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18.03 Problem Set 7: Part II Solutions
Part I points: 26. 6, 27. 10, 28. 12.
I.26. e−t sin(3t) = 1 e(−1+3i)t (−1−3i)t , so L[e−t sin(3t)] =
� 2i − e
�
1 1 1=1 (s + 1 + 3i) − (s + 1 − 3i)= 3.
2i s − (−1+3i) − s − (−1 − 3i) 2i (s + 1)2 + 9 (s + 1)2 + 9
∞ ∞
26. (a) [10] G(s) = f(at)e−st dt. To make this look more like F (s) = f(t)e−st dt,
0 0
make the substitution u = at. Then du = a dt and
∞
f(u)e−su/a du 1 ∞
f(u)e−(s/a)u du =1 � s �
G(s) = = F .
0 a a 0 a a
n! ann!
For example, take f(t) = tn, so F (s) = sn+1 , g(t)=(at)n = antn
, G(s) = sn+1 . Now
� � n+1 n
1 s 1 n! a n! a n!
compute F = = = = G(s).
a a a (s/a)n+1 a sn+1 sn+1
(b) [10] Compute F (s)G(s) = ∞ ∞
f(x)e−sx
g(y)e−sy dxdy = f(x)g(y)e−s(x+y) dxdy,
0 0 R
where R is the first quadrant. The suggested substitution is x = t − τ, y = τ . To convert
∂x ∂x
to these coordinates, note that the Jacobian is det ∂t ∂ τ = det 1 −1 = 1 For
∂y ∂y 0 1
∂t ∂τ
fixed t, τ ranges over numbers between 0 and t, and t ranges over positive numbers. Since
∞ t
x + y = t, F (s)G(s) = f(t − τ)g(τ)e−st dτdt
� �� 0 �0
� �
∞ t ∞ ∞
= f(t − τ)g(τ) dτ e−st dt = (f(t) ∗ g(t)) e−st dt = h(t)e−st dt = H(s).
0 0 0 0
� � 1 �
∞ ∞
(c) [6] F (s) = f(t)e−st dτ = f(t)e−st dτ + 0e−st dτ . The improper integral
0 0 1
converges for any s; the region of convergence is the whole complex plane. Continuing,
�1
1 e−st
�
� 1 − e−s
F (s) = −s �
0
= s . [Why doesn’t this blow up when s → 0? The numerator
goes to zero too, then, and the limit of the quotient (by l’Hopital for example) is 1.]
27. (a) [4] The Laplace transform of the equation aw˙ + bw = δ(t) is asW (s)+bW (s) = 1.
Solve: W (s) = 1= 1/a This is the Laplace transform of w(t) = a
1 u(t)e−bt/a.
as + b s + (b/a)
(b) This is called the “unit ramp response.”
(i) [6] xp = c1t+c0, ac1+b(c1t+c0) = t, c1 = 1
b (as long as b = 0), �ac1+bc0 = 0 so c0 = −b
a
2 ,
xp = 1 t− a x(t) = xp +ce−bt/a, so 0 = x(0) = −b
a
2 +c and x(t) = u(t)(1 t− a (1−e−bt/a)).
b b2 . b b2
If b = 0 then ax˙ = t, which has general solution x(t) = 2
1
a t2 + c. 0 = x(0) = c, so
x(t) = u(t)2
1
a t2
.
t t
1 1
(ii) [6] If b �= 0: w(t) ∗ t =
0 ae−b(t−τ )/a τ dτ = ae−bt/a
0 ebτ/a τ dτ . Do this by parts:
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