1
MA360
Handout 1b (week of Jan 9)
Sp2006 Kallfelz
Assignment: 1,4,6,7,10 (pg5Sheng) /2.(a),(c),(f),(h) (pg 17) [Read pp116, Sheng, and
Review techniques and examples in Handouts1a, Handouts 1b]
I.) The Laplace Transform – Underlying Motivation(s)
Consider an
n
th order nonhomogenerous differential equations, with constant
coefficients, i.e., a differential equation of the form:
( )
( )
t
f
t
y
D
n
=
(Eqn. I.1)
where: a.)
+
+
+
+
=
=



=
∑
0
1
1
1
1
0
...
a
dt
d
a
dt
d
a
dt
d
a
dt
d
a
D
n
n
n
n
n
n
k
k
n
k
k
n
(i.e.,
D
n
is the
nth
order differential operator, with constant coefficients:
a
0,
a
1
,…
a
n
)
b.)
y
(
t
) is the unknown function, with initial conditions
1
:
( )
( )
0
1
1
)
1
(
)
0
(
,
0
,...,
0
c
y
c
y
c
y
n
n
=
=
′
=


c.)
f
(
t
) is a differentiable function to
n
th order
You may be used to solving (Eqn I.1) using methods like Undetermined Coefficients
(UC) or the Variation of Paramaters (VP), etc.
But what you may also recall is the
usefulness of such methods have limited applicability—the algebra can get awfully
messy in cases when the order
n
> 2 of (Eqn I.1).
As an example, consider the particular case when
n
= 2, in the case of the following
ODE:
( )
( )
1
)
(
)
(
1
2
2
2
=
+
′
′
=
+
≡
t
y
t
y
t
y
dt
d
t
y
D
with initial conditions:
y’
(0) = 0,
y
(0) =1
In other words, the above example is rather simple.
Translating in terms of the general
notation of (Eqn. I.1) above, the coefficients are:
a
0
=
1,
a
1
=
0,
a
2
=
1.
Moreover, the
initial constants in this example are:
c
0
=
1,
c
1
=
0, and
f
(
t
) = 1
Anyway, you realize using either UC or VP it’s a straightforward to solve.
In other
words, the degree of
n
= 2 bounds the (algebraic) complexity of the UC or VP procedures
to acceptable limits.
But what about something only slightly more complicated?
Consider, for example the
following 3
rd
order equation:
1
The notation
y
(
k
)
(0) is shorthand for the
k
th derivative of
y
(
t
) at
t
= 0 (for 0
≤
k
≤
n
1, in the above case of
initial conditions.)
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2
( )
( )
( )
( )
t
e
t
y
t
y
t
y
t
y
dt
d
dt
d
t
y
D

=
+
′

=
+

≡
2
2
)
(
3
)
(
2
3
3
3
3
3
with initial conditions:
y
(0) = 1,
y’
(0) = 2,
y”
(0) = 1.
I.e, the coefficients are:
a
0
=
2,
a
1
=
3,
a
2
=
0,
a
3
=
1.
Moreover, the initial constants in this example are:
c
0
=
1,
c
1
=
2,
c
2
=
1 and
f
(
t
) = 2
e

t
.
It’s only slightly more complicated
in form
than that of the first
example, but trying to solve it using methods like UC or VP proves much messier.
Just
the homogeneous part of the solution (i.e., the solution of
( )
( )
0
2
)
(
3
)
(
3
=
+
′

t
y
t
y
t
y
)
involves solving the auxillary equation:
r
3
 3
r
+2 = 0, which involves some nontrivial
algebraic maneuvers.
So the question becomes: Is there a general and systematic way of solving equations of
the form (Eqn I.1) which can bypass some of the algebraic messiness of the more
particular procedures like UC and VP?
The answer is yes!
This turns out to be one of
the most attractive features of the Laplace Transform, as we shall discover in this course.
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 Spring '06
 WILLIAMM.KALLFELZ
 Derivative, DT DT DT, sheng

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