I have the system:
$\left[\begin{array}{@{}c@{}}
x' \\
y'
\end{array} \right]= \left[\begin{array}{@{}c@{}}
3&2 \\
-2 & -1
\end{array} \right]\left[\begin{array}{@{}c@{}}
x' \\
y'
\end{array} \right]+\left[\begin{array}{@{}c@{}}
2e^{-t} \\
e^{-t}
\end{array} \right]$
Which I should solve using the fundamental matrix.
So I start with obtaining the homogenous solution:
I find the eigenvalues;
\begin{pmatrix}
3-\lambda&2 \\
-2 & -1-\lambda
\end{pmatrix}
which gives the determinant: $\lambda^2-2\lambda+1=0$. Thus $\lambda_1=1$. Plugging that in the matrix in the original equation, I get that x=y. So a solution to the homogenous system would be: $y_h=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
1
\end{array} \right]$
Since there is no second solution to the determinant, I would ideally form the fundamental matrix:
\begin{pmatrix}
e^{t} & e^0 \\
e^{t} & e^0
\end{pmatrix}
but this is to no avail. So how do I find the solution of this nonhomogenous system using the fundamental matrix with one eigenvalue?
Thanks
UPDATE:
I set up the generalized eigenvector formula
\begin{equation}
v_2(A-\lambda I)=v_2
\begin{pmatrix}
3-\lambda&2 \\
-2 & -1-\lambda
\end{pmatrix}=v_1
\end{equation}
\begin{equation}
v_2(A-\lambda I)=v_1=
\begin{vmatrix}
3-\lambda&2 & | 1 \\
-2 & -1-\lambda & |-1
\end{vmatrix}
\end{equation}
I now get as given by Moo, with Gaussian elimination, the matrix:
\begin{equation}
\begin{vmatrix}
1 &1 & | 1/2 \\
0 & 0 & |0
\end{vmatrix}
\end{equation}
and have the second eigenvector: $e_2=e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]$
.
So the homogeneous solution is:
\begin{equation}
y_h=e^{\lambda_1 t}e_1+e^{\lambda_2t}e_2=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
-1
\end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]
\end{equation}
At this stage, it remains to find the particular solution. We know that it must be in the form of:
\begin{equation}
y_p=Ce^{-t}
\end{equation}
and thus the general solution is:
\begin{equation}
y_p=y_h+Ce^{-t}=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
-1
\end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]+Ce^{-t}
\end{equation}
But can this be said?
Best Answer
You have what is called a deficient matrix, so you need to find a generalized eigenvector.
We have the system
$$\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]= \left[\begin{array}{@{}c@{}} 3&2 \\ -2 & -1 \end{array} \right]\left[\begin{array}{@{}c@{}} x \\ y \end{array} \right]+\left[\begin{array}{@{}c@{}} 2e^{-t} \\ e^{-t} \end{array} \right]$$
We find a repeated eigenvalue of $\lambda_{1,2} = 1$ and we can find a single eigenvector of
$$v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$
Finding generalized eigenvectors is not a simple topic and requires work to learn the ins and outs, but in this case, we will use this example.
Solve for $v_2$ using the row-reduced-echelon-form (RREF) of $[A-\lambda I]v_2 = [A -I]v_2 = v_1 $
We get the augmented matrix
$$ \left[\begin{array}{rr|r} 1 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] $$
We can choose
$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$
Update For the eigenvalues, we find
$$|A - \lambda I| = \begin{vmatrix} -\lambda +3 & 2 \\ -2 & -\lambda -1 \\ \end{vmatrix} = (-\lambda+3)(-\lambda - 1) -2(-2) = \lambda ^2-2 \lambda +1 = 0$$
This results in
$$\lambda_{1, 2} = 1$$
To find the generalized eigenvector, we solve (you are actually using the eigenvalue $\lambda = 1$ below)
$$[A - \lambda I]v_2 = [A - 1 I]v_2 = \begin{bmatrix} 2 & 2 \\ -2 & -2 \\ \end{bmatrix}v_2 = v_1 = \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}$$
That is
$$\begin{bmatrix} 2 & 2 \\ -2 & -2 \\ \end{bmatrix}v_2 = \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}$$
As an augmented matrix, this is
$$ \left[\begin{array}{rr|r} 2 & 2 & -1 \\ -2 & -2 & 1 \end{array}\right] $$
The RREF (Gaussian Elimination) is
$$ \left[\begin{array}{rr|r} 1 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] $$
From this, we can choose
$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$