How to solve&nbsp;x''(t)=-100-1x(t)What I'm thinking is to consider the first order versionx'(t)=-100-1x(t)which I know how to solve, the solution isx(t)=ce-te-tHow do I use this to solve the second-order equation?

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How to solve&amp;nbsp;x&#039;&#039;(t)=-100-1x(t)What I&#039;m thinking is to consider the first order versionx&#039;(t)=-100-1x(t)which I know how to solve, the solution isx(t)=ce-te-tHow do I use this to solve the second-order equation?

ritmesysv · Accepted Answer

Step 1Let&amp;nbsp;y(t)=(x(t),x′(t)). Then we havey&#039;(t)=0I2-I20y(t).Then once you have the solution&amp;nbsp;y(t)=eAty(0), you can get a solution for x.Step 2This is the general strategy. However, in your case, we have a simpler problem because the matrix is diagonal. Here,x1&#039;&#039;(t)x2&#039;&#039;(t)=-x1(t)-x2(t)and hence&amp;nbsp;x1&amp;nbsp;and&amp;nbsp;x2&amp;nbsp;are solutions to the one-dimensional equation&amp;nbsp;u&amp;nbsp;″=−u. So&amp;nbsp;xi=aicos⁡(t)+bisin⁡(t)&amp;nbsp;for&amp;nbsp;i=1,2.

Zemmiq34 · Accepted Answer

Step 1Notice that the matrix you have at the beginning is the unit matrix of dimention 2, I2. Then you have (x(t),y(t))=(−x(t),−y(t)) so xt)+x(t)=0. The characteristic polynomial for this ODE is r2=−1 which means r1=i and r2=−i hence x(t)=c1⋅eit+c2⋅e−it and hence your solution (by doing the same thing for y(t)) isStep 2(x(t),y(t))=(c1eit+c2e−it,c3eit+c4e−it).

How to solve \textbf{x}''(t)=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\textbf{x}(t) What I'm thinking is to consider the

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