Question 3.12.1: Solve Dx + (D + 2)y = 0 (D - 3)x- 2y = 0. (6)...

Operating on the first equation by D-3 and on the second by D and then subtracting eliminates x from the system. It follows that the differential equation for y is

[(D-3)(D+2)+2 D] y=0 \quad \text { or } \quad\left(D^{2}+D-6\right) y=0.

Since the characteristic equation of this last differential equation is m^{2}+m-6= (m-2)(m+3)=0, we obtain the solution

y(t)=c_{1} e^{2 t}+c_{2} e^{-3 t}.  (7)

Eliminating y in a similar manner yields \left(D^{2}+D-6\right) x=0, from which we find

x(t)=c_{3} e^{2 t}+c_{4} e^{-3 t}.  (8)

As we noted in the foregoing discussion, a solution of (6) does not contain four independent constants. Substituting (7) and (8) into the first equation of (6) gives

\left(4 c_{1}+2 c_{3}\right) e^{2 t}+\left(-c_{2}-3 c_{4}\right) e^{-3 t}=0.

From 4 c_{1}+2 c_{3}=0 and -c_{2}-3 c_{4}=0 we get c_{3}=-2 c_{1} and c_{4}=-\frac{1}{3} c_{2}. Accordingly, a solution of the system is

x(t)=-2 c_{1} e^{2 t}-\frac{1}{3} c_{2} e^{-3 t}, \quad y(t)=c_{1} e^{2 t}+c_{2} e^{-3 t}.