# Solve system of first order differential equations

I have to solve differential systems like this: $$\left\{ \begin{array}{c} x' = 3x - y + z \\ y' = x + 5y - z \\ z' = x - y + 3z \end{array} \right.$$

Until now I computed the eigenvalues $k = \{2,4,5\}$ by solving the equation resulted from this determinant of this matrix: $$\begin{pmatrix} 3 && -1 && 1 \\ 1 && 5 && -1 \\ 1 && -1 && 3 \end{pmatrix} - kI_3 = \begin{pmatrix} 3-k && -1 && 1 \\ 1 && 5-k && -1 \\ 1 && -1 && 3-k \end{pmatrix}$$ I don't know what to do next.

NOTE: This is an example but from it I want to learn the method to solve any system of this kind. I learn better from particular examples than directly from generalization.

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Now you need to find eigenvectors. – Artem May 29 '13 at 20:09

## 2 Answers

To find the eigenvalues of: $$A =\begin{bmatrix}3 & -1 & 1\\1 & 5 & -1\\1 & -1 &3\end{bmatrix},$$

we set up $|A - \lambda I| = 0$ and solve the characteristic polynomial, so we have:

$$|A -\lambda I| = \begin{bmatrix} 3-\lambda & -1 & 1\\1 & 5-\lambda & -1\\1 & -1 &3 -\lambda \end{bmatrix} = 0$$

From this, we get the characteristic polynomial as: $$-\lambda^3+11 \lambda^2-38 \lambda+40 = -(\lambda-5) (\lambda-4) (\lambda-2)= 0$$

This gives us three eigenvalues: $λ_1 = 2, λ_2 = 4$ and $λ_3 = 5$

To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$, so lets do $\lambda_1 = 2$. We have:

$$[A - \lambda_i I]v_i = \begin{bmatrix} 3-2 & -1 & 1\\1 & 5-2 & -1\\1 & -1 &3 -2 \end{bmatrix}v_1 = \begin{bmatrix} 1 & -1 & 1\\1 & 3 & -1\\1 & -1 & 1 \end{bmatrix}v_1 = 0$$

This leads to the row-reduced-echelon form:

$$\begin{bmatrix} 1 & 0 & \dfrac{1}{2}\\0 & 1 & -\dfrac{1}{2}\\0 & 0 & 0 \end{bmatrix}v_1 = 0$$

This gives us a solution of:

$b = \dfrac{1}{2}c \rightarrow c = 2 \rightarrow b = 1$, and $a = -\dfrac{1}{2} c \rightarrow a = -1$.

Thus the eigenvector for the eigenvalue $\lambda_1 = 2$ is $v_1 = (-1, 1, 2)$.

If we repeat this process two more times for the other two eigenvalues, we end up with the eigenvalue/eigenvector pairs:

$$\lambda_1 = 2, ~v_1 = (-1, 1, 2)$$

$$\lambda_2 = 4, ~v_2 = (1, 0, 1)$$

$$\lambda_3 = 5, ~v_3 = (1, -1, 1)$$

Note that different approaches are needed when we get into repeated eigenvalues, but that is for another problem.

Try to derive those last two eigenvectors and report back if they do not work out.

To write out the solution for the three equation, we would use a linear combination of the the eigenvalues, eigenvectors and some unknown constants, as:

$$X(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix} = c_1e^{2t} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_3 e^{5t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$

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+1: excellent write up and presentation! – amWhy May 30 '13 at 0:17
+Amazoti thank you for your response. I will digest it tonight. I will work on what you suggested and also I will try to solve another system from alpha to omega to see if I understood all steps. – Dr.Optix May 30 '13 at 15:52
@Dr.Optix: You are very welcome. If you have any problems with the above or with additional problems, just give a yell! Regards – Amzoti May 30 '13 at 16:02
Wow...you can spot a challenge, indeed! Yes...very slow...not much reviewing available to do, either! – amWhy May 31 '13 at 3:39
+1 for my dear friend. Please have a look at this. What do u think? Thanks – Babak S. May 31 '13 at 7:54

The idea here is just like a single ODE with constant coefficients: Assume the solutions are given by $x = \exp(\alpha t)$, $y = \exp(\beta t)$, $z = \exp(\gamma t)$. Your system is linear, so you can mix and match solutions just like a single ODE. Substitute, and you'll get a set of equations for the $\alpha, \beta, \gamma$, and take it from there.

Or you could differentiate the first, and substitute the derivatives from the others, and keep going until you get a single third order ODE. Solve that, and work backwards. The result will be exactly the same as above, the cubic to solve will be the same too. Just different paths. (Yes, I did prove this by sheer curiosity when I was at school.)

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