Second Order Differential Equations (Part 8. Variation of Parameters)
Posted by Suhaivi Hamdan on 8:06 PM with No comments
Variation of Parameters
In the last section we looked at the method of undetermined
coefficients for finding a particular solution to

and we saw that while it reduced things down to just an
algebra problem, the algebra could become quite messy. On top of that undetermined coefficients will
only work for a fairly small class of functions.
The method of Variation of Parameters is a much more general
method that can be used in many more cases.
However, there are two disadvantages to the method. First, the complementary solution is
absolutely required to do the problem.
This is in contrast to the method of undetermined coefficients where it
was advisable to have the complementary solution on hand, but was not
required. Second, as we will see, in
order to complete the method we will be doing a couple of integrals and there
is no guarantee that we will be able to do the integrals. So, while it will always be possible to write
down a formula to get the particular solution, we may not be able to actually
find it if the integrals are too difficult or if we are unable to find the
complementary solution.
We’re going to derive the formula for variation of
parameters. We’ll start off by acknowledging
that the complementary solution to (1) is



Remember as well that this is the general solution to the
homogeneous differential equation.

Also recall that in order to write down the complementary
solution we know that y_{1}(t)
and y_{2}(t) are a
fundamental set of solutions.
What we’re going to do is see if we can find a pair of
functions, u_{1}(t) and u_{2}(t) so that


will be a solution to (1). We have two unknowns here and so we’ll need
two equations eventually. One equation
is easy. Our proposed solution must
satisfy the differential equation, so we’ll get the first equation by plugging
our proposed solution into (1). The second equation can come from a variety
of places. We are going to get our
second equation simply by making an assumption that will make our work easier. We’ll say more about this shortly.
So, let’s start. If
we’re going to plug our proposed solution into the differential equation we’re
going to need some derivatives so let’s get those. The first derivative is



Here’s the assumption.
Simply to make the first derivative easier to deal with we are going to
assume that whatever u_{1}(t)
and u_{2}(t) are they will
satisfy the following.

(3)

Now, there is no reason ahead of time to believe that this
can be done. However, we will see that
this will work out. We simply make this
assumption on the hope that it won’t cause problems down the road and to make
the first derivative easier so don’t get excited about it.
With this assumption the first derivative becomes.


The second derivative is then,


Plug the solution and its derivatives into (1).



Rearranging a little gives the following.


Now, both y_{1}(t)
and y_{2}(t) are solutions to
(2)
and so the second and third terms are zero.
Acknowledging this and rearranging a little gives us,




We’ve almost got the two equations that we need. Before proceeding we’re going to go back and
make a further assumption. The last
equation, (4),
is actually the one that we want, however, in order to make things simpler for
us we are going to assume that the function p(t) = 1.
In other words, we are going to go back and start working
with the differential equation,



If the coefficient of the second derivative isn’t one divide
it out so that it becomes a one. The
formula that we’re going to be getting will assume this! Upon doing this the two equations that we
want so solve for the unknown functions are


Note that in this system we know the two solutions and so
the only two unknowns here are and . Solving this system is actually quite
simple. First, solve (5)
for and plug this into (6) and
do some simplification.






Next, let’s notice that


Recall that y_{1}(t)
and y_{2}(t) are a
fundamental set of solutions and so we know that the Wronskian won’t be zero!
Finally, all that we need to do is integrate (8)
and (9)
in order to determine what u_{1}(t)
and u_{2}(t) are. Doing this gives,



So, provided we can do these integrals, a particular
solution to the differential equation is


So, let’s summarize up what we’ve determined here.
Variation of
Parameters
Consider the differential equation,
Assume that y_{1}(t)
and y_{2}(t) are a
fundamental set of solutions for
Then a particular solution to the nonhomogeneous
differential equation is,

Depending on the person and the problem, some will find the
formula easier to memorize and use, while others will find the process used to
get the formula easier. The examples in
this section will be done using the formula.
Before proceeding with a couple of examples let’s first
address the issues involving the constants of integration that will arise out
of the integrals. Putting in the constants
of integration will give the following.


The final quantity in the parenthesis is nothing more than
the complementary solution with c_{1}
= c and c_{2} = k and
we know that if we plug this into the differential equation it will simplify
out to zero since it is the solution to the homogeneous differential
equation. In other words, these terms
add nothing to the particular solution and so we will go ahead and assume that c = 0 and k = 0 in all the examples.
One final note before we proceed with examples. Do not worry about which of your two
solutions in the complementary solution is
y_{1}(t) and which one
is y_{2}(t). It doesn’t matter. You will get the same answer no matter which
one you choose to be y_{1}(t)
and which one you choose to be y_{2}(t).
Let’s work a couple of examples now.
Example 1 Find
a general solution to the following differential equation.
Solution
First, since the formula for variation of parameters
requires a coefficient of a one in front of the second derivative let’s take
care of that before we forget. The
differential equation that we’ll actually be solving is
We’ll leave it to you to verify that the complementary
solution for this differential equation is
So, we have
The Wronskian of these two functions is
The particular solution is then,
The general solution is,

Example 2 Find
a general solution to the following differential equation.
Solution
We first need the complementary solution for this
differential equation. We’ll leave it
to you to verify that the complementary solution is,
So, we have
The Wronskian of these two functions is
The particular solution is then,
The general solution is,

This method can also be used on nonconstant coefficient
differential equations, provided we know a fundamental set of solutions for the
associated homogeneous differential equation.
Example 3 Find
the general solution to
given that
form a fundamental set of solutions for the homogeneous
differential equation.
Solution
As with the first example, we first need to divide out by
a t.
The Wronskian for the fundamental set of solutions is
The particular solution is.
The general solution for this differential equation is.

We need to address one more topic about the solution to the
previous example. The solution can be
simplified down somewhat if we do the following.


Now, since is an unknown constant subtracting 2 from it
won’t change that fact. So we can just
write the as and be done with it. Here is a simplified version of the solution
for this example.
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