Download Non Constant Coefficient Initial Value Problems-Differential Equations and Transforms-Handout and more Exercises Differential Equations and Transforms in PDF only on Docsity!
© 2007 Paul Dawkins 44 http://tutorial.math.lamar.edu/terms.aspx
Nonconstant Coefficient IVP’s
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.
Fact
If f(t) is a piecewise continuous function on
[ )
0, • of exponential order then,
lim 0
s
F s
Æ•
A function f(t) is said to be of exponential order a if there exists positive constants T and M such
that
for all
t
f t M t T
a
£ e ≥
Put in other words, a function that is of exponential order will grow no faster than
t
M
a
e
for some M and a and all sufficiently large t. One way to check whether a function is of
exponential order or not is to compute the following limit.
lim
t
t
f t
a
Æ•
e
If this limit is finite for some a then the function will be of exponential order a. Likewise, if the
limit is infinite for every a 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
3
t
f t = e
We can check this by computing the above limit.
3
2
3
lim lim lim
t t t
t
t t
t t
t
a
a
a
Æ• Æ• Æ•
e
e e
e
This is true for any value of a and so the function is not of exponential order.
Do not worry too much about this exponential order stuff. This fact is occasionally needed in
using Laplace transforms with non constant coefficients.
So, let’s take a look at an example.
© 2007 Paul Dawkins 45 http://tutorial.math.lamar.edu/terms.aspx
Example 1 Solve the following IVP.
y 3 ty 6 y 2, y 0 0 y 0 0
Solution
So, for this one we will need to recall that #30 in our table of Laplace transforms tells us that,
d
ty y
ds
d
sY s y
ds
sY s Y s
L L
So, upon taking the Laplace transforms of everything and plugging in the initial conditions we
get,
2
2
2
s Y s sy y sY s Y s Y s
s
sY s s Y s
s
s
Y s Y s
s s
Ê ˆ
Á ˜
Ë ¯
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,
3
2 2
ln
3
3
3 6 6
ds
s
s
s
s s
m s s
Ú
= e = e = e
Multiplying through, integrating and solving for Y(s) gives,
2 2
2 2
2
3
3
3 3
6 6
6 6
6
s s
s s
s
s Y s ds s ds
s Y s c
Y s c
s s
Ê ˆ
Á ˜
Ë ¯
Û
Û
Ù
Ù
ı
ı
e e
e e
e
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
2
6
3 3
lim 0
s
s
c
s s
Æ•
Ê ˆ
Á ˜
Á ˜
Ë ¯
e
© 2007 Paul Dawkins 47 http://tutorial.math.lamar.edu/terms.aspx
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,
y t = 2 + ct
Now, is where we will use the second initial condition. Upon differentiating and plugging in the
second initial condition we can see that c = -4.
So, the solution to this IVP is,
y t = 2 - 4 t
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.
Sometimes Laplace transforms can be used to solve nonconstant differential equations, however,
in general, nonconstant differential equations are still very difficult to solve.
