Solve System with Repeated Eigenvalues

[faradayscat]

Homework Statement

I want to solve this systemx' = \left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)x + \left( \begin{array}\\ t \\ 2t \end{array} \right)

Homework Equations

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?

 

[stevendaryl]

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)

 

[faradayscat]

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]

Homework Statement

I want to solve this systemx' = \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.

Homework Equations

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?

 

[faradayscat]

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) = c₁e^5t\left( \begin{array}\\ -1 \\ 2 \end{array} \right) + c₂e^5t\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]

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) = c₁e^5t\left( \begin{array}\\ -1 \\ 2 \end{array} \right) + c₂e^5t\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]

Homework Statement

I want to solve this systemx' = \left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right) + \left( \begin{array}\\ t \\ 2t \end{array} \right)

Homework Equations

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.

 

[faradayscat]

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

I just checked, and its satisfied. Thanks!

 

[faradayscat]

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.

 

[faradayscat]

Thanks everyone, I have no further questions.