How do I solve the following system of differential equations

  1. $x' = a_{1}x+b_{1}y+c_{1}z$;
  2. $y' = a_{2}x+b_{2}y+c_{2}z$;
  3. $z' = a_{3}x+b_{3}y+c_{3}z$.

Do I need any specific conditions here?

First, I guess a solution, $x(t) = Ae^{rt}$, $y(t) = Be^{rt}$, and $z(t) = Ce^{rt}$.

Then substitute the solution in the system of DEs:

  1. $rAe^{rt} = a_{1}Ae^{rt}+b_{1}Be^{rt}+c_{1}Ce^{rt}$;
  2. $rBe^{rt} = a_{2}Ae^{rt}+b_{2}Be^{rt}+c_{2}Ce^{rt}$;
  3. $rCe^{rt} = a_{3}Ae^{rt}+b_{3}Be^{rt}+c_{3}Ce^{rt}$.

Then the put the system of equations in a matrix:

$$\begin{bmatrix} a_{1}-r &b_{1} &c_{1} \\ a_{2}&b_{2}-1 &c_{2} \\ a_{3}&b_{3} &c_{3}-r \end{bmatrix} \begin{bmatrix} A\\ B\\ C \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$.

The determinant of the 3X3 matrix yields the characteristic equation and after skipping some steps, we get:

$$r^{3} - (a_{1}+b_{2}+c_{3})r^{2} - (-a_{1}b_{2}-a_{1}c_{3}-b_{2}c_{3}+b_{3}c_{2}+a_{3}c_{1}+a_{2}b_{1})r-a_{1}b_{2}c_{3}+a_{1}b_{3}c_{2}+a_{3}b_{2}c_{1}+a_{2}b_{1}c_{3}-a_{3}b_{1}c_{2}-a_{2}c_{1}b_{3} = 0.$$

Now, where do I go from here, or is there a simpler way of doing the problem?

EDIT: Tried doing with the eigenvalues and eigenvalues but get the same thing and can't solve for the eigenvalues.

  • 1
    $\begingroup$ I'd say you're supposed to give the answer as a function of the eigenvalues of the associated matrix (which you'll just label $\lambda_1, \lambda_2, \lambda_3$ or something). Otherwise this is hell. $\endgroup$ – Git Gud Feb 8 '15 at 19:20
  • $\begingroup$ @GitGud We weren't even taught about the eigenvalues. $\endgroup$ – OGC Feb 8 '15 at 19:29
  • 1
    $\begingroup$ You may want to have a look at this. $\endgroup$ – science Feb 8 '15 at 19:45
  • 1
    $\begingroup$ I'm much more interested in the solution your instructor has in mind. In any case my solution would be a small generalization of this. $\endgroup$ – Git Gud Feb 8 '15 at 20:22
  • 2
    $\begingroup$ I'd just let $\mathbf{v}(t)= \exp(t\mathbf{A})\mathbf{v}(0)$, and leave it at that. $\endgroup$ – user14717 Feb 8 '15 at 20:42

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