Eigenvalues of a 5x5 Matrix, continued

[gabriels-horn]

Homework Statement

The https://www.physicsforums.com/showthread.php?t=403476" was to determine the eigenvalues of the following matrix. View attachment 25800

The problem of interest deals with actually finding a solution to the system above without the use of matrix methods.

Homework Equations

The template was to use the following system to solve this problem.
X\prime = AX =>\begin{bmatrix}x_1\prime \\x_2\prime \\x_3\prime \\x_4\prime \\x_5\prime\end{bmatrix}=\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}

The Attempt at a Solution

So multiplying the matrix yields the following.

x_1\prime = 2x_1 + x_2

x_2\prime = 2x_2

x_3\prime = 2x_3

x_4\prime = 2x_4 +x_5

x_5\prime = 2x_5

It was suggested that x_2 and x_5 be eliminated to solve this system. Any pointers on how to best do that? Can you separate the variables? For example, x_5\prime = 2x_5 is separated to

\int dy = \int 2x_5 dx_5

and becomes y = x_5^2

 

[Mark44]

Homework Statement

The https://www.physicsforums.com/showthread.php?t=403476" was to determine the eigenvalues of the following matrix. View attachment 25800

The problem of interest deals with actually finding a solution to the matrix above without the use of matrix methods. https://www.physicsforums.com/attachments/25838

Homework Equations

The hint given was to use the following system to solve this problem.

\begin{bmatrix}x_1\prime \\x_2\prime \\x_3\prime \\x_4\prime \\x_5\prime\end{bmatrix}=\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}

The Attempt at a Solution

So multiplying the matrix yields the following.

x_1\prime = 2x_1 + x_2

x_2\prime = 2x_2

x_3\prime = 2x_3

x_4\prime = 2x_4 +x_5

x_5\prime = 2x_5

It was suggested that x_2 and x_5 be eliminated to solve this system. Any pointers on how to best do that? Can you separate the variables? For example, x_5\prime = 2x_5 is separated to

\int dy = \int 2x_5 dx_5

and becomes y = x_5^2

No it doesn't. The differential equation you solved here does not involve y. It involved a dependent variable, x₅, and an independent variable that is not shown, that I will call t.

There is a big difference between these two differential equations:
y' = 2x (solution y = x² + C)
x' = 2x (solution x = Ae^2t)

 

[Landau]

@Mark: why would you quote the whole message, especially if you're the first who replies?

@gabriels-horn: I have no idea what the exercise is. What is "a solution to the matrix"?

 

[gabriels-horn]

@gabriels-horn: I have no idea what the exercise is. What is "a solution to the matrix"?

I rephrased to "a solution to the system" instead. I noticed the attachment was not working but I uploaded it again and it seems to be working fine.

 

[Mark44]

@Mark: why would you quote the whole message, especially if you're the first who replies?

This is the recommended policy on this forum. Occasionally a poster will go back and edit the initial post or even delete its content. Quoting the whole message ensures that there is a record of the initial post.

@gabriels-horn: I have no idea what the exercise is. What is "a solution to the matrix"?

 

[gabriels-horn]

No it doesn't. The differential equation you solved here does not involve y. It involved a dependent variable, x₅, and an independent variable that is not shown, that I will call t.

There is a big difference between these two differential equations:
y' = 2x (solution y = x² + C)
x' = 2x (solution x = Ae^2t)

Agreed. I mixed up the variables before and changed y to t for convention. So separating the equation \frac{dx_5}{dt} = 2x_5 yields

\int 2dt=\int \frac{dx_5}{x_5}

which simplifies to x_5=e^2^t

My question is how to proceed from this point to solve the system.

 

[Mark44]

It actually becomes x₅ = Ae^2t. Do the same for x₂, and then you'll have a system of three DEs in three unknowns.

 

[Tedjn]

Indeed, although don't forget the arbitrary constant. Your solution might be a class of functions (which would constitute the general solution). You should likewise solve for all the variables that are in a differential equation by themselves and then substitute.

 

[gabriels-horn]

It actually becomes x₅ = Ae^2t. Do the same for x₂, and then you'll have a system of three DEs in three unknowns.

Indeed, although don't forget the arbitrary constant. Your solution might be a class of functions (which would constitute the general solution). You should likewise solve for all the variables that are in a differential equation by themselves and then substitute.

Solving for x_2 and x_5 and plugging them in yields the new system

x_1\prime = 2x_1 + C_1e^2^t

x_3\prime = 2x_3

x_4\prime = 2x_4 +C_2e^2^t

I changed A to C since A has been used already. Any tips on how to proceed from here without using matrix methods as the problem says?

 

[Mark44]

The constants should be different, say C₁ and C₂. In the remaining differential equations, two are nonhomogeneous and one is homogeneous. Each equation involves only one dependent variable, so they should all be fairly easy to solve.

 

[gabriels-horn]

The constants should be different, say C₁ and C₂. In the remaining differential equations, two are nonhomogeneous and one is homogeneous. Each equation involves only one dependent variable, so they should all be fairly easy to solve.

Solving the system yields the following solutions,

x_1 = c_1te^2^t

x_3 = c_2e^2^t

x_4 = c_3te^2^t

which is starting to look like the general solution that I need in matrix form, i.e.
X_G = c_1\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t+c_2\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}2e^2^t+c_3\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t

Does this look right? By using this method, how do you fill the entries for the matrices?

 

[Mark44]

Solving the system yields the following solutions,

x_1 = c_1te^2^t

x_3 = c_2e^2^t

x_4 = c_3te^2^t

which is starting to look like the general solution that I need in matrix form, i.e.
c_1\begin{bmatrix}y\end{bmatrix}\begin{bmatrix}te^2^t+c_2\begin{bmatrix}y\end{bmatrix}2e^2^t+\begin{bmatrix}y\begin{bmatrix}?\end{bmatrix}te^2^t

The LaTeX above is screwed up.

Don't forget to put in x₂ and x₅. You can always check you solution by verifying that <x₁', x₂', x₃', x₄', x₅'>^T = A<x₁, x₂, x₃, x₄, x₅>^T.

 

[gabriels-horn]

The LaTeX above is screwed up.

Corrected as above.

 

[vela]

Your solutions for x₁ and x₄ aren't complete. Those are just the particular solutions; they're missing the homogeneous part of the general solution.

 

[Mark44]

Solving the system yields the following solutions,

x_1 = c_1te^2^t

x_3 = c_2e^2^t

x_4 = c_3te^2^t

These (above) are not right. Go back to post #9 and solve those three DEs. Also, you need to include x_2 and x_5.

which is starting to look like the general solution that I need in matrix form, i.e.
X_G = c_1\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t+c_2\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}2e^2^t+c_3\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t

Does this look right? By using this method, how do you fill the entries for the matrices?

This isn't right either. There should be no x on the right side. After all, you're trying to show what x (the vector) is in terms of the independent variable t.

 

[gabriels-horn]

Your solutions for x₁ and x₄ aren't complete. Those are just the particular solutions; they're missing the homogeneous part of the general solution.

I solved x_1\prime = 2x_1 + C_1e^2^t by rearranging to

x_1\prime - 2x_1 = C_1e^2^t

Finding p(t) = -2 and h(t) = e^2^t allowed me to solve for \mu(t), which gave the above solution. I tacked on the constant at the end to appear more like the usual solutions do for systems of linear ODE's.

These (above) are not right. Go back to post #9 and solve those three DEs. Also, you need to include x_2 and x_5.

This isn't right either. There should be no x on the right side. After all, you're trying to show what x (the vector) is in terms of the independent variable t.

Sorry for the confusion as to putting x's in the general solution above, they were meant to be constant entries and in no way related to the x values that I solved for earlier.

 

[vela]

Well, you're making mistakes somewhere because the solutions you came up with aren't complete solutions. It sounds like you're using the integrating factor approach, which if you apply it correctly, should yield the complete solution. Show us your calculations.

 

[Mark44]

But if x₁' - 2x₁ = c₁e^2t, the general solution is NOT x₁(t) = Cte^2t, which is what you have. There is another term. The same is true for x₄.

 

[gabriels-horn]

Well, you're making mistakes somewhere because the solutions you came up with aren't complete solutions. It sounds like you're using the integrating factor approach, which if you apply it correctly, should yield the complete solution. Show us your calculations.

But if x₁' - 2x₁ = c₁e^2t, the general solution is NOT x₁(t) = Cte^2t, which is what you have. There is another term. The same is true for x₄.

You're right, I ignored the second term. Here is the work,

\mu(t) = e^\int^p^(^t^)^d^t \to e^-^2^t which gets put into the integrating factor equation, yielding

\mu(t)=\frac{\int e^-^2^t e^2^tdt+ C}{e^-^2^t} which simplifies to

\mu(t)=te^2^t + Ce^2^t

 

[Mark44]

So, what do you have now for x_1(t), x_2(t), ... , x_5(t)?

 

[gabriels-horn]

So, what do you have now for x_1(t), x_2(t), ... , x_5(t)?

The system becomes after combining the constants,

x_1 = c_1(t+1)e^2^t

x_2 = c_2e^2^t

x_3 = c_3e^2^t

x_4 = c_4(t+1)e^2^t

x_5 = c_5e^2^t

 

[Mark44]

For the third time, where are x_2 and x_5? I'll hold off commenting on your general solution.

 

[vela]

Don't forget to keep track of the arbitrary constants. You should have

x_1(t) = C_1 te^{2t} + Ce^{2t}

where C₁ is the constant that appears in x₂.

EDIT: I see you posted while I was typing. You can't combine the constants the way you did.

 

[gabriels-horn]

For the third time, where are x_2 and x_5? I'll hold off commenting on your general solution.

I guess you read that post before I was able to put them in. I think I'm bogging down the server with all of the LaTeX commands. The general solution should be
X_G = c_1\begin{bmatrix}a_1\\a_2\\a_3\\a_4\\a_5\end{bmatrix}(t+1)e^2^t+c_2\begin{bmatrix}b_1\\b_2\\b_3\\b_4\\b_5\end{bmatrix}e^2^t+c_3\begin{bmatrix}d_1\\d_2\\d_3\\d_4\\d_5\end{bmatrix}e^2^t+c_4\begin{bmatrix}e_1\\e_2\\e_3\\e_4\\e_5\end{bmatrix}(t+1)e^2^t+c_5\begin{bmatrix}f_1\\f_2\\f_3\\f_4\\f_5\end{bmatrix}e^2^t

 

[Mark44]

You show 30 arbitrary constants. In my solution there are 7, divided up between two vectors.

 

[gabriels-horn]

Anyone care to enlighten me as to the correct form of the general solution?

 

[Mark44]

I get
x_1 = c_1 e^(2t) + Ate^(2t)
x_2 = c_2 e^(2t)
x_3 = c_3 e^(2t)
x_4 = c_4 e^(2t) + Bte^(2t)
x_5 = c_5 e^(2t)

 

[vela]

Your expression

X_G = c_1\begin{bmatrix}a_1\\a_2\\a_3\\a_4\\a_5\end{bmatrix}(t+1)e^2^t+c_2\begin{bmatrix}b_1\\b_2\\b_3\\b_4\\b_5\end{bmatrix}e^2^t+c_3\begin{bmatrix}d_1\\d_2\\d_3\\d_4\\d_5\end{bmatrix}e^2^t+c_4\begin{bmatrix}e_1\\e_2\\e_3\\e_4\\e_5\end{bmatrix}(t+1)e^2^t+c_5\begin{bmatrix}f_1\\f_2\\f_3\\f_4\\f_5\end{bmatrix}e^2^t

is equivalent to

x_1 = c_1 a_1 (t+1)e^{2t}+c_2 b_1 e^{2t} + c_3 d_1 e^{2t} + c_4 e_1 (t+1)e^{2t} + c_5 f_1 e^{2t}

and so on. What values of a₁, b₁, d₁, e₁, and f₁ you need so it matches what your solution for x₁?

But first, you need to fix your expressions for x₁ and x₄, which will in turn require you to rewrite X_G slightly.

 

[gabriels-horn]

Your expression

is equivalent to

x_1 = c_1 a_1 (t+1)e^{2t}+c_2 b_1 e^{2t} + c_3 d_1 e^{2t} + c_4 e_1 (t+1)e^{2t} + c_5 f_1 e^{2t}

and so on. What values of a₁, b₁, d₁, e₁, and f₁ you need so it matches what your solution for x₁?

But first, you need to fix your expressions for x₁ and x₄, which will in turn require you to rewrite X_G slightly.

The problem just asked to write the general solution with matrices, but I got the feeling that the matrices for the system I wrote were incorrect. Maybe my notation is incorrect. I used a, b, d, e and f to represent the actual entries that will go into them. I don't necessarily know what those are, but they were alluded to in the previous thread related to this one. Thanks for the help with this one.

 

[vela]

You should be able to determine the values of a, b, d, e, and f by inspection once you have the solution.

Here's a simpler example. Suppose you had a 2x2 system with the solution

\begin{align*}<br /> x_1 &amp;= c_1 te^t + c_2e^{2t} \\<br /> x_2 &amp;= c_2 e^{2t}<br /> \end{align*}

You can write that as

\begin{pmatrix}x_1\\x_2\end{pmatrix} = c_1\begin{pmatrix}te^t\\0\end{pmatrix}+c_2\begin{pmatrix}e^{2t}\\e^{2t}\end{pmatrix}=\begin{pmatrix}te^t &amp; e^{2t} \\ 0 &amp; e^{2t}\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}