In this section we will take a look at the first method that
can be used to find a particular solution to a nonhomogeneous differential
equation.
One of the main advantages of this method is that it reduces
the problem down to an algebra problem.
The algebra can get messy on occasion, but for most of the problems it
will not be terribly difficult. Another
nice thing about this method is that the complementary solution will not be
explicitly required, although as we will see knowledge of the complementary
solution will be needed in some cases and so we’ll generally find that as well.
There are two disadvantages to this method. First, it will only work for a fairly small
class of g(t)’s. The class of g(t)’s for which the method works, does include some of the more
common functions, however, there are many functions out there for which
undetermined coefficients simply won’t work.
Second, it is generally only useful for constant coefficient differential
equations.
The method is quite simple.
All that we need to do is look at g(t)
and make a guess as to the form of Y_{P}(t)
leaving the coefficient(s) undetermined (and hence the name of the
method). Plug the guess into the
differential equation and see if we can determine values of the
coefficients. If we can determine values
for the coefficients then we guessed correctly, if we can’t find values for the
coefficients then we guessed incorrectly.
It’s usually easier to see this method in action rather than
to try and describe it, so let’s jump into some examples.
Example 1 Determine
a particular solution to
Solution
The point here is to find a particular solution, however
the first thing that we’re going to do is find the complementary solution to
this differential equation. Recall
that the complementary solution comes from solving,
The characteristic equation for this differential equation
and its roots are.
The complementary solution is then,
At this point the reason for doing this first will not be
apparent, however we want you in the habit of finding it before we start the
work to find a particular solution. Eventually,
as we’ll see, having the complementary solution in hand will be helpful and
so it’s best to be in the habit of finding it first prior to doing the work
for undetermined coefficients.
Now, let’s proceed with finding a particular
solution. As mentioned prior to the
start of this example we need to make a guess as to the form of a particular
solution to this differential equation.
Since g(t) is an exponential
and we know that exponentials never just appear or disappear in the
differentiation process it seems that a likely form of the particular
solution would be
Now, all that we need to do is do a couple of derivatives,
plug this into the differential equation and see if we can determine what A needs to be.
Plugging into the differential equation gives
So, in order for our guess to be a solution we will need
to choose A so that the
coefficients of the exponentials on either side of the equal sign are the
same. In other words we need to choose
A so that,
Okay, we found a value for the coefficient. This means that we guessed correctly. A particular solution to the differential
equation is then,

Before proceeding any further let’s again note that we
started off the solution above by finding the complementary solution. This is not technically part the method of
Undetermined Coefficients however, as we’ll eventually see, having this in hand
before we make our guess for the particular solution can save us a lot of work
and/or headache. Finding the
complementary solution first is simply a good habit to have so we’ll try to get
you in the habit over the course of the next few examples. At this point do not worry about why it is a
good habit. We’ll eventually see why it
is a good habit.
Now, back to the work at hand. Notice in the last example that we kept
saying “a” particular solution, not “the” particular solution. This is because there are other possibilities
out there for the particular solution we’ve just managed to find one of
them. Any of them will work when it
comes to writing down the general solution to the differential equation.
Speaking of which…
This section is devoted to finding particular solutions and most of the
examples will be finding only the particular solution. However, we should do at least one full blown
IVP to make sure that we can say that we’ve done one.
Example 2 Solve
the following IVP
Solution
We know that the general solution will be of the form,
and we already have both the complementary and particular solution
from the first example so we don’t really need to do any extra work for this
problem.
One of the more common mistakes in these problems is to
find the complementary solution and then, because we’re probably in the habit
of doing it, apply the initial conditions to the complementary solution to
find the constants. This however, is
incorrect. The complementary solution
is only the solution to the homogeneous differential equation and we are
after a solution to the nonhomogeneous differential equation and the initial
conditions must satisfy that solution instead of the complementary solution.
So, we need the general solution to the nonhomogeneous
differential equation. Taking the
complementary solution and the particular solution that we found in the
previous example we get the following for a general solution and its
derivative.
Now, apply the initial conditions to these.
Solving this system gives c_{1} = 2 and c_{2}
= 1. The actual solution is then.

This will be the only IVP in this section so don’t forget
how these are done for nonhomogeneous differential equations!
Let’s take a look at another example that will give the
second type of g(t) for which
undetermined coefficients will work.
Example 3 Find
a particular solution for the following differential equation.
Solution
Again, let’s note that we should probably find the
complementary solution before we proceed onto the guess for a particular
solution. However, because the
homogeneous differential equation for this example is the same as that for
the first example we won’t bother with that here.
Now, let’s take our experience from the first example and
apply that here. The first example had
an exponential function in the g(t)
and our guess was an exponential. This
differential equation has a sine so let’s try the following guess for the
particular solution.
Differentiating and plugging into the differential
equation gives,
Collecting like terms yields
We need to pick A
so that we get the same function on both sides of the equal sign. This means that the coefficients of the
sines and cosines must be equal. Or,
Notice two things.
First, since there is no cosine on the right hand side this means that
the coefficient must be zero on that side.
More importantly we have a serious problem here. In order for the cosine to drop out, as it
must in order for the guess to satisfy the differential equation, we need to
set A = 0, but if A = 0, the sine will also drop out and
that can’t happen. Likewise, choosing A to keep the sine around will also
keep the cosine around.
What this means is that our initial guess was wrong. If we get multiple values of the same
constant or are unable to find the value of a constant then we have guessed
wrong.
One of the nicer aspects of this method is that when we
guess wrong our work will often suggest a fix. In this case the problem was the cosine
that cropped up. So, to counter this
let’s add a cosine to our guess. Our
new guess is
Plugging this into the differential equation and
collecting like terms gives,
Now, set the coefficients equal
Solving this system gives us
We found constants and this time we guessed
correctly. A particular solution to
the differential equation is then,

Notice that if we had had a cosine instead of a sine in the
last example then our guess would have been the same. In fact, if both a sine and a cosine had
shown up we will see that the same guess will also work.
Let’s take a look at the third and final type of basic g(t) that we can have. There are other types of g(t) that we can have, but as we will see they will all come back
to two types that we’ve already done as well as the next one.
Example 4 Find
a particular solution for the following differential equation.
Solution
Once, again we will generally want the complementary
solution in hand first, but again we’re working with the same homogeneous
differential equation (you’ll eventually see why we keep working with the
same homogeneous problem) so we’ll again just refer to the first example.
For this example g(t)
is a cubic polynomial. For this we
will need the following guess for the particular solution.
Notice that even though g(t) doesn’t have a t^{2}
in it our guess will still need one!
So, differentiate and plug into the differential equation.
Now, as we’ve done in the previous examples we will need
the coefficients of the terms on both sides of the equal sign to be the same
so set coefficients equal and solve.
Notice that in this case it was very easy to solve for the
constants. The first equation gave A.
Then once we knew A the
second equation gave B, etc.
A particular solution for this differential equation is then

Now that we’ve gone over the three basic kinds of functions
that we can use undetermined coefficients on let’s summarize.
Notice that there are really only three kinds of functions
given above. If you think about it the
single cosine and single sine functions are really special cases of the case
where both the sine and cosine are present.
Also, we have not yet justified the guess for the case where both a sine
and a cosine show up. We will justify
this later.
We now need move on to some more complicated functions. The more complicated functions arise by
taking products and sums of the basic kinds of functions. Let’s first look at products.
Example 5 Find
a particular solution for the following differential equation.
Solution
You’re probably getting tired of the opening comment, but
again finding the complementary solution first really a good idea but again
we’ve already done the work in the first example so we won’t do it again
here. We promise that eventually
you’ll see why we keep using the same homogeneous problem and why we say it’s
a good idea to have the complementary solution in hand first. At this point all we’re trying to do is
reinforce the habit of finding the complementary solution first.
Okay, let’s start off by writing down the guesses for the
individual pieces of the function. The
guess for the t would be
while the guess for the exponential would be
Now, since we’ve got a product of two functions it seems
like taking a product of the guesses for the individual pieces might
work. Doing this would give
However, we will have problems with this. As we will see, when we plug our guess into
the differential equation we will only get two equations out of this. The problem is that with this guess we’ve
got three unknown constants. With only
two equations we won’t be able to solve for all the constants.
This is easy to fix however. Let’s notice that we could do the following
If we multiply the C
through, we can see that the guess can be written in such a way that there
are really only two constants. So, we
will use the following for our guess.
Notice that this is nothing more than the guess for the t with an exponential tacked on for
good measure.
Now that we’ve got our guess, let’s differentiate, plug
into the differential equation and collect like terms.
Note that when we’re collecting like terms we want the
coefficient of each term to have only constants in it. Following this rule we will get two terms
when we collect like terms. Now, set
coefficients equal.
A particular solution for this differential equation is
then

This last example illustrated the general rule that we will follow
when products involve an exponential.
When a product involves an exponential we will first strip out the
exponential and write down the guess for the portion of the function without
the exponential, then we will go back and tack on the exponential without any
leading coefficient.
Let’s take a look at some more products. In the interest of brevity we will just write
down the guess for a particular solution and not go through all the details of
finding the constants. Also, because we
aren’t going to give an actual differential equation we can’t deal with finding
the complementary solution first.
Example 6 Write
down the form of the particular solution to
for the following g(t)’s.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
So, we have an exponential in the function. Remember the rule. We will ignore the exponential and write
down a guess for 16 sin(10t) then
put the exponential back in.
The guess for the sine is
Now, for the actual guess for the particular solution
we’ll take the above guess and tack an exponential onto it. This gives,
One final note before we move onto the next part. The 16 in front of the function has
absolutely no bearing on our guess.
Any constants multiplying the whole function are ignored.
[Return to Problems]
(b)
We will start this one the same way that we initially
started the previous example. The
guess for the polynomial is
and the guess for the cosine is
If we multiply the two guesses we get.
Let’s simplify things up a little. First multiply the polynomial through as
follows.
Notice that everywhere one of the unknown constants occurs
it is in a product of unknown constants.
This means that if we went through and used this as our guess the
system of equations that we would need to solve for the unknown constants
would have products of the unknowns in them.
These types of systems are generally very difficult to solve.
So, to avoid this we will do the same thing that we did in
the previous example. Everywhere we
see a product of constants we will rename it and call it a single
constant. The guess that we’ll use for
this function will be.
This is a general rule that we will use when faced with a
product of a polynomial and a trig function.
We write down the guess for the polynomial and then multiply that by a
cosine. We then write down the guess
for the polynomial again, using different coefficients, and multiply this by
a sine.
[Return to Problems]
(c)
This final part has all three parts to it. First we will ignore the exponential and
write down a guess for.
The minus sign can also be ignored. The guess for this is
Now, tack an exponential back on and we’re done.
Notice that we put the exponential on both terms.
[Return to Problems]

There a couple of general rules that you need to remember
for products.
 If g(t) contains an exponential,
ignore it and write down the guess for the remainder. Then tack the exponential back on
without any leading coefficient.
 For
products of polynomials and trig functions you first write down the guess
for just the polynomial and multiply that by the appropriate cosine. Then add on a new guess for the
polynomial with different coefficients and multiply that by the
appropriate sine.
If you can remember these two rules you can’t go wrong with
products. Writing down the guesses for
products is usually not that difficult.
The difficulty arises when you need to actually find the constants.
Now, let’s take a look at sums of the basic components
and/or products of the basic components.
To do this we’ll need the following fact.
Fact
If Y_{P1}(t)
is a particular solution for
and if Y_{P2}(t)
is a particular solution for
then Y_{P1}(t)+
Y_{P2}(t) is a particular solution for

This fact can be used to both find particular solutions to
differential equations that have sums in them and to write down guess for
functions that have sums in them.
Example 7 Find
a particular solution for the following differential equation.
Solution
This example is the reason that we’ve been using the same homogeneous
differential equation for all the previous examples. There is nothing to do with this
problem. All that we need to do it go
back to the appropriate examples above and get the particular solution from
that example and add them all together.
Doing this gives

Let’s take a look at a couple of other examples. As with the products we’ll just get guesses
here and not worry about actually finding the coefficients.
Example 8 Write
down the form of the particular solution to
for the following g(t)’s.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
(g) [Solution]
Solution
(a)
This first one we’ve actually already told you how to
do. This is in the table of the basic
functions. However we wanted to
justify the guess that we put down there.
Using the fact on sums of function we would be tempted to write down a
guess for the cosine and a guess for the sine. This would give.
So, we would get a cosine from each guess and a sine from
each guess. The problem with this as a
guess is that we are only going to get two equations to solve after plugging
into the differential equation and yet we have 4 unknowns. We will never be able to solve for each of
the constants.
To fix this notice that we can combine some terms as
follows.
Upon doing this we can see that we’ve really got a single
cosine with a coefficient and a single sine with a coefficient and so we may
as well just use
The general rule of thumb for writing down guesses for
functions that involve sums is to always combine like terms into single terms
with single coefficients. This will
greatly simplify the work required to find the coefficients.
[Return to Problems]
(b)
For this one we will get two sets of sines and
cosines. This will arise because we
have two different arguments in them.
We will get one set for the sine with just a t as its argument and we’ll get another set for the sine and
cosine with the 14t as their
arguments.
The guess for this function is
[Return to Problems]
(c)
The main point of this problem is dealing with the
constant. But that isn’t too bad. We just wanted to make sure that an example
of that is somewhere in the notes. If
you recall that a constant is nothing more than a zeroth degree polynomial
the guess becomes clear.
The guess for this function is
[Return to Problems]
(d)
This one can be a little tricky if you aren’t paying
attention. Let’s first rewrite the
function
All we did was move the 9.
However upon doing that we see that the function is really a sum of a
quadratic polynomial and a sine. The
guess for this is then
If we don’t do this and treat the function as the sum of
three terms we would get
and as with the first part in this example we would end up
with two terms that are essentially the same (the C and the G) and so
would need to be combined. An added
step that isn’t really necessary if we first rewrite the function.
Look for problems where rearranging the function can
simplify the initial guess.
[Return to Problems]
(e)
So, this look like
we’ve got a sum of three terms here.
Let’s write down a guess for that.
Notice however that if we were to multiply the exponential
in the second term through we would end up with two terms that are
essentially the same and would need to be combined. This is a case where the guess for one term
is completely contained in the guess for a different term. When this happens we just drop the guess
that’s already included in the other term.
So, the guess here is actually.
Notice that this arose because we had two terms in our g(t) whose only difference was the
polynomial that sat in front of them.
When this happens we look at the term that contains the largest degree
polynomial, write down the guess for that and don’t bother writing down the guess
for the other term as that guess will be completely contained in the first
guess.
[Return to Problems]
(f)
In this case we’ve got two terms whose guess without the
polynomials in front of them would be the same. Therefore, we will take the one with the
largest degree polynomial in front of it and write down the guess for that
one and ignore the other term. So, the
guess for the function is
[Return to Problems]
(g)
This last part is designed to make sure you understand the
general rule that we used in the last two parts. This time there really are three terms and
we will need a guess for each term.
The guess here is
We can only combine guesses if they are identical up to
the constant. So we can’t combine the
first exponential with the second because the second is really multiplied by
a cosine and a sine and so the two exponentials are in fact different
functions. Likewise, the last sine and
cosine can’t be combined with those in the middle term because the sine and
cosine in the middle term are in fact multiplied by an exponential and so are
different.
[Return to Problems]

So, when dealing with sums of functions make sure that you
look for identical guesses that may or may not be contained in other guesses
and combine them. This will simplify
your work later on.
We have one last topic in this section that needs to be
dealt with. In the first few examples we
were constantly harping on the usefulness of having the complementary solution
in hand before making the guess for a particular solution. We never gave any reason for this other that
“trust us”. It is now time to see why
having the complementary solution in hand first is useful. This is best shown with an example so let’s
jump into one.
Example 9 Find
a particular solution for the following differential equation.
Solution
This problem seems almost too simple to be given this late
in the section. This is especially
true given the ease of finding a particular solution for g(t)’s that are just
exponential functions. Also, because
the point of this example is to illustrate why it is generally a good idea to
have the complementary solution in hand first we’ll let’s go ahead and recall
the complementary solution first. Here
it is,
Now, without worrying about the complementary solution for
a couple more seconds let’s go ahead and get to work on the particular
solution. There is not much to the
guess here. From our previous work we
know that the guess for the particular solution should be,
Plugging this into the differential equation gives,
Hmmmm…. Something
seems wrong here. Clearly an
exponential can’t be zero. So, what
went wrong? We finally need the
complementary solution. Notice that the second term in the complementary
solution (listed above) is exactly our guess for the form of the particular
solution and now recall that both portions of the complementary solution are
solutions to the homogeneous differential equation,
In other words, we had better have gotten zero by plugging
our guess into the differential equation, it is a solution to the homogeneous
differential equation!
So, how do we fix this?
The way that we fix this is to add a t to our guess as follows.
Plugging this into our differential equation gives,
Now, we can set coefficients equal.
So, the particular solution in this case is,

So, what did we learn from this last example. While technically we don’t need the
complementary solution to do undetermined coefficients, you can go through a
lot of work only to figure out at the end that you needed to add in a t to the guess because it appeared in
the complementary solution. This work is
avoidable if we first find the complementary solution and comparing our guess
to the complementary solution and seeing if any portion of your guess shows up
in the complementary solution.
If a portion of your guess does show up in the complementary
solution then we’ll need to modify that portion of the guess by adding in a t
to the portion of the guess that is causing the problems. We do need to be a little careful and make
sure that we add the t in the correct
place however. The following set of
examples will show you how to do this.
Example 10 Write
down the guess for the particular solution to the given differential
equation. Do not find the
coefficients.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
In these solutions we’ll leave the details of checking the
complementary solution to you.
(a)
The complementary solution is
Remembering to put the “1” with the 7t gives a first guess for the particular solution.
Notice that the last term in the guess is the last term in
the complementary solution. The first
two terms however aren’t a problem and don’t appear in the complementary
solution. Therefore, we will only add
a t onto the last term.
The correct guess for the form of the particular solution
is.
[Return to Problems]
(b)
The complementary solution is
A first guess for the particular solution is
Notice that if we multiplied the exponential term through
the parenthesis that we would end up getting part of the complementary
solution showing up. Since the problem
part arises from the first term the whole
first term will get multiplied by t. The second and third terms are okay as they
are.
The correct guess for the form of the particular solution
in this case is.
So, in general, if you were to multiply out a guess and if any term in the result shows up in the complementary solution, then the whole
term will get a t not just the
problem portion of the term.
[Return to Problems]
(c)
The complementary solution is
A first guess for the particular solution is
In this case both the second and third terms contain
portions of the complementary solution.
The first term doesn’t however, since upon multiplying out, both the
sine and the cosine would have an exponential with them and that isn’t part
of the complementary solution. We only
need to worry about terms showing up in the complementary solution if the
only difference between the complementary solution term and the particular
guess term is the constant in front of them.
So, in this case the second and third terms will get a t while the first won’t
The correct guess for the form of the particular solution
is.
[Return to Problems]
(d)
To get this problem we changed the differential equation
from the last example and left the g(t)
alone. The complementary solution this
time is
As with the last part, a first guess for the particular
solution is
This time however it is the first term that causes
problems and not the second or third.
In fact, the first term is exactly the complementary solution and so
it will need a t. Recall that we will only have a problem
with a term in our guess if it only differs from the complementary solution
by a constant. The second and third
terms in our guess don’t have the exponential in them and so they don’t
differ from the complementary solution by only a constant.
The correct guess for the form of the particular solution
is.
[Return to Problems]
(e)
The complementary solution is
The two terms in g(t)
are identical with the exception of a polynomial in front of them. So this means that we only need to look at
the term with the highest degree polynomial in front of it. A first guess for the particular solution
is
Notice that if we multiplied the exponential term through
the parenthesis the last two terms would be the complementary solution. Therefore, we will need to multiply this
whole thing by a t.
The next guess for the particular solution is then.
This still causes problems however. If we multiplied the t and the exponential through, the last term will still be in the
complementary solution. In this case,
unlike the previous ones, a t
wasn’t sufficient to fix the problem.
So, we will add in another t
to our guess.
The correct guess for the form of the particular solution
is.
Upon multiplying this out none of the terms are in the
complementary solution and so it will be okay.
[Return to Problems]

As this last set of examples has shown, we really should
have the complementary solution in hand before even writing down the first
guess for the particular solution. By
doing this we can compare our guess to the complementary solution and if any of
the terms from your particular solution show up we will know that we’ll have
problems. Once the problem is identified
we can add a t to the problem term(s)
and compare our new guess to the complementary solution. If there are no problems we can proceed with
the problem, if there are problems add in another t and compare again.
Can you see a general rule as to when a t will be needed and when a t^{2}
will be needed for second order differential equations?