# Repeated eigenvalues

## Homework Statement

I want to solve this system

x' = $\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)$x + $\left( \begin{array}\\ t \\ 2t \end{array} \right)$

## The Attempt at a Solution

i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?

Last edited:

stevendaryl
Staff Emeritus
Science Advisor

## Homework Statement

I want to solve this system

x' = [(7,1),(-4,3)] + t[(1),(2)]

I apologize for being dense, but I don't understand this notation. Is the expression on the right a 2x2 matrix? If so, it would make it more readable if you used Tex notation:

$\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)$

I apologize for being dense, but I don't understand this notation. Is the expression on the right a 2x2 matrix? If so, it would make it more readable if you used Tex notation:

$\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)$

Yes, that's right. I'm not really familiar with Latex, I'll have to read up on it. I've edited my post, thanks

Mark44
Mentor

## Homework Statement

I want to solve this system

x' = $\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)$ + $\left( \begin{array}\\ t \\ 2t \end{array} \right)$
As written, this makes no sense. You can't add a 2 x 2 array to a 2 x1 array (column matrix).

This would make more sense if it were something like this:
##\vec{x'} = \begin{bmatrix} 7 & 1 \\ -4 & 3 \end{bmatrix}\vec{x} + \begin{bmatrix} 1 \\ 2 \end{bmatrix}##

Here ##\vec{x}## means ##\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}## and similar for its derivative.
faradayscat said:

## The Attempt at a Solution

i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?

As written, this makes no sense. You can't add a 2 x 2 array to a 2 x1 array (column matrix).

This would make more sense if it were something like this:
##\vec{x'} = \begin{bmatrix} 7 & 1 \\ -4 & 3 \end{bmatrix}\vec{x} + \begin{bmatrix} 1 \\ 2 \end{bmatrix}##

Here ##\vec{x}## means ##\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}## and similar for its derivative.

Yes that's that I meant, sorry for the confusion. So could i set up a fundamental matrix with the homogeneous solutions and solve the non-homogeneous system with, say, variation of parameters? I tried and this is what I get:

x(t) = c1e5t$\left( \begin{array}\\ -1 \\ 2 \end{array} \right)$ + c2e5t$\left( \begin{array}\\ -t-1/2 \\ 2t \end{array} \right)$ - (t/25)$\left( \begin{array}\\ 1 \\ 18 \end{array} \right)$ + (1/125)$\left( \begin{array}\\ 3 \\ -26 \end{array} \right)$

I checked my solution on wolfram and it's slightly different, which annoys me.

Mark44
Mentor
Yes that's that I meant, sorry for the confusion. So could i set up a fundamental matrix with the homogeneous solutions and solve the non-homogeneous system with, say, variation of parameters? I tried and this is what I get:

x(t) = c1e5t$\left( \begin{array}\\ -1 \\ 2 \end{array} \right)$ + c2e5t$\left( \begin{array}\\ -t-1/2 \\ 2t \end{array} \right)$ - (t/25)$\left( \begin{array}\\ 1 \\ 18 \end{array} \right)$ + (1/125)$\left( \begin{array}\\ 3 \\ -26 \end{array} \right)$

I checked my solution on wolfram and it's slightly different, which annoys me.
Just check that your solution satisfies the system of diff. equations.

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed

## Homework Statement

I want to solve this system

x' = $\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)$ + $\left( \begin{array}\\ t \\ 2t \end{array} \right)$

## The Attempt at a Solution

i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?

Isn't there a factor ##x## missing on the right? Should you not be dealing with the system
$$\pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ?$$
You can either plug in the matrix exponential in the solution
$${\mathbf{x}} = e^{At} \int_0^t e^{-A \tau} {\mathbf{f}}(\tau) \, d \tau$$
to your equation ##{\mathbf{x}}'(t) = A {\mathbf{x}}(t) + {\mathbf{f}}(t)##, or else use the Laplace-transform method.

As for the matrix exponential: you have ##A = P J P^{-1}##, where ##J## is the Jordan canonical form of ##A##:
$$J = \pmatrix{5 & 1 \\0 & 5}$$
Furthermore, for any scalar ##x## we have ##e^{Ax} = P e^{Jx} P^{-1}##, and ##e^{Jx}## is easy to determine; see
webpages on matrix exponentials.

Just check that your solution satisfies the system of diff. equations.

I just checked, and its satisfied. Thanks!

Isn't there a factor ##x## missing on the right? Should you not be dealing with the system
$$\pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ?$$
You can either plug in the matrix exponential in the solution
$${\mathbf{x}} = e^{At} \int_0^t e^{-A \tau} {\mathbf{f}}(\tau) \, d \tau$$
to your equation ##{\mathbf{x}}'(t) = A {\mathbf{x}}(t) + {\mathbf{f}}(t)##, or else use the Laplace-transform method.

As for the matrix exponential: you have ##A = P J P^{-1}##, where ##J## is the Jordan canonical form of ##A##:
$$J = \pmatrix{5 & 1 \\0 & 5}$$
Furthermore, for any scalar ##x## we have ##e^{Ax} = P e^{Jx} P^{-1}##, and ##e^{Jx}## is easy to determine; see
webpages on matrix exponentials.

Yes, I forgot the vector 'x' next to the coefficient matrix. Thanks for your tips, I actually did the variation of parameters method and everything worked out after I checked my solution as Mark44 suggested.

Thanks everyone, I have no further questions.