Math 215/255 HW6 Solutions
Section 6.3
#12 (a)
(2,2)
(5,2)
@
@
@
@
@
(0,0)
(7,0)
(b) f (t) can be decomposed as the sum of the following functions
t(1 − u2 (t)),
2(u2 (t) − u5 (t)),
(7 − t)(u5 (t) − u7 (t)),
0
each of which is nonzero only in one of the 4 intervals. Their sum gives
f (t) = t + (2 − t)u2 (t) + (5 − t)u5 (t) + (t − 7)u7 (t).
#14
We can rewrite
f (t) = u1 (t)(t2 − 2t + 2) = u1 (t)[(t − 1)2 + 1]
which can be identified as u1 (t)g(t − 1) with g(t) = t2 + 1. Since L[g(t)] =
2
1
−s
Lf (t) = e
+
.
s3
s
#17
To apply the translation in t formula, we rewrite
f (t) = [(t − 2) − 1]u2 (t) − [(t − 3) + 1]u3 (t),
and we have
L[f (t)] = e−2s [
#20
1
1
1
1
− ] − e−s [ 2 + ].
s2
s
s
s
Let
G(s) =
s2
1
1
=
.
+s−2
(s + 2)(s − 1)
By partial fraction
1 1
1
G(s) = [
−
].
3 s−1 s+2
Thus L−1 [G(s)] = 13 [et − e−2t ], and
1
L−1 [e−2s G(s)] = [et−2 − e−2(t−2) ]u2 (t).
3
#21
Let
G(s) =
2(s − 1)
2(s − 1)
=
.
s2 − 2s + 2
(s − 1)2 + 1
1
2
s3
+ 1s , we have
Thus L−1 [G(s)] = 2et cos t, and
L−1 [e−2s G(s)] = 2et−2 cos(t − 2)u2 (t).
Section 6.4
#2a Let L[y(t)] = Y (s). Then L[y 0 ] = sY (s) − 0 and L[y 00 ] = s(sY (s) − 0) − 1. We can write
h(t) = uπ (t) − u2π (t). Taking the Laplace transform of both sides of the ODE, we get
(s2 + 2s + 2)Y (s) − 1 = L[h(t)] =
Thus
Y (s) =
e−πs e−2πs
−
.
s
s
e−πs
e−2πs
1
+
−
.
s2 + 2s + 2 s(s2 + 2s + 2) s(s2 + 2s + 2)
Note s2 + 2s + 2 = (s + 1)2 + 1 and, by partial fraction,
1 1 1 (s + 1) + 1
1
1
1 −t
−t
=
−
= L 1 − e cos t − e sin t .
s(s2 + 2s + 2)
2 s 2 (s + 1)2 + 1
2
2
Applying the translation in t formula,
h
i
1
y(t) = e−t sin t + uπ (t) 1 − e−(t−π) [cos(t − π) + sin(t − π)]
2
h
i
1
− u2π (t) 1 − e−(t−2π) [cos(t − 2π) + sin(t − 2π)]
2
1
1
−t
= e sin t + uπ (t) 1 + eπ−t (cos t + sin t) + u2π (t) −1 + e2π−t (cos t + sin t) .
2
2
#4a
get
Taking the Laplace transform of both sides of the ODE and using the initial conditions, we
(s2 + 4)Y (s) = L[sin t + uπ (t) sin(t − π)] =
Thus
Y (s) =
1
e−πs
+
.
s2 + 1 s2 + 1
e−πs
1
+
.
(s2 + 1)(s2 + 4) (s2 + 1)(s2 + 4)
By partial fraction,
1/3
1/3
1
1
1
=
−
=
L
sin
t
−
sin
2t
.
(s2 + 1)(s2 + 4)
s2 + 1 s2 + 4
3
6
Applying the translation in t formula,
e−πs
1
1
1
1
−1
L
= uπ (t)
sin(t − π) − sin 2(t − π) = uπ (t) − sin t − sin 2t .
(s2 + 1)(s2 + 4)
3
6
3
6
Thus
1
1
1
1
Y (s) = sin t − sin 2t − uπ (t)
sin t + sin 2t .
3
6
3
6
2
#5a
get
Taking the Laplace transform of both sides of the ODE and using the initial conditions, we
(s2 + 3s + 2)Y (s) = L[f (t)].
Since f (t) = 1 − u10 (t),
L[f (t)] =
Thus
Y (s) =
1 e−10s
−
.
s
s
e−10s
1
−
.
s(s2 + 3s + 2) s(s2 + 3s + 2)
Note s(s2 + 3s + 2) = s(s + 1)(s + 2). By partial fraction,
1
1/2
1
1/2
=
−
+
.
s(s + 1)(s + 2)
s
s+1 s+2
Thus
1
1
1
] = − e−t + e−2t .
s(s + 1)(s + 2)
2
2
Applying the translation in t formula,
1
1 −2t
1 −2t+20
1
−t
−t+10
y(t) = − e + e
− u10 (t)
−e
+ e
.
2
2
2
2
L−1 [
Section 6.5
#2a Taking the Laplace transform of both sides of the ODE, we get
s2 Y (s) − sy(0) − y 0 (0) + 4Y (s) = e−πs − e−2πs .
Applying the initial conditions, the left side is (s2 + 4)Y (s). Thus
Y (s) =
e−2πs
e−πs
−
.
s2 + 4 s2 + 4
Applying the translation in t formula,
y(t) =
#5a
1
1
1
sin(2t − 2π)uπ (t) − sin(2t − 4π)u2π (t) = sin(2t)[uπ (t) − u2π (t)].
2
2
2
Taking the Laplace transform of both sides of the ODE, we get
(s2 + 2s + 3)Y (s) =
so
Y (s) =
s2
1
+ e−3πs ,
+1
1
e−3πs
+
.
(s2 + 1)(s2 + 2s + 3) (s2 + 2s + 3)
Note s2 + 2s + 3 = (s + 1)2 + 2. By partial fraction,
Y (s) =
1
1
− 14 s
1
4
4 (s + 1)
+
+
+
e−3πs .
2
2
2
s + 1 s + 1 (s + 1) + 2 (s + 1)2 + 2
Thus, using the table and the translation in t formula,
√
√
1
1
1
1
y(t) = − cos t + sin t + e−t cos 2t + √ u3π (t)e−(t−3π) sin 2(t − 3π).
4
4
4
2
3