In this section we are going to see how Laplace
transforms can be used to solve some differential equations that do not have
constant coefficients. This is not
always an easy thing to do. However,
there are some simple cases that can be done.
To do this we will need a quick fact.
A function f(t) is
said to be of exponential order α if there exists positive constants T and M such that
Put in other words, a function that is of exponential order
will grow no faster than
for some M and α and all sufficiently large t.
One way to check whether a function is of exponential order or not is to
compute the following limit.
If this limit is finite for some α then the function will be of exponential order
Likewise, if the limit is infinite for every α then the function is not of exponential order.
Almost all of the functions that you are liable to deal with
in a first course in differential equations are of exponential order. A good example of a function that is not of
exponential order is
We can check this by computing the above limit.
This is true for any value of α and so the function is not of exponential
Do not worry too much about this exponential order
stuff. This fact is occasionally needed
in using Laplace transforms with non constant
So, let’s take a look at an example.
Example 1 Solve
the following IVP.
So, for this one we will need to recall that #30 in our table of Laplace
transforms tells us that,
So, upon taking the Laplace transforms of everything and
plugging in the initial conditions we get,
Unlike the examples in the previous section where we ended up with a transform for
the solution, here we get a linear first order differential equation that
must be solved in order to get a transform for the solution.
The integrating factor for this differential equation is,
Multiplying through, integrating and solving for Y(s) gives,
Now, we have a transform for the solution. However that second term looks unlike
anything we’ve seen to this point.
This is where the fact about the transforms of exponential order
functions comes into play. We are
going to assume that whatever our solution is, it is of exponential
order. This means that
The first term does go to zero in the limit. The second term however, will only go to
zero if c = 0. Therefore, we must have c = 0 in order for this to be the
transform of our solution.
So, the transform of our solution, as well as the solution
I’ll leave it to
you to verify that this is in fact a solution if you’d like to.
Now, not all nonconstant differential equations need to use (1). So, let’s take a look at one more example.
Example 2 Solve
the following IVP.
From the first example we have,
We’ll also need,
Taking the Laplace
transform of everything and plugging in the initial conditions gives,
Once again we have a linear first order differential
equation that we must solve in order to get a transform for the
solution. Notice as well that we never
used the second initial condition in this work. That is okay, we will use it eventually.
Since this linear differential equation is much easier to
solve compared to the first one, we’ll leave the details to you. Upon solving the differential equation we
Now, this transform goes to zero for all values of c and we can take the inverse
transform of the second term.
Therefore, we won’t need to use (1)
to get rid of the second term as did in the previous example.
Taking the inverse transform gives,
Now, is where we will use the second initial condition. Upon differentiating and plugging in the
second initial condition we can see that c
So, the solution to this IVP is,
So, we’ve seen how to use Laplace
transforms to solve some nonconstant coefficient differential equations. Notice however that all we did was add in an
occasional t to the
coefficients. We couldn’t get too
complicated with the coefficients. If we
had we would not have been able to easily use Laplace
transforms to solve them.
transforms can be used to solve nonconstant differential equations, however, in
general, nonconstant differential equations are still very difficult to solve.