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Linear Nonhomogeneous Differential Equations and Variation of Parameters: Example 6
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Solve
What type of differential equation is this?
It is a linear nonhomogeneous differential equation with constant coefficients.
How can we solve it?
We can use the
Method of Variation of Parameters
.
How do we start?
We find
, the homogeneous solution, which takes care of the case of 0 on the right-hand side, and
, a particular solution, which handles the actual right-hand side.
How do we find the homogeneous solution?
We find the
characteristic equation
.
Do that.
The characteristic equation is
Now write the homogeneous solution..
This is the
Homogeneous Solution
.
What is the form of the particular solution?
We use the same functions as are in the homogeneous solution. However, each is multiplied by a parameter instead of a constant.
Let’s set that up in terms of
from the
Method of Variation of Parameters
.
How many terms do we need?
Two, because of the two terms in
Write the general form of the particular solution.
What are
here?
We use
from the homogeneous solution.
Substitute these into the particular solution.
What do
represent?
They are functions of
x
.
How do we determine them?
We use the equations from
Method of Variation of Parameters
How many equations are needed?
Two, in order to determine our two functions,
Which equations do we use from step 3 of the Method of Variation of Parameters?
We use the first and the last equations.
What is the value of
n
in the last equation?
It is the same as the number of terms in our particular solution.
n
= 2 in this case.
What is
here?
Write the first equation using these results.
Eq. 1.
Set up the second equation.
Eq. 2.
So we have 2 equations in the 2 unknowns
. How do we solve for them?
We can add a multiple of Eq. 1 to a multiple of Eq. 2.
What are the factors to use so that
eliminates.
The factors are
for Eq. 1 and
for Eq.2.
Do those multiplications.
We get
, and
Do the addition.
Factor and use trig identities to simplify.
We get
Solve for
.
We get
, or
Now substitute this result for
in Equation 1 to find
.
We get
Solve for
.
We now have the derivatives
. Solve for
.
Determine
Combine the results for
to get
.
Distribute the products and then factor the 2nd and third terms.
What trig identity can we use here?
The Pythagorean identity.
Do the simplification.
We get
This is the
particular solution.
Combine this with the homogeneous solution to get the full solution.
This is the
full solution.
How can we check the particular solution?
We can determine its first and second derivatives and substitute them into the original differential equation.
Find the first derivative of the particular solution,
Do the multiplication and simplify.
We get
Find the second derivative of the particular solution.
Now substitute these derivatives in the original differential equation,
We get
The first and third terms eliminate, and the second and fourth combine to
, which checks.
The end.
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General Contents
Detailed Contents
Index
