Math 2250-4
Mon Mar 26
10.1-10.3 Laplace transform, and application to DE IVPs, including Chapter 5.
Today we'll continue to fill in the Laplace transform table (on page 2), and to use the table entries to solve
linear differential equations.
Exercise 1) (to review) Use the table to compute
1a) L 4 K 5 cos 3 t C 2eK4 t sin 12 t s
2
1
1b) LK1
C 2
t .
sK2
s C2 sC5
Exercise 2) (to review) Use Laplace transforms to solve the IVP we didn't get to on Friday, for an
underdamped, unforced oscillator DE:
x## t C 6 x# t C 34 x t = 0
x 0 =3
x# 0 = 1
N
f t , with
f t
% CeM t
f t eKs t dt for s O M
Fs d
0
c1 f1 t C c2 f2 t
Y
verified
c1F1 s C c2F2 s
1
1
s
t
A
A
(s O 0
1
s2
2
s3
n!
t2
tn, n 2 ;
sn C 1
1
ea t
(s O R a
sKa
s
cos k t
k
sin k t
sinh k t
s2 C k2
k
ea tcos k t
s2 C k2
sKa
kt
sKa
f
n
t
f t dt
0
(s O a
sOa
C k2
s F s Kf 0
s K s f 0 K f# 0
sn F s K sn K 1f 0 K...Kf n K 1 0
Fs
s
s2F
tf t
t2 f t
tn f t , n 2 Z
f t
t
KF# s
F## s
K1 n F n s
t cos k t
s2 K k2
1
t sin k t
2k
1
sin k t K k t cos k t
2 k3
A
(s O k
k
2
A
(s O k
2 C k2
sKa
f# t
f ## t
t , n2;
(s O 0
s2 C k2
s
cosh k t
ea tsin
(s O 0
s2 C k2
A
N
F s ds
s
s2 C k2
s
2
s2 C k2
1
2
s2 C k2
2
A
A
A
A
1
sKa 2
n!
s K a nC1
t ea t
tn e a t, n 2 Z
eat f t
F sKa
eKa s
s
K
a
s
e F s
u tKa , aO0
f tKa u tKa
more later!
Laplace transform table
work down the table ...
3a) L cosh k t
s =
3b) L sinh k t
s =
s
s2 K k2
k
s2 K k2
Exercise 4) Use L f# t
a) L f## t s = s2 F s
b) L f### t s = s3 F s
c) L f n t s = sn F s
t
F
d) L
f t dt s =
0
.
s = sL f t s K f 0 to show
K s f 0 K f# 0 ,
K s2 f 0 K s f# 0 K f## 0 ,
K sn K 1 f 0 K sn K 2 f# 0 K ... Kf
s
.
s
nK1
0 ,n2;.
Exercise 5) Find LK1
1
t
2
s s C4
a) using the result of 4d.
b) using partial fractions.
Exercise 6) Show L tn
s =
n!
s
nC1
, n 2 ; , using the results of 4.
A harder table entry to understand - go through this computation and see why it seems reasonable, even
though there's one step that we don't completely justify. The table entry is
tf t
KF# s
Here's how we get it:
N
F s =L f t
d
d
0
F s =
ds
ds
0
N
Ks t
f t e
0
f t eKs t dt
s d
N
dt =
0
d
f t eKs t dt .
ds
It's this last step which is true, but needs more justification. We know that the derivative of a sum is the
sum of the derivatives, and the integral is a limit of Riemann sums, so this step does at least seem
reasonable. The rest is straightforward:
N
N
d
Ks t
f t e dt =
f t Kt eKs t dt =KL t f t s
□.
ds
0
0
For resonance and other applications ...
Exercise 7) Use L t f t s =KF# s to show
s2 K k2
a) L t cos k t s =
2
s2 C k2
1
s
b) L
t sin k t s =
2
2k
s2 C k2
c) Then use a and the identity
1
s2 C k2
=
2
1
s2 C k2
2 k2
s2 C k2
2
K
s2 Kk2
s2 C k2
2
to verify the table entry
LK1
1
2 2
s2 C k
t =
1
2
2k
1
sin k t K t cos k t
k
.
Notice how the Laplace transform table is set up to use partial fraction decompositions. And be amazed at
how it lets you quickly deduce the solutions to important DE IVPs, like this resonance problem:
Exercise 8) Use Laplace transforms to write down the solution to
2
x## t C w0 x t = F0 sin w0 t
x 0 = x0
x# 0 = v0 .