18.03 Problem Set 7: Part II Solutions
Part I points: 26. 6, 27. 10, 28. 12.
� (−1+3i)t
�
1
−t
(−1−3i)t
I.26.
e
sin(3t)
=
e
−
e
, so L[e−t sin(3t)] =
2i
�
�
1
1
1
1 (s + 1 + 3i) − (s + 1 − 3i)
3
−
=
=
.
2
2i s − (−1 + 3i) s − (−1 − 3i)
2i
(s + 1) + 9
(s + 1)2 + 9
� ∞
� ∞
−st
26. (a) [10] G(s) =
f (at)e 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
� ∞
�
1 ∞
1 �s�
−su/a du
G(s) =
f (u)e
=
f (u)e−(s/a)u du = F
.
a
a 0
a
a
0
n!
an n!
For example, take f (t) = tn , so F (s) = n+1 , g(t) = (at)n = an tn , G(s) = n+1 . Now
s
s
1 �s� 1
n!
an+1 n!
an n!
compute F
=
=
= n+1 = G(s).
a
a
a (s/a)n+1
a sn+1
s
� ∞� ∞
��
−sx
−sy
(b) [10] Compute F (s)G(s) =
f (x)e g(y)e dxdy =
f (x)g(y)e−s(x+y) dxdy,
0
0
R
where R is the first quadrant. The suggested substitution
� ∂x is∂xx�= t − τ ,�y = τ . To
� convert
1
−1
∂t
∂τ
to these coordinates, note that the Jacobian is det ∂y
= det
= 1 For
∂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 dτ dt
0 �0
� ∞ �� t
� ∞
�
−st
−st
=
f (t − τ )g(τ ) dτ e dt =
(f (t) ∗ g(t)) e dt =
0
0
0
�
(c) [6] F (s) =
0
∞
f (t)e−st dτ =
1
�
f (t)e−st dτ +
0
∞
h(t)e−st dt = H(s).
0
�
∞
0e−st dτ . The improper integral
1
converges for any
of convergence is the whole complex plane. Continuing,
�1 s; the region
1 −st ��
1 − e−s
F (s) =
e � =
. [Why doesn’t this blow up when s → 0? The numerator
−s
s
0
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 aẇ +bw = δ(t) is asW (s)+bW (s) = 1.
1
1/a
Solve: W (s) =
=
This is the Laplace transform of w(t) = a1 u(t)e−bt/a .
as + b
s + (b/a)
(b) This is called the “unit ramp response.”
(i) [6] xp = c1 t+c0 , ac1 +b(c1 t+c0 ) = t, c1 = 1b (as long as b �= 0), ac1 +bc0 = 0 so c0 = − ba2 ,
xp = 1b t− ba2 . x(t) = xp +ce−bt/a , so 0 = x(0) = − ba2 +c and x(t) = u(t)( 1b t− ba2 (1−e−bt/a )).
If b = 0 then aẋ = t, which has general solution x(t) = 21a t2 + c. 0 = x(0) = c, so
x(t) = u(t) 21a t2 .
� t
�
1 −b(t−τ )/a
1 −bt/a t bτ /a
e τ dτ . Do this by parts:
(ii) [6] If b �= 0: w(t) ∗ t =
e
τ dτ = e
a
0
0 a
1
u = τ , du = dτ , dv = e dτ , v =
w(t) ∗ t = e−bt/a
a
�
�
2
1 −tb/a
e
t ab ebt/a − ab2 (ebt/a − 1) = 1b t − ba2 (1 − e−bt/a ).
a
� t
1
1 t2
If b = 0, w(t) ∗ t =
τ dτ =
.
a2
0 a
bτ /a
a bτ /a
e ,
b
�
�
� t
a bτ /a
a bτ /a ��t
τ e � −
e
dτ =
b
0
0 b
A B
C
+ 2+
+ b)
s
s
as + b
b
1
2
Coverup: Multiply by s and set s = 0 to get B = b . Multiply by as + b and set s = − a
2
to get C = ab2 . Here’s a clean way to get A: multiply through by s and then take s very
a/b2
a/b2 1/b
large in size. You find 0 = A + Ca , or A = − ba2 . So X = −
+ 2 +
, which
s
s
s + b/a
is the Laplace transform of x = − ba2 + 1b t + ba2 e−bt/a .
1
11
If b = 0, aẋ = t has Laplace transform asX = 2 so X =
, and x = u(t) a1 21 t2
s
a s3
(iii) [6] aẋ+bx = t has Laplace transform asX+bX =
28. (a) [6] w(t) has Laplace transform W (s) =
1
,
s2 +1
1
,
s2
so X =
1
s2 (as
=
1
1
1
=
. L[sin t] =
3s2 + 6s + 6
3 (s + 1)2 + 1
so by s-shift w(t) = 31 u(t)e−t sin t.
1
. To use partial fractions we need to factor s4 + 1, which is to say
+1
we need to find its roots. They are the fourth roots of −1, which are r, r, −r, and −r where
1
1
a
b
c
d
=
r = √12 (1+i). Now 4
=
+
+
+
.
s +1
(s − r)(s − r)(s + r)(s + r)
s−r s−r s+r s+r
Coverup or cross-multiplication will lead to the coefficients. This is not pretty, and (per
the web) I don’t expect more.
(b) [14] W (s) =
s4
[What I intended to ask was for the weight function for D4 −I. Now the roots are ±1 and
1
a
b
c
d
±i, so we can write 4
=
+
+
+
. Coverup gives easily a =
s −1
s−1 s+1 s−i s+i
b = 41 , c = 4i , d = − 4i . So w(t) = u(t) 14 (et + e−t + ieit − ie−it ) = u(t) 12 (sinh(t) − sin(t)).
I apologize for the mistake.]
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18.03 Differential Equations����
Spring 2010
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