Euler-Cauchy ODE
The general form of a homogeneous Euler-Cauchy ODE is
where p and q are constants. The coefficients of y' and y are discontinuous at t=0. This means that the solution to the differential equation may not be defined for t=0. The existence and uniqueness theory states that a solution exists on any interval (a,b) not containing t=0.
Recall that the general solution of the ode is
where C1 and C2 are constants and y1(t) and y2(t) are two linearly independent solutions of the ode.
Solution Procedure
We first multiply the differential equation by t^2 to get
We now make the guess:
where r is a constant to be determined. The reason for this guess will become clear below. With this guess we have
Substituting these expressions into the ode, we have
This equation is satisfied if t^r=0 or r(r-1)+pr+q=0. The condition t^r=0 is satisfied if t=0 or t=infinity. These are considered as degenerate cases and are neglected. Hence, r must satisfy
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 |t|^a and |t|^b 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(r-1) - 2r -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(t)=|t|^a 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 t^a log(|t|) is also a solution of the ode. The general solution is
Consider the following example:
The characteristic polynomial is r(r-1) + 7r + 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.
Note that
Here we have used Euler's identity. Hence,
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 r(r-1) - 6r + 13. Using the quadratic formula, we find that the roots are 3 + 2i and 3 - 2i. Hence, the general solution can be written:
