Exponential Matrix Solutions to Dynamic System

[TeaDrinkingGuy]

Consider the dynamical system dX/dt=AX, where X=[x,y,z] and A is a 3x3 matrix. Using the exponential matrix method in the most appropriate manner you see fit, find the general solution in each of the following cases:

i) A=[math]\begin{vmatrix} 2 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0& 2 \end{vmatrix}[/math]
ii) A=[math]\begin{vmatrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ -1 & 0& 2 \end{vmatrix}[/math]
Then, find the solutions to these similar adapted variations.

In case i) dX/dt = AX +[imath]\begin{vmatrix} 1 \\ t \\ t^2 \end{vmatrix}[/imath] and x(0)=2 y(0)=-1 z(0)=-4

In case ii) dX/dt = AX +[imath]\begin{vmatrix} 0 \\ 0 \\ e^t \end{vmatrix}[/imath] and x(0)=0 y(0)=0 z(0)=6


I know there is a lot here, but I would sincerely appreciate any help. I've been trying to crack this all day and I feel like I'm banging my head against a brick wall. Thank you in advance.

 

[mmm4444bot]

It will help tutors to see how you began part (i) or what you were thinking, before you started banging your head.

Please see the forum guidelines for more information. Thanks!

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[TeaDrinkingGuy]

It will help tutors to see how you began part (i) or what you were thinking, before you started banging your head.

Please see the forum guidelines for more information. Thanks!

?

I had a crack at part i) by transforming it into a diagonal matrix, getting it in the form PDP^-1 and multiplying it out. This seems to have worked okay, but I don't think this works for the ii), or if it does, it seems to get very messy very quickly.

I just tried to edit the post, but as it was after 30 minutes it didn't let me. apologies if the post is a little sloppy, it's my first time posting on a forum like this.

 

[mmm4444bot]

tried to edit the post

You may add information to your conversation by posting it, including stuff like changes, clarifications and corrections.

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[HallsofIvy]

Consider the dynamical system dX/dt=AX, where X=[x,y,z] and A is a 3x3 matrix. Using the exponential matrix method in the most appropriate manner you see fit, find the general solution in each of the following cases:

i) A=[math]\begin{vmatrix} 2 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0& 2 \end{vmatrix}[/math]

Frankly, I wouldn't use matrices here. Do you see that "dX/dt= AX" is the same as
dx/dt= 2x- z
dy/dt= 2y and
dz/dt= -x+ 2z
where I have taken X= (x, y, z)?
We can immediately solve dy/dt= 2y as \(\displaystyle y(t)= Be^{2t}\).
Adding the other two equations, dx/dt+ dz/dt= d(x+z)/dt= x+ z.
Let p= x+ z to get dp/dt= p, \(\displaystyle p(t)= x(t)+ z(t)= Ae^t\).
\(\displaystyle z(t)= Ae^t- x(t)\) so that \(\displaystyle dx/dt= 2x- Ae^t+ x= 3x- Ae^t\)
The "associated homogeneous equation" is dx/dt= 3x which has solution \(\displaystyle x= Ce^{3t}\). For a solution to the entire equation try \(\displaystyle x= De^t\). Then \(\displaystyle dx/dt= De^t= 3De^t- Ae^t\) so D= 3D- A. 2D= A, = A/2. The general solution to the entire equation is \(\displaystyle x(t)= Ce^{3t}+ (A/2)e^t\). Then \(\displaystyle z(t)= Ae^t- x(t)= Ae^t- C^{3t}- (A/2)e^t= (A/2)e^t- Ce^{3t}\).

Putting all that together
\(\displaystyle X(t)= \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix}=\)\(\displaystyle \begin{bmatrix}Ce^{3t}- (A/2)e^t \\ Be^{2t} \\(A/2)e^t- Ce^{3t}\end{bmatrix}\)

ii) A=[math]\begin{vmatrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ -1 & 0& 2 \end{vmatrix}[/math]
Then, find the solutions to these similar adapted variations.

In case i) dX/dt = AX +[imath]\begin{vmatrix} 1 \\ t \\ t^2 \end{vmatrix}[/imath] and x(0)=2 y(0)=-1 z(0)=-4

In case ii) dX/dt = AX +[imath]\begin{vmatrix} 0 \\ 0 \\ e^t \end{vmatrix}[/imath] and x(0)=0 y(0)=0 z(0)=6


I know there is a lot here, but I would sincerely appreciate any help. I've been trying to crack this all day and I feel like I'm banging my head against a brick wall. Thank you in advance.