Section 5.2

1. Problem Show whether or not the given function is exponential order. If it is, determine a suitable set of values for K, c, and T in (2).
(a) 5 e4 t
(g) cos t3
Solution
(a) Yes. K = 4, c = 5 for example.
(d) Note that cosh 3t = (e3 t + e−3 t)/2 so K = 1/2, c = 3 for example.
(g) Note that |cos t3| ≤ 1 so practically any values of K and c will do.
8. Problem Derive L{ea tsin b t } two ways: using integration by parts and using sin bt = ℑeibt
Solution
Using integration by parts is too messy. Don't bother. Instead, take the second approach, i.e.
I
=
L (ea t sin b t)
=
L (ℑ(ea t ei b t ))
=
ℑ( L( e(a + i b) t) )
=
ℑ⎛
⎝ 		⌠
⌡ 		∞

0 
 e (a + i b)t e−s t dt⎞
⎠
=
ℑ⎛
⎝ 		⌠
⌡ 		∞

0 
 e (a + i b − s)t dt⎞
⎠
=
ℑ⎛
⎝ 		⎡
⎣ 		e (a + i b −s) t
a − s + i b
⎤
⎦ 		∞

0 
⎞
⎠
=
ℑ⎛
⎝ 		−1
a − s + i b
⎞
⎠
=
ℑ⎛
⎝ 		1
s − a − i b
⎞
⎠
=
ℑ⎛
⎝ 		(s − a) + i b
(s − a)2 + b2
⎞
⎠
=
b
(s − a)2 + b2
 .
where ℑ(z) is the imaginary part of a complex number, z.

Section 5.3

  1. Problem Find the inverse of the given transform two different ways: using partial fractions and using the convolution theorem. Cite any entries used from Appendix C.
(a)
3
s (s + 8)
Solution
Use
1
(s − a)(s − b)
 = 1
b − a
 ⎛
⎝ 		1
s − b
 − 1
s − a
⎞
⎠ 		.
(a)
L−1⎛
⎝ 		3
s (s + 8)
⎞
⎠
=
L−1 ⎛
⎝ 		3
8
 ( 1
s
 − 1
s + 8
 )⎞
⎠
=
3
8
 (1 − e− 8 t).
(Convolution theorem)
Noting that L(1) = 1/s and L(e− 8 t) = 1/(s + 8),
L−1 ⎛
⎝ 		3 1
s
 1
s + 8
⎞
⎠
=
3 L−1 ( 1
s
 ) * L−1 ( 1
s + 8
 )
=
3 1 * e− 8 t
=
3 ⌠
⌡ 		t

0 
 e−8 (t − τ) dτ
=
3 e−8 t ⌠
⌡ 		t

0 
 e8 τ dτ
=
3
8
 (1−e− 8 t).
3 Problem Use entry 9 in Appendix C to evaluate the inverse of each. If necessary, use entry 10 as well. NOTE: See the Comment in Example 2.
Solution
(a) Note that s2 + 8 s = (s + 4)2 − 16 = (s+4)2 + (4 i)2. So a = −4, b = 4i and
1
s2 + 8 s
 = 1
4 i
 4 i
(s + 4)2 + (4 i)2
 .
Therefore,

L−1 ⎛
⎝ 		1
s2 + 8 s
⎞
⎠
=
L−1 ⎛
⎝ 		1
4i
 4 i
(s + 4)2 + (4 i)2
⎞
⎠
=
1
4i
 L−1 ⎛
⎝ 		4i
(s + 4)2 + (4 i)2
⎞
⎠
=
1
4i
 e−4t sin (4 i t)
=
1
4i
 e−4 t 1
2i
 (e i (4it) − e−i (4it))
=
1
8
 (1 − e− 8 t).
6. Problem Prove that L{f * g * h} = F(s) G(s) H(s) or, equivalently, that L−1{F(s) G(s) H(s)} = f * g * h. NOTE: Does f * g * h mean (f * g) * h or f * (g * h) ? According to the associative property (22b) it does not matter;they are equal.
Solution

L( f* (g * h))=
f
 
g * h
 
 =
f
 
g
 
h
 
 .

7. Problem To illustrate the result stated in Exercise 6, find the inverse of 1/s3 as L−1{[1/(s3)]} = L−1{[1/(s)][1/(s)][1/(s)]}=1*1*1, and show that the result agrees with that given in Appendix C.
Solution

L−1( 1
s3
 ) = 1*1*1 = 1*(1*1) = ⌠
⌡ 		t

0 
 1 ( ⌠
⌡ 		t

0 
 1 ×1 dτ′) dτ = ⌠
⌡ 		t

0 
 1 ×(t−τ) dτ = (t τ−τ2/2)|0t = t2
2
 .

10. Problem Evaluate the transform of each;
(a)
⌠
⌡ 		t

0 
 et−τ sin 2τdτ
(d)
⌠
⌡ 		t

0 
 cosh 3(t−τ) dτ
Solution
Convolution !!
(a)
L(⌠
⌡ 		t

0 
 et−τ sin2τdτ) = L(et * sin2t) = L(et) L(sin2t) = 1
s−1
 2
s2+4
 .

(d)
L⎛
⎝ 		⌠
⌡ 		t

0 
 cosh 3(t−τ) dτ⎞
⎠ 		= L(cosh 3 t * 1) = L(cosh 3 t ) L(1) = s
s2−32
 1
s
 = 1
s2−9
 .

Section 5.4

1. Problem Use the Laplace transform to find the general solution, or the particular solution if initial conditions are given.
(a) x′+ 2x = 4 t2
(d) x" = 6t;     x(0)=2, x′(0)=−1
Solution
(a) Apply the Laplace transforms on the equation to get

s
x
 
 (s) − x(0) +2
x
 
 (s)
=
L(4t2)
=
8
s3
 ,
or

(s+2)
x
 
 (s) = x(0)+ 8
s3
 ,
which can be solved as

x
 
 (s)
=
x(0)
s+2
 + 8
s3 (s+2)
=
x(0)
s+2
 + ⎛
⎝ 		4
s3
 − 2
s2
 + 1
s
 − 1
s+2
⎞
⎠ 		,
which can be inverted as

x(t) = x(0) e−2t + (1−e−2t −2t + 2t2).
(d) Applying the Laplace transforms on the equation yields

s2
x
 
 (s) − s x(0) −
⋅
x
 
 (0) = 6
s2
 ,
or

s2
x
 
 (s) − 2s +1 = 6
s2
 ,
which can be solved as

x
 
 (s) = 2
s
 − 1
s2
 + 6
s4
 ,
so

x(t) = 2 − t + t3.
5. Problem (Variable-coefficient equation) Consider the problem

t x" + x′+ t x = 0     (0 ≤ t < ∞)
(1)

x(0)=1,     x′(0)=0
where our special interest lies in seeing whether or not we can solve (1) by the Laplace transform method even though differential equation has nonconstant coefficients.
(a) Take the Laplace transform of the differential equation. Note that the transforms of t x"(t) and t x(t),

L{tx"(t)}=⌠
⌡ 		∞

0 
 tx"e−stdt

L{tx(t)}=⌠
⌡ 		∞

0 
 txe−stdt
present a difficulty in that we cannot express them in terms of X(s) the way we can express L{x′(t)}=sX(s)−x(0) L{x"(t)}=s2X(s)−sx(0)−x′(0). Nevertheless, terms can be handled as follows. Observe that

L{tx"(t)}
=
⌠
⌡ 		∞

0 
 tx"e−stdt
=
−⌠
⌡ 		∞

0 
 d
ds
 (x"e−st)dt
=
− d
ds
 ⌠
⌡ 		∞

0 
 x"e−stdt
=
− d
ds
 [s2X(s)−sx(0)−x′(0)]
=
− d
ds
 [s2X(s)−s],

   
(2)
if we assume that the unknown x(t) is sufficiently well behaved for the third equality (where we have interchanged the order of two limit processes, the s differentiation and the t integration) to be justified. Handling the L{tx(t)} term in the same way, show that the application of the Laplace transform to (1) leads to the equation

(s2 + 1) dX
ds
 + s X = 0
(3)
on X(x). Note that whereas the Laplace transform method reduces constant-coefficient differential equations to linear algebraic equations on X(s), here the nonconstant coefficients result in the equation on X(s) being itself a linear differential equation! However, it is a simple one. Solving (3), show that

X(s)= C

√
s2+1
(4)
(b) From Appendix C, we find the inverse as x(t)=C J0(t), where J0 is the Bessel function of the first kind, of order zero. Applying the initial condition once again gives x(0) = 1 = C J0(0) = C, so C=1, and the desired solution of (1) is x(t) = J0(t). Here,however, we ask you to proceed as though you do not know about Bessel functions. Specifically, re-express (4) as

X(s)= C
s
√
1+(1/s2)
 = C
s
 ⎛
⎝ 		1− 1
2
 1
s2
 + …⎞
⎠
(5)
where the last equality amounts to the Taylor expansion of √{1+r} in the quantity r, about r=0, where r=1/s2. Carry that expansion further; invert the resulting series term by term (assuming that that step is valid), and thus show that

x(t)=C⎡
⎣ 		1− t2
22
 + 1
(2!)2
 t4
24
 − 1
(3!)2
 t6
26
 +...⎤
⎦
Setting x(0)=1 gives C=1, and the result is that we have obtained the solution in power series form. Of course, that power series is the Taylor series of the Bessel function J0(t). NOTE: Observe that rather than pulling an s out of the square root in (5), and then expanding 1/√{1+(1/s2)} in powers of 1/s2, we could have expanded (4) directly in powers of s as X(s)=C (1−[1/2]s2+...). However, positive powers of s are not invertible, so this form is of no use. [We will see, in Theorem 5.7.6, that to be invertible a transform must tend to zero as s→∞. Positive powers of s do not satisfy this condition, but negative powers do.] Also, observe that the degree to which nonconstant-coefficient differential equations are harder than constant-coefficient ones can be glimpsed from the fact that coefficients proportional to t cause the equation on X(s) to be a first-order differential equation; coefficients proportional to t2 will cause the equation on X(s) to be a second-order differential equation,and so on.
Solution
(a) Just follow the hint.
(s2+1) dX
ds
 + sX=0
is equivalent to
dX
X
 + s
s2+1
 ds = 0
which can be integrated to yield
logX + 1
2
 log(s2+1) = C1
or
log⎛
⎝ 		X
√
 
1+s2
 
 ⎞
⎠ 		= C1
or
X = C

√
1+s2
 .
(b) Follow the hint until you feel comfortable.


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On 20 Nov 2022, 18:03.