Math 315 - Homework #5 Solutions
Note: B&D refers to our course textbook written by Boyce and DiPrima.
Section 6.1
1. Show that L{cos at} =
s2
s
, s > 0. (Hint: Apply integration by parts twice.)
+ a2
Solution: Let F (s) = L{cos at}. Then
∞
Z
cos(at)e−st dt,
F (s) =
s > 0.
0
Integration by parts (u = cos(at), dv = e−st dt) yields
∞
Z
1 −st
a ∞
F (s) = − e
sin(at)e−st dt
cos(at) −
s
s
0
0
|
{z
}
1/s
=
1 a
−
s
s
∞
Z
sin(at)e−st dt
0
Integrating the above integral by parts (u = sin(at), dv = e−st dt) yields
∞
Z ∞
Z
1
a ∞
sin(at)e−st dt = − e−st sin(at) +
cos(at)e−st dt
s
s
0
0
0
|
{z
}
0
=
=
a
s
Z
∞
cos(at)e−st dt
0
a
F (s)
s
Therefore,
F (s) =
a2
1
F (s) =
2
s
s
2
s + a2
1
F (s) =
s2
s
i
1 a ha
−
F (s)
s
s s
=⇒
=⇒
=⇒
F (s) +
F (s) =
s
s2 + a2
2. (B&D Problem 22) Find the Laplace transform of the following function:
(
t, 0 ≤ t < 1
f (t) =
0, 1 ≤ t < ∞
Solution: There are two approaches that we could take. The first approach would be to directly use
the definition of the Laplace transform:
Z ∞
Z 1
L{f (t)} =
f (t) e−st dt =
te−st dt
0
0
1
which can be done by integration by parts. Let’s try the other approach which requires us to first
express f (t) in terms of the Heaviside function, u1 (t), and then rely on one of our shifting theorems
(particularly Theorem 6.3.1). Note that
f (t) = t − t u1 (t).
We need to cleverly manipulate f (t) in such a way that Theorem 6.3.1 can be used directly. Notice
that the presence of u1 (t) calls for a shift of the variable t (the one right next to u1 (t)) by a factor of
1. Therefore, let’s express f (t) as
f (t) = t − t u1 (t)
= t − [(t − 1) + 1] u1 (t)
= t − (t − 1)u1 (t) − u1 (t).
Notice that we can use Theorem 6.3.1 to find the Laplace transform of (t − 1)u1 (t):
L{(t − 1)u1 (t)} = e−s L{t} =
e−s
.
s2
Going back to f (t), we see that
1
e−s
e−s
−
−
.
s2
s2
s
L{f (t)} =
Remark: The first approach (which is to use the definition of the Laplace transform) would probably
be easier since it doesn’t require any clever manipulation of the problem in order to obtain the final
result. Integrating by parts in this case isn’t too difficult.
3. The gamma function is denoted by Γ(p) and is defined by the integral
Z ∞
Γ(p + 1) =
e−x xp dx.
0
The integral converges as x → ∞ for all p. For p < 0, it is also improper at x = 0 because the integrand
becomes unbounded as x → 0. However, the integral can be shown to converge at x = 0 for p > −1.
(a) Show that, for p > 0,
Γ(p + 1) = pΓ(p).
Hint: Integrate Γ(p) by parts.
Solution: From the definition of the gamma function, we have
Z ∞
Γ(p) =
e−x xp−1 dx.
0
Integrating Γ(p) by parts (u = e
−x
, dv = x
p−1
dx) yields
Z
∞
xp 1 ∞ −x p
Γ(p) = e−x ·
+
e x dx
p 0
p 0
|
{z
}
|
{z
}
0
=
1
Γ(p + 1).
p
Thus, Γ(p + 1) = pΓ(p).
2
Γ(p+1)
(b) Show that Γ(1) = 1.
Solution: To obtain Γ(1), we let p = 0 in the above definition. It follows that
∞
Z ∞
e−x dx = −e−x = 0 − (−1) = 1.
Γ(p) =
0
0
(c) If p is a positive integer n, show that
Γ(n + 1) = n!.
Solution: We can show this informally by using the identity in part (a). If n is a positive integer,
we have
Γ(n + 1) = n Γ(n)
= n (n − 1) Γ(n − 1)
= n(n − 1)[(n − 2) Γ(n − 2)]
= n(n − 1)(n − 2)[(n − 3) Γ(n − 3)]
= n(n − 1)(n − 2)(n − 3)(n − 4) · · · 2 Γ(2)
= n(n − 1)(n − 2)(n − 3)(n − 4) · · · 2 · 1 Γ(1)
= n!
Formal Approach: Mathematical Induction
We first show that the base case holds, that is, the formula is true for the smallest positive integer,
n = 1:
Γ(1 + 1) = 1 Γ(1) = 1 = 1!.
Since the formula holds for n = 1, we proceed to the inductive step. Assume that the formula
holds for 1 ≤ k < n where k is an integer, i.e. Γ(k + 1) = k! (this is our inductive hypothesis).
Using this assumption, we wish to show that the formula holds for k + 1. It follows that
Γ (k + 1) + 1 = Γ(k + 2)
= (k + 1) Γ(k + 1)
since Γ(p + 1) = p Γ(p)
= (k + 1) k!
[by our inductive hypothesis]
= (k + 1)!
Thus, if k = n − 1, then k + 1 = n, and so the formula holds even for the integer n. We now
deduce that if n is any integer, then Γ(n + 1) = n!.
(d) If n is a positive integer, show that
L{tn } =
n!
sn+1
, s > 0.
Hint: Make a suitable change of variable to express L{tn } in terms of the gamma function.
Solution: From the definition of the Laplace transform, we have
Z ∞
L{tn } =
tn e−st dt, s > 0.
0
3
Let x = st. Then dx = s dt. By making this change of variable, we obtain
Z ∞ n
x
1
n
L{t } =
e−x · dx
s
s
0
Z ∞
1
xn e−x dx
= n
s ·s 0
=
=
1
Γ(n + 1)
sn+1
n!
sn+1
.
Section 6.2
4. (B&D Problems 4 and 10) In each of Problems (a) and (b), find the inverse Laplace transform of the
given function.
(a) F (s) =
3s
s2 − s − 6
3s
. Since the denominator is a product of distinct linear
(s − 3)(s + 2)
factors, the partial fraction decomposition of F (s) is
Solution: Note that F (s) =
3s
A
B
=
+
.
(s − 3)(s + 2)
s−3 s+2
Let’s solve for the constants A and B:
A
B
3s
=
+
(s − 3)(s + 2)
s−3 s+2
=⇒
3s = A(s + 2) + B(s − 3)
= (A + B) s + (2A − 3B)
| {z }
| {z }
3
Solving the following system
A+ B=3
2A − 3B = 0
yields A = 9/5 and B = 6/5. Therefore
F (s) =
9/5
6/5
+
.
s−3 s+2
It follows that
L−1 {F (s)} =
4
9 3t 6 −2t
e + e
.
5
5
0
(b) F (s) =
s2
2s − 3
+ 2s + 10
Solution: Note that the denominator is not factorable over the field of real numbers. Let’s try
completing the square:
F (s) =
s2
2s − 3
2s − 3
=
.
+ 2s + 10
(s + 1)2 + 9
Let’s modify the numerator so that we can express F (s) as a Laplace transform that we recognize.
Note that
2(s + 1) − 5
s+1
5
3
F (s) =
=2
−
.
(s + 1)2 + 9
(s + 1)2 + 9
3 (s + 1)2 + 9
It follows that
5
L−1 {F (s)} = 2e−t cos(3t) − e−t sin(3t) .
3
5. (B&D Problems 12 and 16) In each of Problems (a) and (b), use the Laplace transform to solve the
given initial value problem.
(a) y 00 + 3y 0 + 2y = 0;
y(0) = 1, y 0 (0) = 0
Solution: Applying the Laplace transform on both sides of the equations yields
s2 Y (s) − s y(0) − y 0 (0) + 3[sY (s) − y(0)] + 2Y (s) = 0
|{z}
|{z} | {z }
1
0
=⇒
(s2 + 3s + 2)Y (s) − s − 3 = 0
=⇒
Y (s) =
1
s+3
(s + 1)(s + 2)
By partial fraction decomposition, we have
A
B
s+3
=
+
(s + 1)(s + 2)
s+1 s+2
=⇒
s + 3 = A(s + 2) + B(s + 1)
= (A + B) s + (2A + B)
| {z }
| {z }
1
Solving the following system
A+B = 1
2A + B = 3
yields A = 2 and B = −1. Thus,
Y (s) =
2
1
−
.
s+1 s+2
It follows that
y(t) = L{Y (s)} = 2e−t − e−2t .
5
3
(b) y 00 + 2y 0 + 5y = 0;
y(0) = 2, y 0 (0) = −1
Solution:
Applying the Laplace transform on both sides of the equation yields
s2 Y (s) − s y(0) − y 0 (0) + 2[sY (s) − y(0)] + 5Y (s) = 0
|{z}
|{z} | {z }
2
−1
=⇒
(s2 + 2s + 5)Y (s) − 2s − 3 = 0
=⇒
Y (s) =
2
2s + 3
s2 + 2s + 5
Note that the denominator is not factorable over the field of real numbers. Let’s try completing
the square:
2s + 3
2s + 3
=
s2 + 2s + 5
(s + 1)2 + 4
Let’s modify the numerator so that we can express F (s) as a Laplace transform that we recognize.
Note that
2(s + 1) + 1
s+1
2
1
Y (s) =
=
2
+
.
(s + 1)2 + 4
(s + 1)2 + 4
2 (s + 1)2 + 4
It follows that
1
y(t) = L−1 {Y (s)} = 2e−t cos(2t) + e−t sin(2t) .
2
Section 6.3
6. Express the following function in terms of the Heaviside function uc (t).

sin t,



 t3 ,
f (t) =

ln t,



π,
0≤t<π
π≤t<5
5≤t<8
t≥8
Solution: We’ll need three switches here: uπ (t), u5 (t), and u8 (t). It follows that
f (t) = sin t + (t3 − sin t)uπ (t) + (ln t − t3 )u5 (t) + (π − ln t)u8 (t) .
7. (B&D Problem 15) Find the Laplace transform of the following function.


t<π
 0,
f (t) = t − π, π ≤ t < 2π


0,
t ≥ 2π
Solution: Let’s express f (t) in terms of uπ (t) and u2π (t):
f (t) = (t − π)uπ (t) − (t − π)u2π (t).
6
We will need to express f (t) in such a way that Theorem 6.3.1 is applicable. Let’s express f (t) as
f (t) = (t − π)uπ (t) − [(t − π − π) + π]u2π (t)
= (t − π)uπ (t) − (t − 2π)u2π (t) − πu2π (t).
Therefore, by Theorem 6.3.1 and the Laplace transform of the Heaviside function, we have
L{f (t)} =
e−2πs
πe−2πs
e−πs
−
−
.
s2
s2
s
Note that we could also use the definition of the Laplace transform to obtain this result.
8. (B&D Problem 23) Find the inverse Laplace transform of the following function.
F (s) =
(s − 2)e−s
s2 − 4s + 3
s−2
. Although the denominator is factorable over the field of real
s2 − 4s + 3
numbers, it is better to complete the square:
Solution: Let H(s) =
H(s) =
s−2
.
(s − 2)2 − 1
Thus,
h(t) = L−1 {H(s)} = e2t cosh t.
Therefore, by Theorem 6.3.1, we have
L−1 {F (s)} = L−1 {H(s)e−s } = h(t − 1)u1 (t) = e2(t−1) cosh(t − 1)u1 (t) .
Section 6.4
9. (B&D Problems 5 and 11) For each of Problems (a) and (b), find the solution of the given initial value
problem. Draw the graphs of the solution and of the forcing function; explain how they are related.
(
00
0
(a) y + 3y + 2y = f (t);
0
y(0) = 0, y (0) = 0;
f (t) =
1,
0,
0 ≤ t < 10
t ≥ 10
Solution: Note that f (t) = 1 − u10 (t). Applying the Laplace transform on both sides of the
equation yields
1 e−10s
s2 Y (s) − s y(0) − y 0 (0) + 3[sY (s) − y(0)] + 2Y (s) = −
|{z} | {z }
|{z}
s
s
0
0
0
=⇒
(s2 + 3s + 2)Y (s) =
=⇒
Y (s) =
1 e−10s
−
s
s
1
e−10s
−
s(s + 1)(s + 2) s(s + 1)(s + 2)
7
Let H(s) =
1
. Let’s decompose H(s):
s(s + 1)(s + 2)
1
A
B
C
= +
+
s(s + 1)(s + 2)
s
s+1 s+2
=⇒
1 = A(s + 1)(s + 2) + Bs(s + 2) + Cs(s + 1)
= (A + B + C) s2 + (3A + 2B + C) s + |{z}
2A
|
{z
}
|
{z
}
0
0
Solving the following system
A+
B +C = 0
3A + 2B + C = 0
2A
=1
yields A = 1/2, B = −1, and C = 1/2. Thus,
H(s) =
1
1/2
1/2
−
+
.
s
s+1 s+2
It follows that
h(t) = L−1 {H(s)} =
1
1
− e−t + e−2t .
2
2
Since Y (s) = H(s) − H(s)e−10s , we have
−1
y(t) = L
1 −2t
1
1 −2(t−10)
1
−t
−(t−10)
−
−e
+ e
u10 (t)
{Y (s)} = − e + e
2
2
2
2
Figure 1: Plot of the applied forcing function.
8
1
Figure 2: Plot of the solution.
The solution appears to be increasing and approaching the value y = 1/2 on 0 < t < 10. This is
due to the forcing term which takes the value 1 on 0 < t < 10. When t ≥ 10, the forcing function
takes the value zero and the solution decays on this interval.
(b) y 00 + 4y = uπ (t) − u3π (t);
y(0) = 0, y 0 (0) = 0
Solution: Applying the Laplace transform on both sides of the equation yields
s2 Y (s) − s y(0) − y 0 (0) +4Y (s) =
|{z} | {z }
0
Let H(s) =
e−πs
e−3πs
−
s
s
0
=⇒
(s2 + 4)Y (s) =
=⇒
Y (s) =
e−πs
e−3πs
−
s
s
e−3πs
e−πs
−
s(s2 + 4) s(s2 + 4)
1
. Let’s decompose H(s):
s(s2 + 4)
1
A Bs + C
= + 2
+ 4)
s
s +4
=⇒
s(s2
1 = A(s2 + 4) + (Bs + C)s
= (A + B) s2 + |{z}
Cs + |{z}
4A
| {z }
0
Therefore, A = 1/4, B = −1/4, and C = 0, and so we have
1/4
s/4
− 2
s
s +4
1 1
s
=
− 2
.
4 s s +4
H(s) =
9
0
1
It follows that
h(t) = L−1 {H(s)} =
1
1 − cos(2t) .
4
Since Y (s) = H(s)e−πs − H(s)e−3πs , we have
y(t) = L−1 {Y (s)} =
1
1
1 − cos(2(t − π)) uπ (t) − 1 − cos(2(t − 3π)) u3π (t) .
4
4
Figure 3: Plot of the applied forcing function.
Figure 4: Plot of the solution.
Since the system is undamped, all contributions to the behavior of the solution will come directly
from the applied force. Note that the solution of the associated homogeneous IVP is y(t) = 0.
This means that the solution of the given IVP will be zero unless the forcing term is nonzero
(which is evident from the graph of the solution). For π ≤ t < 3π, the jump of the forcing term is
one, which produces an oscillation (a cosine function) in the graph of the solution with amplitude
1/4 about the line y = 1/4.
10