# Finding the general solution of a system of differential equations

by chris_0101
Tags: differential, equations, solution
 P: 71 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
 PF Patron Sci Advisor Thanks Emeritus P: 38,424 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 .
 P: 71 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

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