I have an ODE of the form X" + aX' + (bc)^2 X = 0.
Is the correct way of solving this to try an exponential solution of the form Ae^rx?
Yes, a,b and c are constants, sorry I forgot to put that in.
It becomes a quadratic equation:
r^2 + ar + (bc)^2 = 0.
The quadratic formula gives
r = [-a + sqrt(a^2 - 4(bc)^2)]/2 and
r = [-a - sqrt(a^2 - 4(bc)^2)]/2.
I note that there will be two independent solutions when the discrimant is positive, that is, when
a^2 - 4(bc)^2 > 0, or
a^2 > 4(bc)^2, or
|a| > |2bc|