# Finding the general solution of a system of differential equations

1. Apr 6, 2011

### chris_0101

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

2. Relevant equations
I really don't know what to put here but my method is:
-Find det(A-$$\lambda$$I)
-Find the roots of the determinant - which are the eigenvalues
-Solve for (X -($$\lambda$$)I)

I am stuck at this part

3. The attempt at a solution

So I managed to find the determinant, which is:
$$\lambda$$^3 -5$$\lambda$$^2 -3$$\lambda$$ +9

The Eigenvalues are:
$$\lambda$$1 = 1
$$\lambda$$2 = -3
$$\lambda$$3 = -3

Now I'm trying to solve for the eigenvector of eigenvalue 1, however manipulating the matrix does not yield anything that I can use to solve for k1 k2 and k3

Any help with this will be greatly appreciated,

Thanks

2. Apr 6, 2011

### HallsofIvy

Staff Emeritus
An eigenvector $$\begin{pmatrix}x \\ y \\ z\end{pmatrix}$$ corresponding to eigenvalue 1 must, of course, satisfy
$$\begin{pmatrix}1 & -1 & 0 \\ 0 & -1 & -2 \\ 0 & 2 & -5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z\end{pmatrix}$$

$$\begin{pmatrix}x - y \\ -y- 2z \\ 2y- 5z \end{pmatrix}= \begin{pmatrix}x \\ y \\ z\end{pmatrix}$$
Which means we must solve x- y= x, -y- 2z= y, 2y- 5z= z. From the first, equation, subtracting x from both sides, -y= 0 so y= 0. Putting that into both of the other equations, z= 0. Therefore, any eigenvector must be of the form <x, 0, 0>.

3. Apr 6, 2011

### chris_0101

Hey, thanks for the response, I now understand what you have mentioned. Now I have run into another issue, finding the third eigenvector. I managed to find the second eigenvector for eigenvalue2, which is <1,4,4>. Now the third eigenvalue is the same as the second so I would assume that the eigenvector will be the same however this is not the case. Using Wolfram alpha, the third eigenvector is <0,0,0> and I don't know how to get this eigenvector. Any help would be great,
Thanks