A final note on complex numbers: if p(z) = a_n z^n + ... + a_1 z + a_0 is a polynomial whose coefficents are real, then the conjugate of p(z) is p(bar-z) (since conjugation commutes with sum and product, and the conjugate of a_k is a_k since it's real). Thus if r is a root of p, then so is its conjugate bar-r. Thus roots of p are either real or come in conjugate pairs. If the degree of p (which is n if a_n is not zero) is odd, then there must be at least one real root.
Remember that we've decided that we should accept complex-valued solutions to our differential equations, if they come along. (The natural consequence is that we might as well consider ODEs with complex coefficients, but we won't do that much in this course.) If the coefficients are real, then the conjugate of a solution is again a solution - the same if we started with a real solution, but a different one if not. But the real and imaginary parts of a complex number (or function) are linear combinations of the number and its conjugate, so they are also solutions, but now real.
We have a swinging door, controlled by the ODE
y" + py' + q = 0. (*)
The roots of its characteristic polynomial r^2 + pr + q are
r1, r2, given by r = -p/2 +- sqrt((p/2)^2-q) .
Let omega = sqrt(abs((p/2)^2 - q).
If omega is nonzero then the roots are distinct, and we have the linearly independent exponential solutions ("fundamental modes") e^{r1 x} and e^{r2 x}. The general solution is
a e^{r1 x} + b e^{r2 x} for (possibly complex) a, b constants.
If p^2 is greater than q, then the roots are real and we are happy.
If p^2 is less than q, then the roots are not real and we are less happy. We still have our two fundamental modes but they are complex valued functions:
y1 = e^{-px/2} (cos(omega x) + i sin(omega x))
y2 = e^{-px/2} (cos(omega x) - i sin(omega x))
They are complex conjugates. The real and imaginary parts of the first,
e^(-px/2} cos(omega x) and e^{-px/2} sin(omega x) ,
are again a pair of linearly independent solutions, but now are real. The general solution is thus
e^{-px/2} (a cos(omega x) + b sin(omega x)).
For example, perhaps we have y" + 4y + 5 = 0. The roots are -2 + i and -2 - i so our basic real solutions are
e^{-2x} cos(x) and e^{-2x} sin(x).
I graph them and point out that we are getting damped oscillation, as expected.
There is the final possibility that p^2 = 4q, in which case we have a single root -p/2 which is real. We get only one exponential solution, e^{-px/2}, and must look elsewhere for a second linearly independent solution. The example y" = 0 shows the way: the exponential solution is 1 , but x is another (nonexponential) solution. In general, we have the two solutions
e^{-px/2} and x e^{-px/2},
and so the general solution is
e^{-px/2} (a + bx) .
Summary: the behavior of solutions of (*) is controlled by the position of the roots r1 and r2 of the characteristic polynomial. Notice that
r^2 + pr + q = (r-r1)(r-r2) gives p = -(r1+r2), q = r1 r2 .
If p and q are both positive then the real parts of both roots are positive: the "damped" situation, in which case all solutions tend to zero as x goes to infinity. This divides further:
p^2 less than 4q : nonreal roots; "underdamped"; damped oscillation.
p^2 greater than 4q : real roots; "overdamped"; nonscillatory decay.
p^2 = 4q : negative real repeated root: "critically damped": nonoscillatory decay.
If p is negative and q is positive, then all nonzero solutions take on arbitrarily large values as x goes to infinity. This case divides further:
p^2 less than 4q : nonreal roots; "anti-damped" oscillation.
p^2 greater than 4q : real roots: nonoscillatory growth.
p^2 = 4q : positive real repeated root: nonoscillatory growth.
If q is negative then there are two real roots of opposite sign. Multiples of one of the fundamental modes go exponentially to 0, but all other solutions go exponentially to infinity.
If q = 0, one root is zero and we have constant solutions.