# First-Order Linear Differential Problem

1. May 31, 2012

### Hiche

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

Solve the following IVP:

$X' = \begin{pmatrix}2 & -1\\3 & -2\end{pmatrix}X + \begin{pmatrix}0\\t\end{pmatrix}$ with $X(0) = \begin{pmatrix}1\\0\end{pmatrix}$

2. Relevant equations

3. The attempt at a solution

The eigenvalue corresponding to $\begin{pmatrix}2 & -1\\3 & -2\end{pmatrix}$ is $\lambda = 0$. We find that $X_c = c_1\begin{pmatrix}1\\2\end{pmatrix} e^{0t}$. Now in order to find $X_p$, how exactly is the right way? I took $X_p = \begin{pmatrix}a_1\\b_1\end{pmatrix}t$ and wanted to find $a_1$ and $b_1$. Right or wrong?

2. May 31, 2012

### tiny-tim

Hi Hiche!
Nooo

3. May 31, 2012

### Hiche

oh, crap! Apparently, 3 * 1 = 4 -_-

So, again, the eigenvalues are $\lambda_1 = -1$ and $\lambda_2 = 1$. I hope this is correct. So the solution of $X_c = c_1\begin{pmatrix}1\\1\end{pmatrix}e^t + c_2\begin{pmatrix}1\\3\end{pmatrix}e^{-t}$.

Now about $X_p$. Is my method correct (first post)?

4. May 31, 2012

### tiny-tim

(i'm not sure, but…) i'd be inclined to go for $X_p = \begin{pmatrix}a_0\\b_0\end{pmatrix} + \begin{pmatrix}a_1\\b_1\end{pmatrix}t$

5. May 31, 2012

### Hiche

Okay, so upon a little work, $X_p = \begin{pmatrix}1\\2\end{pmatrix}t$ and the general solution is $X = X_c + X_p$

Thank you, tiny-tim.

6. Jun 1, 2012

### tiny-tim

But $X_p = \begin{pmatrix}1\\2\end{pmatrix}t$ isn't a solution …

Xp' = (1,2), and Xp(0) = (0,0)