Solving x'=Ax+g using undetermined coefficients

[Jamin2112]

Homework Statement

Problem #4 from here: http://www.math.washington.edu/~jtittelf/ExtraCredit.pdf

Homework Equations

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The Attempt at a Solution

There's only one example of using undetermined coefficients in my textbook. I'm not sure where to start. x = t²a + tb + c ? That's just a ball park guess, since then x' will have just have constant vectors and vectors multiplied by t, like in the problem.

 

[gabbagabbahey]

I'd start by finding the eigenvalues and eigenvectors of \begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}...

 

[Jamin2112]

I'd start by finding the eigenvalues and eigenvectors of \begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}...

Fine. det(A-ßI)=0 ----> (2-ß)*(3-ß)-2*1 = 0 ----> ß² - 5ß + 4 = 0 ----> (ß-4)(ß-1)=0 ----> ß=4, 1

-----> ß⁽¹⁾ = (1 1)^T, B⁽²⁾=(-2 1)^T

Ah-ha! So I know part of the solution is e^4t(1 1)^T + e^t(-2 1)^T.

Maybe I assume the full solution is e^4t(1 1)^T + e^t(-2 1)^T + at² + bt + c?

 

[gabbagabbahey]

You don't get the zero vector for your first eigenvector...show your calculations for that part

 

[Jamin2112]

You don't get the zero vector for your first eigenvector...show your calculations for that part

(see edit)

 

[gabbagabbahey]

Maybe I assume the full solution is e^4t(1 1)^T + e^t(-2 1)^T + at² + bt + c?

Close...I think you want to assume the solution is of the form

\textbf{x}(t)=c_1e^{4t}\begin{pmatrix} 1 \\ 1\end{pmatrix}+c_2e^{t}\begin{pmatrix} -2 \\ 1\end{pmatrix}+\textbf{a}t+\textbf{b}

The c_1 and c_2 are needed since any linear combination of your two independent homogeneous solutions will also be a solution to the homogeneous equation. And I don't think you need a t^2 term for your particular solution.

 

[Jamin2112]

Close...I think you want to assume the solution is of the form

\textbf{x}(t)=c_1e^{4t}\begin{pmatrix} 1 \\ 1\end{pmatrix}+c_2e^{t}\begin{pmatrix} -2 \\ 1\end{pmatrix}+\textbf{a}t+\textbf{b}

The c_1 and c_2 are needed since any linear combination of your two independent homogeneous solutions will also be a solution to the homogeneous equation. And I don't think you need a t^2 term for your particular solution.

If you just have at + b then you'll just have a when you take the derivative. Ultimately I'm trying up an x whose derivative looks something like it does in the problem. Right? (t t)^T = t(1 1)^T.

 

[gabbagabbahey]

Right, but \begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}\textbf{x}(t) will also have a t in it which should cancel the at for an appropriate choice of a.

 

[Jamin2112]

Right, but \begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}\textbf{x}(t) will also have a t in it which should cancel the at for an appropriate choice of a.

a and b have constant entries, right?

 

[gabbagabbahey]

a and b have constant entries, right?

They'd better!

 

[gabbagabbahey]

Keep in mind, that if you get an equation like Ma=-a for some matrix M, either -1 is one of its eigenvalues or a=0

 

[Jamin2112]

Keep in mind, that if you get an equation like Ma=-a for some matrix M, either -1 is one of its eigenvalues or a=0

Thanks, buddy. One more quick question!

On problem 3, I have a 3x3 matrix eigenvalues -2,-1 and corresponding eigenvectors (1 0 0)^T and (0 1 0)^T. Now, I know what with a 2x2 matrix, if I have only one eigenvector, giving me a solution x=e^at$, then I guess a second solution x=te^at$ + e^at#. With a 3x3 matrix, do I try that with each of my two eigenvectors? Doing so seems to give me a solution for one but not the other.

 

[gabbagabbahey]

The easiest way to solve problem 3 is to not do it in matrix form...just let \textbf{x}(t)=\begin{pmatrix}x_1(t) \\ x_2(t) \\ x_3(t) \end{pmatrix} and solve the 3 DEs you get.