How do I solve the following system of differential equations
- $x' = a_{1}x+b_{1}y+c_{1}z$;
- $y' = a_{2}x+b_{2}y+c_{2}z$;
- $z' = a_{3}x+b_{3}y+c_{3}z$.
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:
- $rAe^{rt} = a_{1}Ae^{rt}+b_{1}Be^{rt}+c_{1}Ce^{rt}$;
- $rBe^{rt} = a_{2}Ae^{rt}+b_{2}Be^{rt}+c_{2}Ce^{rt}$;
- $rCe^{rt} = a_{3}Ae^{rt}+b_{3}Be^{rt}+c_{3}Ce^{rt}$.
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.