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:

It is best to describe the procedure with a concrete
example. Consider the

linear ode

We know that a solution to this problem is y_1=exp(-3t).
To obtain the

general solution we need a second linearly
independent solution to the

problem.

We find the second solution by assuming

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

and

Substituting these expressions into (*), we obtain

Many terms cancel in the above expression. We obtain

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

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 y_1=exp(-3t).

Suppose we set A=0.
Then y_2=Bexp(-3t). In this case, y_1 and y_2 are

multiples
of each other, and are linearly *dependent*. On the
other hand,

suppose we choose B=0. Then y_2=Atexp(-3t).
In this case y_1=exp(-3t)

and y_2=Atexp(-3t) are indeed
linearly independent. As long as A is

nonzero, y_1 and
y_2 are linearly independent. It makes sense to set
B=0,

since if

the term Bexp-3t is the same as y_1=exp(-3t) and adds
*nothing new*.

To make a long story short(!), we choose A=1 and B=0.
The second linearly

independent solution to (*) is
y_2=texp(-3t) and the general solution to (*) is

Let us now consider a general homogeneous linear ode:

and suppose f(t), which is known, is one solution of the ode. That means

We assume the second solution has the form:

where v(t) is unknown. We now substitute y_2 into the
original ode (**).

We have

Substituting these expressions into (**), we have

Rearranging terms, we have

The term in parentheses is 0, since f(t) is a solution to
(**). Hence, we

are left with

Dividing by f, we have

This is a linear second-order ode where v'' depends on v'
and t only and

can be solved. It can be shown that

where

It follows that

Consider the ode:

It can be shown that y_1(t)=f(t)=t is a solution. We can
apply formulas

(****) and (***) to find the second solution.
Here p(t)=-t. We have

Substituting this into (***), we have

This integral cannot be expressed in terms of
elementary functions. The

second linearly
independent solution is

The general solution to the ode is

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