MAT 274
c
Bin
Cheng
HW 6 Solutions
MAT 274 HW 6 Solutions
Due 11:59pm, M 10/10, 2011.
( 70 points total )
Notation: the step function ua (t) is defined as
(
0,
ua (t) := u(t − a) =
1,
t<a
t≥a
1. (10’) Find the Laplace transform of the solution Y (s) = L[y(t)] for the DE
d4
y(t) − 3y 000 (t) + 3y 00 (t) − 3y 0 (t) + 2y(t) = t sin(2t)
dt4
with initial data
y(0) = y 0 (0) = 0,
y 00 (0) = y 000 (0) = 4.
Solution. Let Y (s) = L{y(t)}. Then,
d4
4
3
2 0
00
000
4
L{ dt
4 y(t)} = s Y (s) − s y(0) − s y (0) − sy (0) − y (0) = s Y (s) − 4s − 4
Likewise, L{y 000 (t)} = s3 Y (s) − 4, L{y 00 (t)} = s2 Y (s), L{y 0 (t)} = sY (s).
2
For the t sin(2t) term on the RHS, we can start with L{sin(2t)} = s2 +2
2 and then
d
n
n
apply the differetiation rules L{t f (t)} = (−1) ds L{f (t)} to obtain L{t sin(2t)} =
d
2
d
L{sin(2t)} = − ds
− ds
= (s24s
s2 +22
+4)2
So, apply Laplace transform on both sides the original DE
s4 Y (s) − 4s − 4 − 3(s3 Y (s) − 4) + 3s2 Y (s) − 3sY (s) + 2Y (s) =
4s
(s2 + 4)2
Gather all terms with Y (s) on the LSH
(s4 − 3s3 + 3s2 − 3s + 2)Y (s) =
i.e.
Y (s) =
4s
(s2 +4)2
s4 − 3s3 +
(s2
4s
+ 4s − 8
+ 4)2
+ 4s − 8
3s2 − 3s + 2
.
2. (5’×6=30’) Find the Laplace transforms of the following functions.
a) (t − 2)2 · e−3t ,
Solution. Rewrite (t − 2)2 · e−3t = t2 e−3t − 4te−3t + 4e−3t and use
L{tn eat } =
n!
(s − a)n+1
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MAT 274
to obtain the answer
c
Bin
Cheng
HW 6 Solutions
2
4
4
−
+
(s + 3)3 (s + 3)2 s + 3
b) (t − 2)2 · e−3(t−2) , (hint: use part a))
Solution. Since (t − 2)2 · e−3(t−2) = (t − 2)2 · e−3t+6 = (t − 2)2 · e−3t · e6 , the answer is
simply that of part (a) times e6 ,
2
4
4
· e6
−
+
(s + 3)3 (s + 3)2 s + 3
c) u(t − 2) · (t − 2)2 · e−3(t−2) , (hint: do not use part a) or b))
Solution. This function is a cut-off translation of t2 e−3t . So, we first transform the
un-translated version
2
L{t2 e−3t } =
(s + 3)3
and apply the translation rules L{u(t − a)f (t − a)} = e−as L{f (t)}
L{u(t − 2) · (t − 2)2 · e−3(t−2) } = e−2s · L{t2 e−3t } =
2e−2s
(s + 3)3
For each of the following problems, in addition to finding the Laplace transform,
also plot the graph of the function vs. t (not its Laplace transform vs. s)
t
d) e− π · sin(6t)
Solution. First transform
6
+ 62
at
Then, apply the translation rules L{e f (t)} = F (s − a) to obtain
L{sin(6t)} =
t
L{e− π sin(6t)} =
s2
6
.
(s + πt )2 + 62
t−2π
e) u(t − 2π) · e− π · sin(6(t − 2π)), (hint: use part d))
Solution. This a cut-off translation of part d), so the answer is that of part d) times
e−2πs ,
6
e−2πs ·
t 2
(s + π ) + 62


t<3

1,
f) g(t) =
t−3
e 2 , 3≤t≤5


2,
t>5
Solution. Merge g(t) into one expression
g(t) = (1 − u1 (t)) · 1 + (u3 (t) − u5 (t))e
t−3
2
+ u5 (t) · 2.
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MAT 274
c
Bin
Cheng
HW 6 Solutions
The first term is transformed as
1
L{(1 − u1 (t)) · 1} = (1 − e−s )
s
The third term is transformed as
2
L{u5 (t) · 2} = e−5s
s
For the second term above, we split it into u3 (t)e
respectively
L{u3 (t)e
L{−u5 (t)e
t−3
2
t−3
2
} = L{u(t − 3)e
} = −L{u(t − 5)e
t−5+2
2
t−3
2
t−3
2
and −u5 (t)e
t
} = e−3s L{e 2 } = e−3s
} = −e−3s L{e
t+2
2
t−3
2
1
s−
, and transform
1
2
t
} = −e−3s L{e 2 · e} = −e−3s
e
s−
1
2
The answer will the sum of all 4 terms on the RHS above.
o
n
3. (10’) Knowing L √1πt = √1s , find the inverse Laplace transform
−1
L
Solution.
d2 1
√
ds2 s
and
−1
L
e−3s
√
s
d2 1
1
t2
2 −1
√
√
√
L
=
(−t)
L
=
ds2 s
s
πt
−3s e
1
L−1 √
.
= u(t − 3) · p
s
π(t − 3)
−1
4. (20’) Use Laplace transform to solve for y(t) in
y 00 − 6y 0 + 9y = 12(t − 2) · u(t − 2), y(0) = 1, y 0 (0) = 6.
Show all your work.
Solution. L{LHS} = (s2 Y − s − 6) − 6(sY − 1) + 9Y = (s2 − 6s + 9)Y − s and
L{RHS} = e−2s L{12t} = e−2s 12
. So, the original DE is transformed as
s2
(s2 − 6s + 9)Y − s = e−2s
12
s2
e−2s 12
+s
e−2s 12
s
s2
s2
=⇒ Y (s) = 2
= 2
+
= Y1 (s) + Y2 (s)
s − 6s + 9
s − 6s + 9 s2 − 6s + 9
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MAT 274
c
Bin
Cheng
HW 6 Solutions
We’ll inverse transform Y1 , Y2 respectively.
For Y1 , first, set s−2s aside and perform partial fractions on
12
s2
s2
− 6s + 9
=
A B
C
12
D
= + 2+
+
2
− 3)
s
s
s − 3 (s − 3)2
(1)
s2 (s
=⇒ 12 = As(s − 3)2 + B(s − 3)2 + C(s − 3)s2 + Ds2
Set s = 0 to have 12 = B(−3)2 =⇒ B = 34 .
Set s = 3 to have 12 = D32 =⇒ D = 43 .
Set s = 1 to have 12 = 4A + 4B − 2C + D =⇒ 4A − 2C = − 20
3
Compare the coefficients of s3 on both sides to have 0 = A + C =⇒ A = −C. Plug it
10
into the above equation to have −4C − 2C = − 20
3 . Thus, C = −A = 9 .
Therefore, the partial fractions for (1) is
12
s2
s2 − 6s + 9
=
−10/9 4/3
10/9
4/3
+ 2 +
+
s
s
s − 3 (s − 3)2
and the corresponding inverse transform is
)
(
12
s2
−1
L
s2 − 6s + 9
=−
10 4
10
4
+ t + e3t + e3t t
9
3
9
3
Then, add the e−2s term back to have
(
L−1 {Y1 (s)} =
e−2s
12
s2
)
s2 − 6s + 9
10 4
10 3(t−2) 4 3(t−2)
= u(t − 2) − + (t − 2) + e
+ e
(t − 2)
9
3
9
3
For Y2 (s), a quick way to perform partial fractions is
Y2 (s) =
s2
s
s
s−3+3
1
3
=
=
=
+
2
2
− 6s + 9
(s − 3)
(s − 3)
s − 3 (s − 3)2
So
L−1 {Y2 (s)} = e3t + 3e3t t
And the final answer will be y(t) = L−1 {Y (s)} = L−1 {Y1 (s) + Y2 (s)}, i.e.
10 4
10 3(t−2) 4 3(t−2)
y(t) = u(t − 2) − + (t − 2) + e
+ e
(t − 2) + e3t + 3e3t t
9
3
9
3
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