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© 2007 Paul Dawkins 48 http://tutorial.math.lamar.edu/terms.aspx
IVP’s With Step Functions
In this section we will use Laplace transforms to solve IVP’s which contain Heaviside functions
in the forcing function. This is where Laplace transform really starts to come into its own as a
solution method.
To work these problems we’ll just need to remember the following two formulas,
1 1
where
where
c
c
c s
c s
u t f t c F s F s f t
F s u t f t c f t F s
e
e
L L
L L
In other words, we will always need to remember that in order to take the transform of a function
that involves a Heaviside we’ve got to make sure the function has been properly shifted.
Let’s work an example.
Example 1 Solve the following IVP.
4 2
2
t
y y y u t y y
¢¢ - ¢ + = + e = ¢ = -
Solution
First let’s rewrite the forcing function to make sure that it’s being shifted correctly and to identify
the function that is actually being shifted.
( )
2 2
2
t
y y y u t
¢¢ - ¢+ = + e
So, it is being shifted correctly and the function that is being shifted is
e. Taking the Laplace
transform of everything and plugging in the initial conditions gives,
2
2
2
2
s
s
s Y s sy y sY s y Y s
s s
s s Y s s
s s
e
e
Now solve for Y(s).
2
2
2 2
2
2
2
2 2
2
s
s
s
s
s s Y s s
s s
s s
s s Y s
s s
s s
Y s
s s s s s s
Y s F s G s
e
e
e
e
Notice that we combined a couple of terms to simplify things a little. Now we need to partial
fraction F(s) and G(s). We’ll leave it to you to check the details of the partial fractions.
© 2007 Paul Dawkins 49 http://tutorial.math.lamar.edu/terms.aspx
2
2 2
2 2
s s s
F s
s s s s s s
s
G s
s s s s s s
Ê ˆ
Á ˜
Ë ¯
Ê ˆ
Á ˜
+ - + Ë ¯
We now need to do the inverse transforms on each of these. We’ll start with F(s).
2 2
1 1
2 2
2
1 19
2 4
19 2 1
2 2 19
2 2
1 19 1 19
2 4 2 4
4 6 cos sin
t t
s
F s
s s
s
s
s s
f t t t
Ê ˆ
= Á + ˜
Á ˜
Ë ¯
Ê ˆ
= Á + - ˜
Á ˜
Ë ¯
Ê ˆ
Ê ˆ Ê ˆ
Á ˜
Á ˜ Á ˜
Á ˜ Á ˜
Á ˜
Ë ¯ Ë ¯
Ë ¯
e e
Now G(s).
2 2
1 1
2 2
2
1 19
2 4
5 19
1
2
19 2
2 2
1 19 1 19
2 4 2 4
2
cos sin
t t
t
s
G s
s
s
s
s
s s
g t t t
Ê ˆ
Á ˜
Á ˜
Ë ¯
Ê ˆ
= Á - + ˜
Á ˜
Ë ¯
Ê ˆ
Ê ˆ Ê ˆ
= Á - + ˜
Á ˜ Á ˜
Á ˜ Á ˜
Á ˜
Ë ¯ Ë ¯
Ë ¯
e e e
Okay, we can now get the solution to the differential equation. Starting with the transform we
get,
2
2
s
Y s F s G s
y t f t u t g t
e
where f(t) and g(t) are the functions shown above.
There is can be a fair amount of work involved in solving differential equations that involve
Heaviside functions.
Let’s take a look at another example or two.
© 2007 Paul Dawkins 51 http://tutorial.math.lamar.edu/terms.aspx
Example 3 Solve the following IVP.
3 1
y 5 y 14 y 9 u t 4 t 1 u t y 0 0, y 0 10
Solution
Let’s take the Laplace transform of everything and note that in the third term we are shifting 4 t.
3 2 2 3 2 2
s s
s s
s Y s sy y sY s y Y s
s s s
s s Y s
s s
e e
e e
Now solve for Y(s).
3 2 2 3 2 3
s s
s s
s s
s s Y s
s s
Y s
s s s s s s s s
Y s F s G s H s
e e
e e
e e
So, we have three functions that we’ll need to partial fraction for this problem. I’ll leave it to you
to check the details.
7 2
t t
F s
s s s s s s
f t
= - + e + e
2 2
7 2
t t
G s
s s s s s s s
g t t
= - + e - e
7 2
t t
H s
s s s s
h t
= e - e
Okay, we can now get the solution to the differential equation. Starting with the transform we
get,
3
3 1
s s
Y s F s F s G s H s
y t f t u t f t u t g t h t
e e
where f(t), g(t) and h(t) are given above.
Let’s work one more example.
© 2007 Paul Dawkins 52 http://tutorial.math.lamar.edu/terms.aspx
Example 4 Solve the following IVP.
y 3 y 2 y g t , y 0 0 y 0 2
where,
t
g t t t
t
Ï
Ô
Ì
Ô
Ó
Solution
The first step is to get g(t) written in terms of Heaviside functions so that we can take the
transform.
6 10
g t = 2 + t - 2 u t + 4 - t u t
Now, while this is g(t) written in terms of Heaviside functions it is not yet in proper form for us to
take the transform. Remember that each function must be shifted by a proper amount. So,
getting things set up for the proper shifts gives us,
6 10
6 10
g t t u t t u t
g t t u t t u t
So, for the first Heaviside it looks like
f t = t + 4 is the function that is being shifted and for
the second Heaviside it looks like
f t = - 6 - t is being shifted.
Now take the Laplace transform of everything and plug in the initial conditions.
2 6 10
2 2
2 6 10
2 2
s s
s s
s Y s sy y sY s y Y s
s s s s s
s s Y s
s s s s s
Ê ˆ Ê ˆ
Á ˜ Á ˜
Ë ¯ Ë ¯
Ê ˆ Ê ˆ
Á ˜ Á ˜
Ë ¯ Ë ¯
e e
e e
Solve for Y(s).
2 6 10
2 2
6 10 6 10
2
2
6 10 6 10
2
6 10 6 10
s s
s s s s
s s s s
s s s s
s s Y s
s s s s s
s s Y s
s s
Y s
s s s s s s s s
Y s F s G s H s
Ê ˆ Ê ˆ
Á ˜ Á ˜
Ë ¯ Ë ¯
e e
e e e e
e e e e
e e e e
Now, in the solving process we simplified things into as few terms as possible. Even doing this,
it looks like we’ll still need to do three partial fractions.
I’ll leave the details of the partial fractioning to you to verify. The partial fraction form and
© 2007 Paul Dawkins 54 http://tutorial.math.lamar.edu/terms.aspx
- t ≥ 10
2 2
y 3 y 2 y 4, y 10 y 10 y 10 y 10
where, y 2
(t) is the solution to the second IVP. The solution to this IVP, with some work,
can be made to look like,
3
y t = 2 f t - h t + 4 f t - 6 + g t - 6 - 6 f t - 10 - g t - 10
There is a considerable amount of work required to solve all three of these and in each of these
the forcing function is not that complicated. Using Laplace transforms saved us a fair amount of
work.
