\\ -{x}_{2}\left(t\right)\end{array}\right)$

and hence $x}_{1}\text{}\text{and}\text{}{x}_{2$ are solutions to the one-dimensional equation $u{}^{\u2033}=-u$. So ${x}_{i}={a}_{i}\mathrm{cos}\left(t\right)+{b}_{i}\mathrm{sin}\left(t\right)\text{}\text{for}\text{}i=1,2$.

Zemmiq34

Beginner2022-04-27Added 11 answers

Step 1

Notice that the matrix you have at the beginning is the unit matrix of dimention 2,$\mathbb{I}}_{2$ . Then you have $\left(x\text{(t),y}\left(t\right)\right)=(-x\left(t\right),-y\left(t\right))$ so $xt)+x\left(t\right)=0$ . The characteristic polynomial for this ODE is ${r}^{2}=-1$ which means ${r}_{1}=i\text{}\text{and}\text{}{r}_{2}=-i$ hence $x\left(t\right)={c}_{1}\cdot {e}^{it}+{c}_{2}\cdot {e}^{-it}$ and hence your solution (by doing the same thing for y(t)) is

Step 2

$(x\left(t\right),y\left(t\right))=({c}_{1}{e}^{it}+{c}_{2}{e}^{-it},{c}_{3}{e}^{it}+{c}_{4}{e}^{-it})$ .

Notice that the matrix you have at the beginning is the unit matrix of dimention 2,

Step 2

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