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{\Large\bf Reduction of Order for Linear Second-Order ODE}

\vspace{.3in}

The Reduction of Order technique is a method for
determining a second linearly independent solution to a
homogeneous second-order linear ode given a first
solution. This section has the following: 

\begin{itemize}
\item Example 
\item General Solution Procedure
\end{itemize}

\vspace{.3in}

{\bf Example}

\vspace{.2in}

It is best to describe the procedure with a concrete
example. Consider the linear ode
$$
y''(t)+6y'(t)+9y(t)=0 \hspace{.2in}(*)
$$
We know that a solution to this problem is y1=exp(-3t).
To obtain the general solution we need a second linearly
independent solution to the problem.

We find the second solution by assuming 
$$
y_2(t)=v(t)e^{-3t}
$$
where v(t) is an unknown function. We now substitute
this into the original ode (*) and derive a new ode
for v(t). We have

$$
y'_2=v'e^{-3t}-3e^{-3t}v 
$$

and

$$
y''_2=v''e^{-3t}-6v'e^{-3t}+9ve^{-3t}
$$

Substituting these expressions into (*), we obtain

$$
(v''e^{-3t}-6e^{-3t}v'+9e^{-3t}v)+6(v'e^{-3t}-3e^{-3t}v) 
+6e^{-3t}v=0
$$

Many terms cancel in the above expression. We obtain

$$
v''e^{-3t} = 0 
$$

Either v''=0 or exp(-3t)=0. The second possibility only
occurs if t=infinity and this is a degenerate case.
We must have v''=0. In words: the second derivative of
v is 0. This means that v is a line: v''=At+B, where 
A and B are constants. Hence,
a second solution to the original ode (*) is

$$
y_2(t)=v(t)e^{-3t}=(At+B)e^{-3t}
$$

How do we choose A and B? Recall, our goal is determine a
second linearly independent solution to the original ode (*).
The first solution is y1=exp(-3t). Suppose we set A=0.
Then y2=Bexp(-3t). In this case, y1 and y2 are multiples
of each other, and are linearly {\it dependent}. On the
other hand, suppose we choose B=0. Then y2=Atexp(-3t).
In this case y1=exp(-3t) and y2=Atexp(-3t) are indeed
linearly independent. As long as A is nonzero, y1 and
y2 are linearly independent. It makes sense to set
B=0, since if

$$
y_2=Ate^{-3t}+Be^{-3t}
$$

the term Bexp{-3t} is the same as y1=exp(-3t) and adds
{\it nothing new}. 

To make a long story short(!), we choose A=1 and B=0.
The second linearly independent solution to (*) is
y2=texp(-3t) and the general solution to (*) is

$$
y(t)=C_1e^{-3t}+C_2te^{-3t}
$$

\vspace{.3in} 

{\bf General Solution Procedure}

\vspace{.2in}

Let us now consider a general homogeneous linear ode:
$$
y''(t)+p(t)y'(t)+q(t)y(t)=0 \hspace{.2in} (**)
$$
and suppose f(t), which is known, is one solution of the ode.
That means
$$
f''(t)+p(t)f'(t)+q(t)f(t)=0
$$
We assume the second solution has the form:
$$
y_2=v(t)f(t)
$$
where v(t) is unknown. We now substitute y2 into the
original ode (**). We have
$$
y'_2=v'f+vf' \hspace{.2in} y''_2=v''f+2v'f'+vf''.
$$

Substituting these expressions into (**), we have
$$
v''f+2v'f'+vf''+p(v'f+vf')+qvf=0
$$
Rearranging terms, we have
$$
v(f''+pf'+qf)+v''f+2v'f'+pv'f=0
$$
The term in parentheses is 0, since f(t) is a solution to
(**). Hence, we are left with 
$$
v''f+2v'f'+pv'f=0
$$
Dividing by f, we have
$$
v''(t)=-v'(t)(2\frac{f'(t)}{f(t)}+p(t))
$$
This is a linear second-order ode where v'' depends on v'
and t only and can be solved. It can be shown that 
$$
v=\int\frac{u(t)}{[f(t)]^2}dt \hspace{.2in} (***)
$$
where 
$$
u(t)=e^{-\int p(t)dt}\hspace{.2in}(****)
$$
It follows that  
$$
y_2=f(t)\int\frac{u(t)}{[f(t)]^2}dt 
$$

\vspace{.3in}

{\bf Example}

\vspace{.2in}

Consider the ode:
$$
y''-ty'+y=0
$$
It can be shown that y1(t)=f(t)=t is a solution. We can
apply formulas (****) and (***) to find the second solution.
Here p(t)=-t. We have
$$
u(t)=e^{-\int (-t)dt}=e^{\int t dt}=e^{t^2/2}
$$
Substituting this into (***), we have
$$
v=\int\frac{e^{t^2/2}}{t^2}dt 
$$
This integral cannot be expressed in terms of
elementary functions. The second linearly
independent solution is
$$
y_2(t)=t\int\frac{e^{t^2/2}}{t^2}dt  
$$
The general solution to the ode is
$$
y(t)=C_1y_1(t)+C_2y_2(t)
$$

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