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Reduction of Order in a Differential Equation: Example 3
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Given that
is a solution of
,
find the second independent solution,
.
What technique can be used here?
We can use Reduction of Order.
Express
in terms of
for this method.
We use another function,
v
:
Will we need derivatives of
v
?
Yes.
Substitute the expression for
into
.
We get
What do we do next?
We need to find
and substitute
into the original differential equation,
.
What rules of differentiation do we need for finding
?
To differentiate
, we need the
Product Rule
and
Power Rule
.
Apply the Product Rule to set up finding
.
Complete the differentiation.
We get
,
or in the usual order we have
Use the Product and Power Rules again to set up finding
.
Complete the differentiation.
Collect like terms and write them in the usual order for derivatives.
Substitute these results for
into our original differential equation,
We get
Distribute the multiplication.
Collect like terms.
We get
Divide by
We get
What is the order of this differential equation?
Second order.
What order do we need in order to be able to solve a differential equation?
So far, we know how to solve only first order differential equations.
How can we convert to a first order equation?
We can define a new function,
.
We will also need the first derivative of
w
. Find
.
Since
.
Substitute these results in our differential equation for
v
:
We get
Is this a first order differential equation?
Yes.
How can we solve it?
We can separate variables.
Set this up.
Separate the variables
What do we do next?
We can integrate both sides.
Set that up.
What is the result on the left-hand side?
What is the result on the right-hand side?
Why do we use ln
C
?
We need a constant, and we can write it in this form as well. It will make simplification easier later.
Combine these results.
We get
How do we solve for
w
?
We use each side as an exponent with
e
as a base.
Do it.
We get
Simplify the left-hand side.
Simplify the right-hand side using the
Properties of Logarithms
.
So we have
.
Use the definition of
to convert to a differential equation involving
v
.
We get
.
How do we solve this?
Again, we can separate variables.
Set that up.
Do the integration.
We get
Now use the definition of
to finally get
.
How can we check this result?
We can check this by determining
from this result for
and substitute in the original differential equation.
First, what rule of differentiation should we use to find
?
The Power Rule.
Use it to find
.
Use the Power Rule again to find
.
Substitute these results into the original differential equation,
We get
Distribute the multiplication and collect the like terms.
We get
It checks!
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General Contents
Detailed Contents
Index
is a solution of
,
.