MAT274
HW10
DUE DECEMBER 7 (OR SUBMIT IN MONDAY CLASS).
READINGS: §7.1—7.3 OF EDWARDS & PENNEY
1. Let f (t) = tn where n is a positive integer. Apply definition and perform a
detailed calculation to find its Laplace transform F (s).
R∞
Solution. By definition L[tn ] = 0 e−st tn dt. Perform integration by parts with
u = tn , dv = e−st and du = ntn−1 , v = − 1s e−st
Z ∞
1
1 −st n t=∞
n
− e−st ntn−1 dt
L[t ] = − e t t=0 −
s
s
0
Z
n ∞ −st n−1
=
e t
dt
s 0
since limt→∞ e−st tn = 0 for any s > 0. Now, by definition of Laplace transform
again, the RHS above amounts to ns L[tn−1 ] and thus
L[tn ] =
n n−1
L[t ]
s
This relation can be applied recursively to
L[tn−1 ] =
n − 1 n−2
L[t ]
s
n − 2 n−3
L[t ]
s
etc. In other words, each time we reduce the power of t by 1, we introduce a factor
of ks where k is the power of t. So
L[tn−2 ] =
L[tn ] =
n n−1 n−2
1
n!
·
·
...... L[t0 ] = n L[1]
s
s
s
s
s
Finally, L[1] = 1/s so
L[tn ] =
1
n!
s+1
MAT 274 HW10
December 9, 2010
2. Use the Table of Laplace Transforms and the property
dn
L{(−t) f (t)} = n L{f (t)}
ds
n
to find
−1
L
s
2
(s + 9)2
d 1
Hint: think about ds
.
s2 +9
Solution. Start with using the following item from the table L[sin kt] =
and set k = 3
3
L[sin 3t] = 2
.
s +9
Divide both sides with 3 and use the linearity of L
k
s2 +k2
1
1
L[ sin 3t] = 2
3
s +9
By the property given in the problem,
L[−t ·
1
d 1
2s
sin 3t] =
=
−
3
ds s2 + 9
(s2 + 9)2
So, again, by linearity of L,
1
s
1
L[ · t · sin 3t] = 2
2
3
(s + 9)2
namely
s
1
L
=
t sin 3t.
(s2 + 9)2
6
3. Consider an undamped spring-mass system with spring constant 64. At time
t = 0, the mass is released at displacement +2 with initial velocity -4. Use
Laplace transform to find the particular solution. You may use the Table of
Laplace Transforms directly.
Solution. For simplicity, let the mass m = 1. Then, the DE is
−1
x00 (t) + 64x(t) = 0,
x(0) = 2, x0 (0) = −4
Let X(s) = L[x(t)]. Then, the Laplace Transform of its second derivative satisfies
L[x00 (t)] = s2 X(s) − sx(0) − x0 (0)
c Bin Cheng (ASU)
P age
2
MAT 274 HW10
December 9, 2010
Then, under Laplace transform, the DE x00 + 64x = 0 becomes
s2 X(s) − sx(0) − x0 (0) + 64X(s) = 0.
Apply the initial conditions x(0) = 2, x0 (0) = −4 and arrive at
s2 X(s) − 2s + 4 + 64X(s) = 0.
From here, we easily solve for X(s)
X(s) =
2s − 4
.
s2 + 64
s = cos 8t and
Now, use the Table of Laplace Transform, we know L−1 s2 +64
8 −1
−1
L
= sin 8t. Also, by the linearity of L ,
s2 +64
8
s
1 −1
2s − 4
−1
−1
−1
= 2L
− L
L [X(s)] = L
s2 + 64
s2 + 64
2
s2 + 64
Therefore,
x(t) = L−1 [X(s)] = 2 cos 8t − 0.5 sin 8t.
c Bin Cheng (ASU)
P age
3