Diff. Eqns.
Problem Set 8
Solutions
1. Find the Laplace transforms F (s) and G(s) of the functions
(a)
f (t) = 4 cosh(3t),
(b)
g(t) = eat sin (bt) .
(1)
where a and b are (real) parameters. Determine the range of s such that the
improper integrals converge.
(a)
∞
Z
L {4 cosh (3t)} =
0
Z
∞
=
0
e−st (4 cosh (3t)) dt
e−st 2e3t + 2e−3t dt
∞
Z
=2
(2)
(Euler’s formula)
e(3−s)t + e(−3−s)t dt
(3)
(4)
0
∞
1 (3−s)t
1
(−3−s)t
=2
e
+
e
3−s
−3 − s
0
1
1
1
1 (3−s)t
(−3−s)t
e
+
e
−2
+
= 2 lim
t→∞ 3 − s
−3 − s
3 − s −3 − s
(5)
(6)
To have this limit to converge, we need the exponents of the exponential terms to be
negative. Thus we will require 3 − s < 0 and −3 − s < 0; i.e. s > 3.
= −2
1
1
+
3 − s −3 − s
=
4s
,
s2 − 9
Z
∞
(7)
s > 3.
(8)
e−st eat sin (bt) dt
(9)
(b)
at
L e sin (bt) =
0
Z
∞
e(a−s)t sin (bt) dt
(10)
0
∞
Z
1
a − s ∞ (a−s)t
(a−s)t
cos (bt)
+
e
cos (bt) dt
(IBP) (11)
= e
−
b
b
0
0
Z
1
1 a − s ∞ (a−s)t
= − lim e(a−s)t cos (bt) + +
e
cos (bt) dt
(12)
t→∞
b
b
b
0
=
For convergence of this limit, the exponential must have a negative exponent, so a − s < 0,
meaning s > a.
Z
1 a − s ∞ (a−s)t
e
cos (bt) dt
(IBP)
= +
b
b
0
∞
Z
1 a−s
a − s ∞ (a−s)t
(a−s)t 1
= +
e
e
sin (bt) dt
sin (bt)
−
b
b
b
b
0
0
Z
a − s 2 ∞ (a−s)t
1
e
sin (bt) dt
= −
b
b
0
1
a − s 2 at
= −
L e sin (bt) .
b
b
(13)
(14)
(15)
(16)
So we have the equation
1
L e sin (bt) = −
b
at
a−s
b
2
L eat sin (bt)
(17)
and we may algebraically solve for the Laplace transform.
1+
a−s
b
2 !
1
L eat sin (bt) =
b
b
L eat sin (bt) =
,
b2 + (a − s)2
(18)
s>a
(19)
(20)
2. Use a Laplace transform to solve the initial-value problem
y 00 − y 0 − 2y = t + 3e−t ,
y(0) = 0,
y 0 (0) = 1.
(21)
First we note that
L y 0 (t) =
Z
∞
e−st y 0 (t) dt
(22)
0
∞
= e−st y (t) 0 + s
Z
∞
e−st y (t) dt
(IBP)
(23)
0
= −y(0) + sL {y (t)}
(24)
= sL {y (t)} ,
(25)
s > 0,
and
L y 00 (t) =
∞
Z
e−st y 00 (t) dt
(26)
0
Z ∞
∞
= e y (t) 0 + s
e−st y 0 (t) dt
0
= −y 0 (0) + sL y 0 (t)
−st 0
= −1 + s2 L {y (t)} ,
s > 0.
(27)
(28)
(29)
Also note that
Z
∞
e−st tdt
Z
1 −st ∞ 1 ∞ −st
= − e t
e dt
+
s
s 0
0
1 −st ∞
= − 2e s
0
1
= 2,
s > 0,
s
L {t} =
(30)
0
(31)
(32)
(33)
and
L e−t =
Z
0
Z
∞
∞
=
e−st e−t dt
(34)
e−(1+s)t dt
(35)
0
1 −(1+s)t ∞
=−
e
1+s
(36)
0
=
1
,
1+s
s > −1.
So we take the Laplace transform of (21) and get
L y 00 − L y 0 − 2L {y} = L {t} + 3L e−t
1
3
−1 + s2 L {y (t)} − (sL {y (t)}) − 2L {y} = 2 +
,
s
1+s
1
3
(s − 2) (s + 1) L {y (t)} = 2 +
+1
s
1+s
(37)
(38)
s>0
(39)
(40)
So we get that the Laplace transform is
3
1
1
+
,
+
2
− 2) (s + 1) (s − 2) (s + 1)
(s − 2) (s + 1)
(s + 1) + 3s2 + s2 (s + 1)
=
s2 (s + 1)2 (s − 2)
s3 + 4s2 + s + 1
=
s2 (s + 1)2 (s − 2)
L {y (t)} =
s2 (s
s>0
Using partial fraction decomposition, we get
s3 + 4s2 + s + 1
A B
C
D
E
= + 2+
+
+
2
2
2
s
s
s
+
1
(s
+
1)
s
−
2
s (s + 1) (s − 2)
s3 + 4s2 + s + 1 = As (s + 1)2 (s − 2) + B (s + 1)2 (s − 2)
+ Cs2 (s + 1) (s − 2) + Ds2 (s − 2) + Es2 (s + 1)2
s3 + 4s2 + s + 1 = (A + C + E) s4 + (B − C + D + 2E) s3
+ (−3A − 2C − 2D + E) s2 + (−2A − 3B) s + (−2B)
(41)
(42)
(43)
Comparing coefficients on the left- and right-hand sides, we get the system of equations
below.
A+C +E =0
(44)
B − C + D + 2E = 1
(45)
−3A − 2C − 2D + E = 4
(46)
−2A − 3B = 1
(47)
−2B = 1
(48)
Solving this system, we get A = 1/4, B = −1/2, C = −1, D = −1 and E = 3/4. So we
have that the Laplace transform of the solution is
L (f (t)) =
1
3/4
1
1/4 1/2
+
− 2 −
−
.
2
s
s
s + 1 (s + 1)
s−2
(49)
Using Table 6.2.1 on pg. 317 of the textbook, we get that the solution is
f (t) =
3
1 1
− t − e−t − te−t + e2t
4 2
4
(50)
3. Find the Laplace transform of f (t) and inverse Laplace transform of G(s),
where
e−3s
0
t<1
G(s)
=
f (t) =
(51)
t2 − 3t + 1
t≥1
s2 − s − 6
Z
∞
e−st t2 − 3t + 1 dt
1
Z ∞
∞
1 −st
1 −st 2
−
− e
= − e
t − 3t + 1
(2t − 3) dt
s
s
1
1
∞ Z ∞ 1 −st
1 −s 1
1 −st
−
− e
=− e +
− e (2t − 3)
· 2dt
s
s
s
s
1
1
1 −s 1
1 −s
2 −st ∞
=− e +
− e − 2e s
s
s
s
1
1
1
2
−s
= − − 2+ 3 e
s s
s
L {f (t)} =
L {g (t)} =
e−3s
.
s2 − s − 6
(52)
(IBP)
(53)
(IBP)
(54)
(55)
(56)
(57)
In Table 6.2.1, we have
L {uc (t) f (t − c)} = e−cs L {f (t)} .
(58)
We can use this formula, with c = 3 and
L {f (t)} =
1
.
s2 − s − 6
(59)
We can match this to a formula in the table if we use partial fraction decomposition.
s2
1
1
=
−s−6
(s − 3) (s + 2)
A
B
=
+
s−3 s+2
1 = A (s + 2) + B (s − 3)
= (A + B) s + (2A − 3B) .
(60)
(61)
(62)
(63)
Matching the left- and right-hand sides gives us the following system of equations:
A+B =0
(64)
2A − 3B = 1.
(65)
We can solve this to get A = 1/5 and B = −1/5. Thus we have
1
1
1
1
−
.
L {f (t)} =
5 s−3
5 s+2
(66)
In Table 6.2.1, we have
L {eas } =
1
,
s−a
s > a.
(67)
Thus g(t) is
1
1
g (t) = e3t − e−2t .
5
5
(68)