# Solving x'=Ax+g using undetermined coefficients

1. May 8, 2010

### Jamin2112

1. The problem statement, all variables and given/known data

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

2. Relevant equations

?

3. 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.

2. May 8, 2010

### gabbagabbahey

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

3. May 8, 2010

### Jamin2112

Fine. det(AI)=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 e4t(1 1)T + et(-2 1)T.

Maybe I assume the full solution is e4t(1 1)T + et(-2 1)T + at2 + bt + c????

4. May 8, 2010

### gabbagabbahey

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

5. May 8, 2010

(see edit)

6. May 8, 2010

### gabbagabbahey

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.

7. May 8, 2010

### Jamin2112

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.

8. May 8, 2010

### 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.

9. May 8, 2010

### Jamin2112

a and b have constant entries, right?

10. May 8, 2010

### gabbagabbahey

They'd better!

11. May 8, 2010

### 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

12. May 8, 2010

### Jamin2112

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=eat$, then I guess a second solution x=teat$ + eat#. 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.

13. May 8, 2010

### 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.