AP Calculus BC
Laplace Transformation – Day 3
Name:____________________________
12 January 2016
Laplace Transform Problems
Example problems using the Laplace Transform.
1. Solve the differential equation
y! − y = et,
with the initial value conditions y(0) = 0.
Ans.
y(t) = tet
Solution:
s2Y (s) − 0.16s + sY(s) − 0.16 + 9Y (s) = 0
(s2 + s + 9)Y = 0.16( s + 1)
The solution is
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Y(s) =
0.16(s + 1) 0.16(s + 12 )
0.08
=
+
2
1 2
35
s + s + 9 (s + 2 ) + 4 (s + 12 ) 2 +
35
4
.
Taking the inverse transform gives the answer above.
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2. Solve the differential equation
0.1i"" + 11i" + 100i(t) = 40000cos( 400t )
with the initial value conditions i(0) = 0, i"(0) = 0.
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Ans.
i(t) = −0.2776e−10t + 2.6144e−100t − 2.3368cos 400t + 0.6467sin 400t
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This problem describes the current in an RLC circuit with resistance R = 11 Ohms,
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capacitance
C = 1/100 Farad, and inductance L = 1/10 Henry, with an imposed sinusoidal
voltage.
Special thanks to Mr. Olsen and Calculus (4th edition) by Deborah Hughes-Hallet, et. al. for most of these problems.
Laplace Transformation – Day 3
Application Problems
Solution:
s
s + 400 2
s
(s2 + 110s + 1000)I(s) = 1000 ⋅ 400 2
s + 400 2
1000 ⋅ 400
s
I(s) =
⋅ 2
(s + 10)(s + 100) s + 400 2
0.1s2 I(s) + 11sI(s) + 100I(s) = 40000
Now
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2
400000s
A
B
Cs + D
=
+
+ 2
,
2
2
(s + 10)(s + 100)(s + 400 ) s + 10 s + 100 s + 400 2
giving
€
400000s = A(s + 100)(s2 + 400 2 ) + B(s + 10)(s2 + 400 2 ) + (Cs + D)(s + 10)(s + 100)
Now, for s = – 10, we get
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−4000000 = 90(10 2 + 400 2 )A, or A = −0.27760
For s = – 100, we have
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−400000000 = −90(100 2 + 400 2 )B, or B = 2.6144
By looking at s3 terms, we get
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0 = A + B + C, or C = −2.3368
By looking at s2 terms, we get
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0 = 100A + 10B + 110C + D, or D = 258.66.
Then
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I(s) =
−0.2776 2.6144 −2.3368s 258.66
400
+
+ 2
+
⋅ 2
2
s + 10
s + 100 s + 400
400 s + 400 2
Taking the inverse transform, we get the solution:
€
€
i(t) = −0.2776e−10t + 2.6144e−100t − 2.3368cos 400t + 0.6467sin 400t
page 2
Laplace Transformation – Day 3
Application Problems
3. Solve the differential equation
y "" + 4 y " + 13y = 145cos2t, with y(0) = 10, and y "(0) = 14
Ans.
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y(t) = e−2t cos 3t + 9cos2t + 8sin2t .
Solution:
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Taking
the Laplace Transform gives
(s2Y (s) −10s −14) + 4(sY(s) −10) + 13Y (s) =
(s2 + 4s + 13)Y (s) = 10s + 14 + 40 +
Y (s) =
145s
s2 + 4
145s
s2 + 4
10s + 54
145s
+ 2
s + 4s + 13 (s + 4s + 13)(s2 + 4)
2
The last term can be written
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145s
As + B
Cs + D
= 2
+ 2
2
(s + 4s + 13)(s + 4) s + 4s + 13 s + 4
2
145s = (As + B)(s2 + 4) + (Cs + D)(s2 + 4s + 13)
145s = s3 (A + C) + s2 (B + 4C + D) + s(4 A + 13C + 4D) + (4B + 13D)
From this we get
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#C = −A
%
%D = 4 A − B
$
% 7A − 4B = 145
%&52A − 9B = 0
This yields A = −9; B = −52; C = 9; D = 16 .
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Thus
(10s − 9s) + (54 − 52) 9s + 16
+ 2
s2 + 4s + 13
s +4
(s + 2)
s
2
Y (s) =
+9 2
+8 2
2
2
2
(s + 2) + 3
s +2
s + 22
Y (s) =
€
€
where in the last line we have completed the square in the first term. Taking the inverse
transform, we obtain the answer above.
page 3
Laplace Transformation – Day 3
Application Problems
page 4
Practice Laplace Transform problems
5.
dy 2
= t + y(t)
dt
F(s) =
2
f (0)
2 2 2 2 + f (0)
+
=− − 2 − 3 +
⇒ y(t) = −2 − 2t − t 2 + (2 + f (0))et
s (s −1) s −1
s s s
s −1
3
6. Find the Laplace Transform for f (t) = tet
1
1
L {tet } =
(this leads to L {te at } =
)
2
(s −1)
(s − a)2
7. Solve y!! + 3y! + 2y = 1 where y = 0 and y! = 0 for t = 0.
F(s) =
1
1
1
1
1
1
2
2
=
−
+
⇒ y(t) = − e−t + e−2t
2
s(s + 3s + 2) s s +1 s + 2
2
2
8. Solve y!! + 4y = 5et +16t where y = 4 and y! = 7 for t = 0. (This might take another piece of paper.)
F(s) =
4s + 7
5
16
3s
2
1
4
+
+ 2 2
= 2 2+ 2 2+
+ 2
2
2
s + 4 (s −1)(s + 4) s (s + 4) s + 2 s + 2 s −1 s
⇒ y(t) = 3cos(2t) + sin(2t) + et + 4t
Laplace Transformation – Day 3
Application Problems
page 5
Application problems
You may solve these using any method – but the last problem requires the Laplace Transform.
9. Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square centimeter
per year. At the same time, these leaves decompose at a continuous rate of 75% per year.
Write a differential equation for the total quantity of dead leaves (per square centimeter) at
time t. Sketch a solution showing that the quantity of dead leaves tends toward an
equilibrium level. What is that equilibrium level?
10. As you know, when a course ends, students start to forget the material they have learned.
One model (called the Ebbinghaus model) assumes that the rate at which a student forgets
material is proportional to the difference between material currently remembered and some
positive constant, a.
a. Let y = f(t) be the fraction of the original material remembered t weeks after the course
has ended. Set up a differential equation for y. Your equation will contain two constants;
the constant a is less than y for all t, and the constant b is the constant of proportionality.
b. Solve the differential equation, using the initial condition y = 1.00 at t = 0.
c. Describe the practical meaning (in terms of the amount remembered) of the constants in
the solution y = f(t).
11. An object of mass m is thrown vertically upward from the surface of the earth with initial
velocity v0. We will calculate the value of v0, called the escape velocity, with which the
object can escape the pull of the gravity and never return to earth. Since the object is moving
far from the surface of the earth, we must take into account the variation of gravity with
altitude. If the acceleration due to gravity at sea level is g, and R is the radius of the earth,
the gravitational force, F , on the object of mass m at an altitude h above the surface of the
earth is given by
mgR 2
F=
(R + h)2
a. Suppose v is the velocity of the object (measured upward) at time t. Using Newton’s
Law of Motion one can show that
dv
gR 2
=−
.
dt
(R + h)2
This equation can be written with h instead of t as the independent variable using the chain
dv
gR 2
dv dv dh
rule
=
⋅ . Hence show that v = −
dh
(R + h)2
dt dh dt
b. Solve the differential equation in part a.
c. Find the escape velocity, the smallest value of v0 such that v is never zero.
Laplace Transformation – Day 3
Application Problems
page 6
12. Use the Laplace Transform on this problem.
In a nuclear reactor, about 6.1 percent of fission reactions produce the isotope Iodine-135.
I-135 then decays to Xenon-135 with a half-life of 6.7 hours, and Xe-135 decays to
Cesium-135 with a half-life of 9.2 hours.
a. When the reactor is shut off, I135 is no longer produced. The amount of I135 present is
then subject to the differential equation
dI
= −a ⋅ I .
dt
Here I(t) is the amount of I135 present at t seconds and a =
1 ln(2)
=
= 2.9 ⋅10 −5 sec−1 .
t I 6.7 hr
Show that this equation has the solution
I(t) = I 0 exp(−a ⋅ t) ,
(1)
where I0 is equal to I(0).
b. The amount of Xe135 present at time t seconds after shutdown is subject to the differential
equation
dX
(2)
= −b ⋅ X + a ⋅ I
dt
where I(t) is the amount of I135 present from part a), and b =
1 ln(2)
=
= 2.1⋅10 −5 sec−1 .
t X 9.2 hr
Substitute the expression (1) for I(t) from part a) into the differential equation (2) and
show that the Laplace transform of the result is
X(s) =
1 "
a ⋅ I0 %
$# X 0 +
'
s+b
s+ a&
(3)
c. Take the inverse Laplace transform of (3) and show that the solution is
X(t) = X 0 exp(−b ⋅ t) +
a ⋅ I0
[exp(−a ⋅ t) − exp(−b ⋅ t)]
b−a
(4)
d. To get a sense of how the amount of Xe135 varies with time, graph X(t) using X0 = 0 and
I0 = 1in equation (4). Set the t-scale to [0, 60 hours] with marks every 10 hrs, and the yscale to [0, 0.5], with marks every 0.1.