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{\Large\bf Euler-Cauchy ODE}

\vspace{.3in}

The general form of a homogeneous Euler-Cauchy ODE is  

$$
y''+\frac{p}{t}y'+\frac{q}{t^2}y=0
$$

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
$$
y(t)=C_1y_1(t)+C_2y_2(t)
$$
where C1 and C2 are constants and y1(t) and y2(t) are
two linearly independent solutions of the ode.

\vspace{.3in}

{\bf Solution Procedure}

\vspace{.2in}

We first multiply the differential equation by t^2 to get
$$
t^2y''+pty'+qy=0
$$
We now make the guess: 
$$
y=t^r
$$
where r is a constant to be determined. The reason for
this guess will become clear below. With this guess
we have
$$
y'=rt^{r-1} \hspace{.2in} y''=r(r-1)t^{r-2}
$$

Substituting these expressions into the ode, we have 
$$
t^2r(r-1)t^{r-2}+ptrt^{r-1}+qt^r=t^r(r(r-1)+pr+q)=0
$$
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
$$
F(r)=r^2+pr+q=0
$$
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: 

\begin{itemize}
\item Characteristic polynomial has distinct roots 
\item Characteristic polynomial has a double root
\item Characteristic polynomial has complex conjugate roots  
\end{itemize}

\vspace{.3in}

{\bf Characteristic Polynomial has Distinct Roots}

\vspace{.2in}

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:
$$
y(t) = C_1|t|^a+C_2t^b
$$

Consider the following example:
$$
y''-2y'-18y=0
$$
The characteristic polynomial is r(r-1) - 2r -18 = (r - 6)(r +3),
which has roots 6 and -3. 
$$
y(t)=C_1t^6+C_2t^{-3}
$$

\vspace{.3in}

{\bf Characteristic Polynomial has a Double Root}

\vspace{.2in}

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 
$$
y(t)=C_1|t|^a+C_2|t|^a\log(|t|)
$$

Consider the following example: 
$$
y'' + 7y' + 9y = 0.
$$
The characteristic polynomial is r(r-1) + 7r + 9 = (r + 3)^2,
which has a double root -3. The general solution is 
$$
y(t) = C_1|t|^{-3}+C_2|t|^{-3}\log(|t|)
$$

\vspace{.3in}

{\bf Complex-Conjugate Roots}

\vspace{.2in}

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:
$$
y(t) = C_1|t|^{a+ib}+C_2t^{a-ib} 
$$
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 
$$
|t|^{ib}=e^{ib\log|t|}=\cos(b\log|t|)+i\sin(b\log|t|)
$$
Here we have used Euler's identity. Hence,
$$
|t|^{a+ib}=|t|^a|t|^{ib}=|t|^a(\cos(b\log|t|)+i\sin(b\log|t|))
$$
Similarly, we have
$$
|t|^{a-ib}=|t|^a|t|^{-ib}=|t|^a(\cos(b\log|t|)-i\sin(b\log|t|))
$$

Substituting, these two expressions into the general solution
we have
$$
y(t) = C_1|t|^a(\cos(b\log|t|)+i\sin(b\log|t|)) + 
C_2|t|^a(\cos(b\log|t|)-i\sin(b\log|t|))
$$
Choose C_1=0.5 and C_2=0.5. This yields the solution:
$$
y(t)=|t|^a}\cos(b\log|t|)
$$
Now choose C_1=i0.5 and C_2=-i0.5. This yields the solution:
$$
y(t)=|t|^a\sin(b\log|t|)
$$
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: 
$$
y(t) = D_1|t|^a\cos(b\log|t|) + D_2|t|^a\sin(b\log|t|)
$$
for real numbers D_1 and D_2 are constants.

Consider the example:

$$
y'' - 6y' + 13 = 0
$$

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:
$$
y(t)=D_1|t|^3\cos(2\log|t|)+D_2|t|^3\sin(2\log|t|)
$$

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