Constant-Coefficient Homogeneous ODE
A constant-coefficient homogeneous second-order ode can be put in the form
where p and q are constants. Recall that the general solution is
where C_1 and C_2 are constants and y_1(t) and y_2(t) are any two linearly independent solutions of the ode. Our goal is to find two linearly independent solutions of the ode.
For reasons that become clear below, we try a solution of the form y=exp(rt), where r is an unknown constant. We must find r. If y=exp(rt), then y'=r(exp(rt)) and y''=r^2(exp(rt)). Substituting these into the ode, we have
This equation is satisfied if exp(rt)=0 or r^2+pr+q=0. The condition exp(rt)=0 is satisfied only if rt is negative infinity. This is considered a degenerate case and is neglected. Hence, r must satisfy the equation
F(r) is called the characteristic polynomial. Solving the original ode is reduced to solving an algebraic equation.
Assuming the coefficients p and q are real numbers, there are three cases to consider:
Characteristic Polynomial has Distinct Roots
Suppose that the characteristic polynomial has two distinct, real roots (call them a and b). Then exp(at) and exp(bt) are linearly independent solutions to the original ode and the general solution to the ode is:
Consider the following example:
The characteristic polynomial is r^2 - 3r -18 = (r - 6)(r +3), which has roots 6 and -3.
Characteristic Polynomial has a Double Root
Suppose that the characteristic polynomial has a double root (call it a). y_1=exp(at) is one solution to differential equation. We need a second linearly independent solution to the ode to get the general solution. Using the technique of reduction of order, it can be shown that texp(at) is also a solution of the ode. The general solution is
Consider the following example:
The characteristic polynomial is r^2 + 6r + 9 = (r + 3)^2, which has a double root -3. The general solution is
Complex-Conjugate Roots
Suppose that the characteristic polynomial has complex roots a+ib and a-ib, where a and b are real. These are distinct roots, so the the general solution can be written:
The problem with writing the solution in this form is that it involves complex-valued functions. It is possible to re-express the general solution in terms of two linearly independent real-valued functions.
Recall, Euler's identity:
Using this identity, we have:
Similarly, we have
Substituting, these two expressions into the general solution we have
Choose C_1=0.5 and C_2=0.5. This yields the solution:
Now choose C_1=i0.5 and C_2=-i0.5. This yields the solution:
Both of these functions are solutions to the original ode. In addition, they are linearly independent, since they are not multiples of each. Hence, any solution to the ode can be expressed in terms of these function. So we can write:
for real numbers D_1 and D_2 are constants.
Consider the example:
The characteristic polynomial is 
. Using the
quadratic formula, we find that the roots are 3 + 2i and 3 - 2i.
Hence, the general solution can be written:
