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University of Guelph

Mathematics

MATH 2270

Matthew Demers

Fall

Description

1
4.1 Laplace Transforms
Now, we turn to a totally di▯erent method of solving ODEs: One that involves
transforming the di▯erential equation into an algebraic equation, whereupon we can
solve it and transform back.
De▯nition:
Consider a function F(t) for t ▯ 0. Then the Laplace transform of F(t), denoted
by LfF(t)g, is given by
Z 1
f(s) = LfF(t)g = e▯stF(t)dt; (4.1)
0
provided that the improper integral converges. The following criteria guarantee
the convergence of this integral:
▯ F(t) must be piecewise-continuous;
▯ F(t) is exponentially bounded; that is,
jF(t)j ▯ Ke at (4.2)
for t ▯ M, where K, a, and M are constants. This condition simply means
that the function F(t) can grow no more quickly than an exponential function.
If these two hypotheses hold, then the Laplace Transform of F(t) exists for s > a
(we will not prove this).
Note that as it is de▯ned by an (improper) integral, the Laplace Transform is a
linear operator; that is,
Lfc 1 1 c f2 2= c L1f g 1 c Lf2 g: 2 (4.3) 2
Laplace Transforms simply take one function F(t) as input, and transform it
into another function f(s) as output. We may take the Laplace Transform of many
di▯erent functions. Here are some examples:
Example:
Laplace transform of a constant
Let F(t) = 1. Then,
Z
1 ▯st
Lf1g = (1)e dt (4.4)
0
1 ▯N
= lim ▯ e▯s▯ (4.5)
N!1 s ▯t=0
▯ ▯ ▯ ▯
= lim ▯ 1 e▯s(N) ▯ ▯ 1e▯s(0) (4.6)
N!1 s s
= 0 ▯ (▯ )1 (4.7)
s
1
= : (4.8)
s
Example:
Find the Laplace transform of ▯3.
Note that since the Laplace transform is linear, multiplicative constants can easily
move inside or outside the transform. Hence,
▯3
Lf▯3g = ▯3Lf1g = : (4.9)
s
Laplace transform of an exponential function 3
Let F(t) = e , where a is a constant. Then,
Z 1
Lfe g = (e )e▯stdt (4.10)
0
Z 1
= e(a▯sdt (4.11)
0
▯N
1 (a▯s▯t
= lim e ▯ (4.12)
▯!1 a ▯ s t▯0 ▯ ▯
1 1
= lim e(a▯s)N ▯ e(a▯s)(0) (4.13)
N!1 a ▯ s a ▯ s
(4.14)
The ▯rst term goes to zero, since s > a and thus (a ▯ s) < 0. We obtain
at 1
Lfe g = 0 ▯ (4.15)
a ▯ s
1
= ;s > a: (4.16)
s ▯ a
Example:
▯t
Find the Laplace transform of e .
Applying the rule, we have a = ▯1, giving
▯ ▯t 1
L e = s + 1: (4.17)
Laplace transform of trig functions
Let F(t) = cos(!t), where ! is a constant. Then,
Z 1
Lfe g = (cos(!t))e▯sdt: (4.18)
0
This integral is not pretty, and requires using integration by parts twice. It is doable,
but instead, let’s take a di▯erent approach. Recall that from Euler’s formula, we have
that
ej!x= cos(!x) + j sin(!x): (4.19) 4
Thus,
Lfe j!xg = Lfcos(!x) + j sin(!x)g (4.20)
= Lfcos(!x)g + jLfsin(!x)g; (4.21)
since L is a linear operator. But, we can use the Laplace transform of an exponential,
given in Equation (4.16), to give us
Lfe j!xg = 1 (4.22)
s ▯ j!
= s + j! ; (4.23)
s + ! 2
multiplying both the numerator and denominator by s + j!. Thus, we have
s + j!
2 2 = Lfcos(!x)g + jLfsin(!x)g; or (4.24)
s + !
s + j ! = Lfcos(!x)g + jLfsin(!x)g: (4.25)
s + ! 2 s + ! 2
From this equation, it is easy to deduce that
s
Lfcos(!x)g = s + ! 2;s > j!j (4.26)
and
!
Lfsin(!x)g = s + ! 2;s > j!j: (4.27)
Example:
Find the Laplace transform of sin(5t).
Applying the rule, we have
5
Lfsin(5t)g = 2 : (4.28)
s + 25
Example:
Find the Laplace transform of cos(2t). 5
Applying the rule, we have
s
Lfcos(2t)g = s + 4 : (4.29)
Laplace transform of derivatives
Given a function F(t), consider its derivative F (t). Then,
Z 1
LfF (t)g = (F (t))e▯stdt: (4.30)
0
Using integration by parts, we have that
Z
0 ▯ ▯st▯N 1 ▯st
LfF (t)g = lim F(t)e ▯0 + s(F(t))e dt: (4.31)
N!1 0 Z
▯ ▯ 1
= lim F(N)e ▯s(N)▯ F(0)e ▯s(0) + s (F(t))e▯stdt: (4.32)
N!1 0
= ▯F(0) + sLfF(t)g; or (4.33)
= sLfF(t)g ▯ F(0): (4.34)
Similarly, it is not hard to prove that
00 2 0
LfF (t)g = s LfF(t)g ▯ sF(0) ▯ F (0); (4.35)
we need only to integrate by parts twice instead. In general, we have that
(n) n n▯1 n▯2 0 (n▯2) (n▯1)
LfF (t)g = s LfF(t)g▯s F(0)▯s F (0)▯:::▯sF (0)▯F (0): (4.36)
Example:
d f
Find the Laplace transform of , where f is some function.
dt3
There isn’t much we can do except apply our new rule. Keeping with the notation,
we have
▯ ▯
d f df d f
L = s Lff(t)g ▯ s f(0) ▯ s (0) ▯ s (0): (4.37)
dt 3 dt dt2 6
Laplace transform of powers
Let F(t) = t . While the Laplace transform could be found using integration by
parts, we take a di▯erent approach to make things simpler.
From the last section, we have
LfF (t)g = sLfF(t)g ▯ F(0): (4.38)
If we substitute F(t) = t into this equation, we obtain
Lf1g = sLftg ▯ (0); or (4.39)
1
Lftg = sLf1g (4.40)
▯ ▯▯ ▯
= 1 1 (4.41)
s s
1
= s2: (4.42)
If we substitute F(t) = t into equation (4.38), we obtain
2
Lf2tg = sLft g ▯ (0) (4.43)
2Lftg = sLft g; 2 or (4.44)
▯ ▯
Lft g = 2 (Lftg) (4.45)
s
▯ ▯▯ ▯
2 2 1
Lft g = s s2 (4.46)
2
= 3: (4.47)
s
3 6
In a similar way, Lft g = 4 (do it yourself), and in general, we ▯nd that
s
n n!
Lft g = n+1;s > 0: (4.48)
s
Example:
5
Find the Laplace transform of the function F(t) = 3t .
5
Remember that the Laplace Transform is linear, so that we know Lf3t g is the
5
same as 3Lft g. So, don’t worry about that 3. Now, we have 7
5! 120 360
3Lft g = 3 = 3 = : (4.49)
s 6 s6 s6
at
Laplace transform of e F(t)
If F(t) is a function, then we know that the Laplace Transform of F is
Z 1
▯st
LfF(t)g = e F(t)dt; (4.50)
0
which de▯nes a function of s. So, let LfF(t)g = f(s). Then, we have that
Z
1 ▯(s▯a)t
f(s ▯ a) = e F(t)dt: (4.51)
0
But, this can be rewritten as
Z 1
▯st at
f(s ▯ a) = e e F(t)dt; (4.52)
0
at
which is the Laplace transform for e F(t).
Thus, we have the property that
at
Lfe F(t)g = f(s ▯ a); (4.53)
where f(s) is the Laplace transform for F(t).
This may sound a little complicated, but an example will clear things up:
Example:
4t
Find the Laplace transform of the function F(t) = te .
4t
Seeing the e present as part of this function, we know to use the property we
have just detailed. Hence, the Laplace transform for this function is the same as the
1
Laplace transform for t (that is, s2), except with an \s ▯ 4" subbed in for the usual
\s". Hence,

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