36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 1
Solving Differential Equations
by the Laplace Transform
and by Numerical Methods
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OBJECTIVES
◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
When you have completed this chapter, you should be able to:
• Find the Laplace transform of a function by direct integration or by
using a table.
• Find the Laplace transform of an expression containing derivatives, and
substitute initial conditions.
• Determine the inverse of a Laplace transform by completing the square
or by partial fractions.
• Solve first-order and second-order differential equations using Laplace
transforms.
• Solve electrical applications using Laplace transforms.
• Solve first-order differential equations using Euler’s method, the modified Euler’s method, or the Runge-Kutta method.
• Solve second-order differential equations using the modified Euler’s
method or the Runge-Kutta method.
◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
The methods for solving differential equations that we learned in Chapters 28 and 29
are often called classical methods. In this chapter we learn other powerful techniques: the Laplace transform and numerical methods. The Laplace transform enables us to transform a differential equation into an algebraic one. We can then solve
the algebraic equation and apply an inverse transform to obtain the solution to the
differential equation.
The Laplace transform is good for finding particular solutions of differential
equations. For example, it enables us to solve initial value problems, that is, when
the value of the function is known at t 0. Thus the equations that we deal with here
are functions of time, such as electric circuit problems, rather than functions of x.
We first use the Laplace transform to solve first- and second-order differential
equations with constant coefficients. We then do some applications and compare our
results with those obtained by classical methods in Chapters 28 and 29. We go on
to solve differential equations using numerical methods. These methods use
1
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
successive approximations to give a solution in the form of numbers, rather than
equations. One great advantage of numerical methods is that they can be programmed on a computer.
1 The Laplace Transform of a Function
Laplace Transform of Some Simple Functions
The Laplace transform is named after
the French analyst, probabilist,
astronomer, and physicist Pierre Laplace
(1749–1827).
If y is some function of time, so that y f(t), the Laplace transform of that function is defined by the following improper integral:
Laplace
Transform
[f(t)] 0
f(t)est dt
1
The transformed expression is a function of s, which we call F(s). Thus
[ f(t)] F(s)
We can write the transform of an expression by direct integration, that is, by writing an integral using Eq. 1 and then evaluating it.
Glance back at Sec. 8 in “Methods of
Integration” if you have forgotten how
to evaluate an improper integral.
◆◆◆
Example 1: If y f(t) 1, then
[ f(t)] [1] 0
(1)est dt
To evaluate the integral, we multiply by s and compensate with 1/s.
y
1
[1] s
y = e−u
0
1
est(s dt) est
s
A
0
u
Figure 1 shows a graph of y e , which will help us to evaluate the limits. Note that
as u approaches infinity, eu approaches zero, and as u approaches zero, eu approaches one. So we take est equal to zero at the upper limit t and est 1 at
the lower limit t 0. Thus
1
0
1
1
[1] (0 1) s
s
u
◆◆◆
FIGURE 1
Rule 2 from our table of integrals in Text Appendix C says that af(x) dx a f(x) dx. It follows then that
[af(t)] a[ f (t)]
The Laplace transform of a constant times a function is equal to the constant
times the transform of the function.
◆◆◆
Example 2: If [1] 1/s, as above, then
5
[5] 5[1] s
◆◆◆
Example 3: If y f(t) t, then
[ f(t)] [t] 0
test dt
◆◆◆
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Section 1
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3
The Laplace Transform of a Function
We evaluate the right side using Text Rule 37 with u t and a s.
est
[t] (st
1)
s2
A
0
At the upper limit, est approaches zero, while st approaches infinity. But est is
shrinking faster than st is growing, so their product approaches zero. This is confirmed by Fig. 2, which shows the product t(est ) first increasing but then quickly
approaching zero. So we get
est
[t] (st
1)
s2
A
0
1
1
0 2 2
s
s
y
0.4
y = te−st
0.3
0.2
0.1
0
2
4
6
t
8
FIGURE 2
◆◆◆
◆◆◆
Example 4: Find [sin at].
Solution: Using Eq. 1, we find that
[sin at] 0
Note that in each case the transformed
expression is a function of s only.
est sin at dt
Using Text Rule 41 gives
est
[sin at] (s sin at a cos at)
s2 a2
A
0
st
At the upper limit of infinity, e approaches zero, causing the entire expression to
have a value of zero at that limit. At the lower limit of zero, est approaches one,
sin at approaches zero, and cos at approaches one. Our expression then becomes
1
a
[sin at] 0 (a) s2 a2
s2 a2
Transform of a Sum
If we have the sum of several terms,
f(t) a g(t) b h(t) • • •
then
[ f(t)] 0
a
[a g(t) b h(t) • • •]est dt
0
g(t)est dt b
0
h(t)est dt • • •
Thus
[a g(t) b h(t) • • •] a[g(t)] b[h(t)] • • •
◆◆◆
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
The Laplace transform of a sum of several terms is the sum of the transforms
of each term.
This property allows us to work term by term in writing the transform of an expression made up of sums or differences.
◆◆◆
Example 5: If y 4 5t 2 sin 3t, then
5
3
4
[ y] 2 2
2
s
s 9
s
◆◆◆
Transform of a Derivative
When we write the Laplace transform of
a function, we often say that we are
“taking the transform” of the function.
To solve differential equations, we must be able to take the transform of a derivative. Let f (t) be the derivative of some function f (t). Then
[ f (t)] 0
f (t)est dt
Integrating by parts, we let u est and d f(t) dt. So du sest dt and
f (t) dt f(t). Then
A
[ f(t)] f(t)est
f(t)est
A
0
0
s
0
f (t)est dt
s[ f(t)]
0 f(0) s[ f(t)]
Thus:
Transform of
a Derivative
[ f (t)] s[ f(t)] f(0)
The transform of a derivative of a function equals s times the transform of the
function minus the function evaluated at t 0.
Thus if we have a function y f (t), then
[ y] s[ y] y(0)
Note that the transform of a derivative contains f(0) or y(0), the value of the function at t 0. Thus to evaluate the Laplace transform of a derivative, we must know
the initial conditions.
Example 6: Take the Laplace transform of both sides of the differential equation y 6t, with the initial condition that y(0) 5.
◆◆◆
If two functions are equal, their transforms are equal. Thus we can take the
transform of both sides of an equation
without changing the meaning of the
equation.
Solution:
y 6t
[y] [6t]
6
s[ y] y(0) 2
s
or, substituting,
6
s[ y] 5 2
s
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Section 1
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5
The Laplace Transform of a Function
since y(0) 5. Later we will solve such an equation for [y] and then find the inverse of the transform to obtain y.
◆◆◆
Similarly, we can find the transform of the second derivative since the second
derivative of f(t) is the derivative of f (t).
[ f (t)] s[ f (t)] f (0)
Substituting, we get
[ f (t)] s{s[ f(t)] f (0)} f (0)
Transform of
the Second
Derivative
[ f (t)] s2[ f(t)] sf(0) f (0)
Another way of expressing this equation, if y f(t), is
[ y] s2[ y] sy(0) y(0)
Example 7: Transform both sides of the second-order differential equation
y 3y 4y 5t if y(0) 6 and y(0) 7.
◆◆◆
Solution:
y 3y 4y 5t
[ y] 3[ y] 4[ y] [5t]
5
s2[y] sy(0) y(0) 3{s[y] y(0)} 4[y] 2
s
Substituting y(0) 6 and y(0) 7, we obtain
5
s2[ y] 6s 7 3s[ y] 3(6) 4[ y] 2
s
or
5
s2[y] 3s[y] 4[y] 6s 11 2
s
Transform of an Integral
Let us now find the transform of the integral
Laplace transform, Eq. 1,
t
0
f(t) dt. By our definition of the
f(t) dt f(t) dt e
t
0
0
t
◆◆◆
st
0
dt
Integrating by parts, we let
u
f(t) dt
t
0
and d est dt
so that
du f(t) dt
and est/s
Integrating yields
f(t) dt (1/s)e f(t) dtA
t
0
st
t
0
0
(1/s)
0
f(t)est dt
At the upper limit of infinity, the quantity in the brackets vanishes because est goes
to zero. At the lower limit of zero, the quantity in the brackets vanishes because the
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
upper and lower limits of the integral are equal. Further, the integral in the second
term is the definition (Eq. 1) of the Laplace transform of f(t). Thus:
f(t) dt (1/s) f(t)
Transform of
an Integral
◆◆◆
t
0
Example 8: The voltage across a capacitor of capacitance C is given by
(1C)
We will need this equation later for
solving electrical problems.
i dt
t
0
Find the transform of this voltage.
Solution: Using the transform of an integral, we get
(1C)
i dt (1Cs)[i]
t
◆◆◆
0
Table of Laplace Transforms
The transforms of some common functions are given in Table 1. Instead of transforming a function step by step, we simply look it up in the table.
TABLE 1
Short table of Laplace transforms. Here n is a positive integer.
Transform Number
1
f(t)
f(t)
[ f (t)] F(s)
F(s) 0
est f(t) dt
2
f (t)
s[ f(t)] f(0)
3
f (t)
s2[ f(t)] sf(0) f(0)
4
a g(t) b h(t) • • •
a[g(t)] b[h(t)] • • •
5
1
1
s
6
t
1
2
s
7
tn
n!
n1
s
8
tn1
(n 1)!
1
n
s
9
eat
1
sa
10
1 eat
a
s(s a)
11
teat
1
2
(s a)
12
eat(1 at)
s
2
(s a)
13
tneat
n!
(s a)n1
14
in1eat
(n 1)!
(s a)n
15
eat ebt
ba
(s a)(s b)
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Section 1
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7
The Laplace Transform of a Function
TABLE 1 Continued
f(t)
[ f (t)] F(s)
16
aeat bebt
s(a b)
(s a)(s b)
17
sin at
a
s2 a2
18
cos at
s
s2 a2
19
t sin at
2as
(s2 a2)2
20
t cos at
s2 a2
(s2 a2)2
21
1 cos at
a2
2
s(s a2)
22
at sin at
a3
2 2
s (s a2)
23
eat sin bt
b
(s a)2 b2
24
eat cos bt
sa
(s a)2 b2
25
sin at at cos at
2a3
2
(s a2)2
26
sin at at cos at
2as2
2
(s a2)2
27
cos at 12 at sin at
s3
(s2 a2)2
28
b at
(e
at 1)
a2
29
f(t) dt
Transform Number
◆◆◆
t
0
b
s2(s a)
[f(t)]
s
Example 9: Find the Laplace transform of t3e2t by using Table 1.
Solution: Our function matches Transform 13, with n 3 and a 2, so
3!
6
[t 3e2t ] 4
(s 2)31
(s 2)
Exercise 1
◆
The Laplace Transform of a Function
Transforms by Direct Integration
Find the Laplace transform of each function by direct integration.
1. f(t) 6
3. f(t) t 2
5. f(t) cos 5t
2. f(t) t
4. f(t) 2t 2
6. f(t) et sin t
Transforms by Table
Use Table 1 to find the Laplace transform of each function.
7. f(t) t2 4
8. f(t) t 3 2t 2 3t 4
◆◆◆
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
9.
11.
13.
15.
f(t) 3et 2et
f(t) sin 2t cos 3t
f(t) 5e3t cos 5t
f(t) 5et t sin 3t
10.
12.
14.
16.
f(t) 5te3t
f(t) 3 e4t
f(t) 2e4t t2
f(t) t3 4t2 3et
Transforms of Derivatives
Find the Laplace transform of each expression, and substitute the given initial
conditions.
17.
18.
19.
20.
21.
22.
23.
24.
y 2y, y(0) 1
3y 2y, y(0) 3
y 4y, y(0) 0
5y 3y, y(0) 2
y 3y y, y(0) 1 and y(0) 3
y y 2y, y(0) 1 and y(0) 0
2y 3y y, y(0) 2 and y(0) 3
3y y 2y, y(0) 2 and y(0) 1
2 Inverse Transforms
Before we can use the Laplace transform to solve differential equations, we must
be able to transform a function of s back to a function of t. The inverse Laplace
transform is denoted by 1. Thus if [ f(t)] F(s), then we have the following
equation:
We’ll see that finding the inverse is often
harder than finding the transform itself.
Inverse
Laplace
Transform
1[F(s)] f(t)
2
We use Table 1 to find the inverse of some Laplace transforms. We put our
given expression in a form that matches a right-hand entry of Table 1 and then read
the corresponding entry at the left.
◆◆◆
Example 10: If F(s) 4/(s2 16), find f(t).
Solution: We search Table 1 in the right-hand column for an expression of similar
form and find Transform 17,
a
[sin at] 2
s a2
which matches our expression if a 4. Thus
f(t) sin 4t
◆◆◆
Example 11: Find y if [y] 5/(s 7)4.
Solution: From the table we find Transform 13,
n!
[t neat ] (s a)n1
◆◆◆
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Section 2
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9
Inverse Transforms
In order to match our function, we must have a 7 and n 3. The numerator then
must be 3! 3(2) 6. So we insert 6 in the numerator and compensate with 6 in
the denominator. We then rewrite our function as
6
5
[ y] 4
6 (s 7)
which now is the same as Transform 13 with a coefficient of 56. The inverse is then
5
y t 3e7t
6
◆◆◆
Completing the Square
To get our given function to match a right-hand table entry, we may have to do
some algebra. Sometimes we must complete the square to make our function match
one in the table.
◆◆◆
s1
Example 12: Find y if [ y] .
s2 4s 20
Solution: This does not now match any functions in our table, but it will if we
complete the square on the denominator.
This is no different from the method we
used earlier when we completed the
square.
s2 4s 20 (s2 4s
) 20
(s2 4s 4) 20 4 (s 2)2 42
The denominator is now of the same form as in Transform 24. However, to use
Transform 24, the numerator must be s 2, not s 1. So let us add 3 and subtract
3 from the numerator, so that s 1 (s 2) 3. Then
s1
(s 2) 3
s2
3
[y] 2
2
2
2
2
s 4s 20
(s 2) 4
(s 2) 4
(s 2)2 42
s2
4
3
(s 2)2 42
4 (s 2)2 42
Now the first expression matches Transform 24 and the second matches Transform
23. We find the inverse of these transforms to be
3
3
y e2t cos 4t e2t sin 4t e2t cos 4t sin 4t
4
4
◆◆◆
Partial Fractions
We used partial fractions in Sec. 5 in “Methods of Integration” to make a given
expression match one listed in the table of integrals. Now we use partial fractions
to make an expression match one listed in our table of Laplace transforms.
◆◆◆
Example 13: Find y if [y] 12/(s2 2s 8).
Solution: We separate [y] into partial fractions.
12
[y] 2
s 2s 8
A
B
s4
s2
so
12 A(s 2) B(s 4)
(A B)s (2A 4B)
We could also complete the square
here, but we would find that the
resulting expression would not match a
table entry. Some trial-and-error work in
algebra may be necessary to get the
function to match a table entry.
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
from which A B 0 and 2A 4B 12. Solving simultaneously gives A 2
and B 2, so
12
[ y] s2 2s 8
2
2
s4
s2
Using Transform 9 twice, we get
y 2e4t 2e2t
Exercise 2
◆
◆◆◆
Inverse Transforms
Find the inverse of each transform.
1
1. s
3
2. 2
s
2
3. 3
s
1
4. 2
s 3s
4
5. 2
4s
s
6. 2
(s 6)
3s
7. s2 2
4
8. s3 9s
s4
9. 2
(s 9)
3s
10. 2
(s 4)2
5
11. (s 2)2 9
s2
12. 2
s 6s 8
2s2 1
13. s(s2 1)
s
14. s2 2s 1
5s 2
15. s2(s 1)(s 2)
2
17. s2 s 2
2s
19. s2 5s 6
1
16. 2
(s 1)(s 2)
s1
18. s2 2s
3s
20. (s2 4)(s2 1)
3 Solving Differential Equations by the Laplace Transform
To solve a differential equation with the Laplace transform:
1.
2.
3.
4.
Take the transform of each side of the equation.
Solve for [y] F(s).
Manipulate F(s) until it matches one or more table entries.
Take the inverse transform to find y f(t).
We start with a very simple example.
◆◆◆
Example 14: Solve the first-order differential equation y y 2 if y(0) 0.
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Section 3
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11
Solving Differential Equations by the Laplace Transform
Solution:
1. We write the Laplace transform of both sides and get
[y] [y] [2]
2
s[y] y(0) [y] s
2. We substitute 0 for y(0) and solve for [y].
2
(s 1)[y] s
2
[y] s(s 1)
3. Our function will match Transform 10 if we write it as
1
[y] 2 s(s 1)
4. Taking the inverse gives
y 2(1 et )
◆◆◆
In our next example we express F(s) as three partial fractions in order to take
the inverse transform.
◆◆◆
Example 15: Solve the first-order differential equation
y 3y 4 9t
if y(0) 2.
Solution:
1. We take the Laplace transform of both sides and get
[y] [3y] [4] [9t]
4
9
s[y] y(0) 3[y] 2
s
s
2. We substitute 2 for y(0) and solve for [y].
4
9
(s 3)[y] 2 2
s
s
9
4
2
[y] 2
s (s 3)
s(s 3)
s3
3. We match our expressions with Transforms 28, 10, and 9.
9
3
1
4
2 [y] s2(s 3)
s3
3 s(s 3)
4. Taking the inverse gives
4
5
1
y e3t 3t 1 (1 e3t) 2e3t e3t 3t 3
3
3
◆◆◆
We now use the Laplace transform to solve a second-order differential
equation.
◆◆◆
Example 16: Solve the second-order differential equation
y 4y 3 0
where y is a function of t and y and y are both zero at t 0.
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
Solution:
1. Taking the Laplace transform of both sides,
[y] [4y] [3] 0
3
s2[y] sy(0) y(0) 4[y] 0
s
Substituting y(0) 0 and y(0) 0 gives
3
s2[y] 0 0 4[y] 0
s
2. Solving for [y], we have
3
(s2 4)[y] s
3
[y] 2
s(s 4)
3. We match it to Transform 21.
4
3
3
[y] s(s2 4)
4 s(s2 22)
4. Taking the inverse gives
3
y (1 cos 2t)
4
◆◆◆
We have seen that the hardest part in solving a differential equation by Laplace
transform is often in doing the algebra necessary to get the given function to match
a table entry. We must often use partial fractions or completing the square, and
sometimes both, as in the following example.
◆◆◆
Example 17: Solve y 4y 5y t, where y(0) 1 and y(0) 2.
Solution:
1. Taking the Laplace transform of both sides and substituting initial conditions,
we find that
1
s2[y] sy(0) y(0) 4s[ y] 4y(0) 5[y] 2
s
1
[ y](s 2 4s 5) 2 s 6
s
2. Solving for [y] gives
1
s6
[y] s2(s2 4s 5)
s2 4s 5
(1)
3. Taking for now just the first fraction in Equation (1), which we’ll call F1(s), we
obtain
1
As B
Cs D
F1(s) 2 2
2
2
s (s 4s 5)
s
s 4s 5
Multiplying by s2(s2 4s 5) gives
1 As3 4As2 5As Bs2 4Bs 5B Cs3 Ds2
(A C)s3 (4A B D)s2 (5A 4B)s 5B
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Section 3
◆
13
Solving Differential Equations by the Laplace Transform
Equating coefficients gives
4
A 25
1
B 5
4
C 25
11
D 25
So
4s/25 1/5
4s/25 11/25
F1(s) s2
s2 4s 5
1 5
4
4s 11
2 2
25 s
s 4s 5
s
We can use Transforms 5 and 6 for 4/s and 5/s2, but there is no match for the
third term in the parentheses. However, by completing the square, we can get
the denominator to match those in Transforms 23 and 24. Thus
s2 4s 5 s2 4s 4 4 5
(s 2)2 12
We now manipulate that third term into the form of Transforms 23 and 24.
4s 11
4s 11
s 11/4 2 2
(4) s2 4s 5
(s 2)2 12
(s 2)2 12
s 2 3/4
s2
1
(4) (4) (3) (s 2)2 12
(s 2)2 12
(s 2)2 12
Combining the third term with the first and second gives
1 5
4
s2
1
F1(s) 2 (4) (3) 25 s
s
(s 2)2 12
(s 2)2 12
Returning to the remaining fraction in Equation (1), which we call F2(s), we
complete the square in the denominator and use partial fractions to get
s226
s6
F2(s) 2
2 (s 2)2 12
(s 2) 1
s2
4
2
2 (s 2) 1
(s 2)2 12
4. Taking the inverse transform of F1(s) and F2(s), we obtain
1
y [5t 4 4e2t cos t 3e2t sin t] e2t cos t 4e2t sin t
25
which simplifies to
1
y (5t 4 29e2t cos t 103e2t sin t)
25
Exercise 3 ◆ Solving Differential Equations
by the Laplace Transform
First-Order Equations
Solve each differential equation by the Laplace transform.
1.
3.
5.
7.
y 3y 0, y(0) 1
4y 2y t, y(0) 0
3y 2y t2, y(0) 3
y 2y cos 2t, y(0) 0
2.
4.
6.
8.
2y y 1, y(0) 3
y 5y e2t, y(0) 2
y 3y sin t, y(0) 0
4y y 3t3, y(0) 0
◆◆◆
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
Second-Order Equations
Solve each differential equation by the Laplace transform.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
y 2y 3y 0, y(0) 0 and y(0) 2
y y y 1, y(0) 0 and y(0) 0
y 3y 3, y(0) 1 and y(0) 2
y 2y 2, y(0) 0 and y(0) 3
2y y 4t, y(0) 3 and y(0) 0
y y 5y t, y(0) 1 and y(0) 2
y 4y 3y t, y(0) 2 and y(0) 2
y 4y 8t 3, y(0) 0 and y(0) 0
3y y sin t, y(0) 2 and y(0) 3
2y y 3, y(0) 2 and y(0) 1
y 2y y et, y(0) 0 and y(0) 0
2y 32y cos 2t, y(0) 0 and y(0) 1
y 2y 3y tet, y(0) 0 and y(0) 0
3y 2y y sin 3t, y(0) 0 and y(0) 0
Motion in a Resisting Medium
23. A 15.0-kg object is dropped from rest through air whose resisting force is equal
to 1.85 times the object’s speed (in m/s). Find its speed after 1.00 s.
Exponential Growth and Decay
24. A certain steel ingot is 1900°F and cools at a rate (in °F/min) equal to 1.25
times its present temperature (°F). Find its temperature after 5.00 min.
25. The rate of growth (bacteria/h) of a colony of bacteria is equal to 2.5 times the
number present at any instant. How many bacteria are there after 24 h if there
are 50
00 at first?
Mechanical Vibrations
26. A weight that hangs from a spring is pulled down 1.00 in. at t 0 and then
is released from rest. The differential equation of motion is x 6.25x 25.8x 0. Write the equation for x as a function of time.
27. An alternating force is applied to a weight such that the equation of motion is
x 6.25x 45.3 cos 2.25t. If and x are zero at t 0, write an equation
for x as a function of time.
4 Electrical Applications
1
Many of the differential equations for motion and electric circuits covered in Chapters 28 and 29 are nicely handled by the Laplace transform. However, since the
Laplace transform is used mainly for electric circuits, we emphasize that application
here. The method is illustrated by examples of several types of circuits with dc and
ac sources.
R
2
E
FIGURE 3 RC circuit.
i
C
Series RC Circuit with dc Source
◆◆◆ Example 18: Capacitor Discharging. A fully charged capacitor, Fig. 3, is
discharged by throwing the switch from position 1 to position 2, at t 0. Write
expressions for the current and the voltage across the capacitor.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 15
Section 4
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15
Electrical Applications
Solution: If the voltage across the capacitor is , then the voltage across the resistor must be , since the sum of the voltages around the loop must be zero. Since
the current through the resistor (/R) must equal the current through the capacitor
(C d/dt) we write
d
C R
dt
or
— 0
RC
Taking the transform of each term
1
s[i] 0 –—
[i] 0
RC
or
1
s[i] –—
[i] E
RC
Solving for [i] we get
E
[i] s 1/RC
Using transform 9 to find the inverse gives
Voltage
across a
Discharging
Capacitor
E e–t /RC
3
We next find the current by dividing the voltage by R.
Current in a
Discharging
Capacitor
E
i — e–t/RC
R
Note that the results of this example and
the two examples to follow are the
same as we obtained by classical
methods for solving differential
equations.
4
Example 19: Capacitor Charging. A fully discharched capacitor, Fig. 4, is
charged by throwing the switch from position 2 to position 1 at t 0. Write
expressions for the current and voltage across the capacitor.
◆◆◆
Solution: Summing voltages around the loop gives
Drop across resistor Drop across capacitor Battery voltage
1
By Text Eq. 1081, the voltage across the capacitor is C
1 t
Ri i dt E
C 0
or
RC i i dt EC
t
0
i dt, so
(1)
t
1
0
2
E
FIGURE 4
R
i
E
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 16
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
Taking the transform of each term
[i]
EC
RC[i] s
s
Solving for [i] and rearranging gives
1
E
[i] R s 1/RC
Using transform 9 to find the inverse transform, we get
Current in
a Charging
Capacitor
E
i — e–tRC
R
5
Returing to (1) to get the voltage across the capacitor,
Drop across capacitor = Battery voltage Drop across resistor
E
E R — e–t RC
R
or
Voltage
across a
Charging
Capacitor
E 1 e–t /RC 6
Series RL Circuit with dc Source
◆◆◆ Example 20: Inductor discharging. A fully charged inductor is discharged by
throwing the switch, Fig. 5, from position 1 to position 2 at t = 0. Write expressions
for the current and the voltage across the inductor.
12 Ω
2
Solution: Summing voltages around the loop we get
1
24 V
FIGURE 5
Battery voltage Drop across resistor Drop across inductor
i
4H
di
E Ri L dt
or
E
R
i i
L
L
Taking the transform of each term
E
R
– [i] s [i] i(0)
Ls
L
Substituting 0 for i(0) and solving for [i],
R/L
E
[i] R s(s R/L)
(1)
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 17
Section 4
◆
17
Electrical Applications
Then by transform 10,
Current in
a Charging
Inductor
E
i 1 e–Rt /L
R
7
Now using (1) to find the voltage across the inductor,
Voltage across inductor Battery voltage Voltage across the resistor
E Ri
E
E R 1 e–Rt /L
R
Voltage
across a
Charging
Inductor
Ee –Rt L
8
Series RL Circuit with ac Source
4Ω
Example 21: A switch (Fig. 6) is closed at t 0 when the applied voltage is
zero and increasing. Find the current.
◆◆◆
Solution: Summing the voltages around the loop gives
di
Ri L dt
Substituting values and rearranging, we find that
di
4i 0.02 80 sin 400t
dt
Taking the transform of each term yields
80(400)
4[i] 0.02[s[i] i(0)] s2 (400)2
Substituting 0 for i(0) and solving for [i], we have
32,000
[i](0.02s 4) s2 (400)2
32,000
1,600,000
[i] [s2 (400)2](0.02s 4)
[s2 (400)2](s 200)
We now use the method of Sec. 5 in “Methods of Integration” for separating into
partial fractions a rational fraction that has quadratic factors in the denominator.
1,600,000
As B
C
2 2
2
2
[s (400) ] (s 200)
s (400)
s 200
80 sin 400t V
0.02 H
FIGURE 6 RL circuit with ac source.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 18
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
Multiplying by [s2 (400)2](s 200) gives
1,600,000 (As B)(s 200) C[s2 (400)2]
which simplifies to
1,600,000 (A C)s2 (200A B)s 200B 4002C
Equating coefficients gives
AC0
200A B 0
200B 400C 2 1,600,000
Solving simultaneously for A, B, and C gives
A 8
B 1600
C8
Substituting back, we get
1600 8s
8
[i] 2 2
s (400)
s 200
400
s
1
4
2 8 2 8 2
2
s (400)
s (400)
s 200
Taking the inverse transform using Transforms 9, 17, and 18, we find that
i 4 sin 400t 8 cos 400t 8e200t
We now combine the first two terms in the form I sin(400t v), where
8
and v tan1 63.4° 1.107 rad
4
I 42
82 8.94
so
i 8.94 sin(400t 1.107) 8e200t A
A graph of this wave, as well as the applied voltage wave, is shown in Fig. 7.
i (A)
10
e (V)
i
100
e
5
0
10
225 Ω
75.4 V
−5
−50
−10
−100
20
30
FIGURE 7
t (ms)
◆◆◆
1.5 H
4.75 F
FIGURE 8 RLC circuit.
Series RLC Circuit with dc Source
◆◆◆ Example 22: A switch (Fig. 8) is closed at t 0, and there is no initial charge
on the capacitor. Write an expression for the current.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 19
Section 4
◆
19
Electrical Applications
Solution:
1. The differential equation for this circuit is
di
1
Ri L dt
C
i dt E
t
0
2. Transforming each term and substituting values gives
75.4
106[i]
225[i] 1.5{s [i] i(0)} s
4.75s
3. Setting i(0) equal to zero and solving for [i] yields
50.3
[i] s2 150s 140,000
4. We now try to match this expression with those in the table. Let us complete
the square in the denominator.
s2 150s 140,000 (s2 150s 5625) 140,000 5625
(s 75)2 (134,000)
(s 75)2 j 2(367)2
(s 75 j367)(s 75 j367)
after replacing 1 by j 2 and factoring the difference of two squares. So
50.3
[i] s2 150s 140,000
50.3
(s 75 j367)(s 75 j367)
We can now find partial fractions.
50.3
A
B
(s 75 j367)(s 75 j367)
s 75 j367
s 75 j367
Multiplying through by s 75 j367 gives
B(s 75 j367)
50.3
A s 75 j367
s 75 j367
Setting s 75 j367, we get
50.3
50.3
A j0.0685
75 j367 75 j367
j734
We find B by multiplying through by s 75 j 367 and then setting s j36775, and we get B j0.0625 (work not shown). Substituting gives us
j0.0685
j0.0685
[i] s 75 j367
s 75 j367
This still doesn’t match any table entry. Let us now combine the two fractions
over a common denominator.
j 0.0685
j 0.0685
[i] s 75 j367
s 75 j367
j0.0685(s 75 j367) j0.0685(s 75 j367)
(s 75 j367)(s 75 j367)
367
j 250.4
0.137 2
(s 75)2 (367)2
s 150s 140,000
We see that most of the work when
using the Laplace transform is in algebra,
rather than in calculus.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 20
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
5. We finally have a match, with Transform 23. Taking the inverse transform gives
i 137e75t sin 367t mA
which agrees with the result found in Chapter 29, Example 19.
Exercise 4
◆
◆◆◆
Electrical Applications
1. The current in a certain RC circuit satisfies the equation 172i 2750i 115.
If i is zero at t 0, show that
i 41.81 e16t mA
2. In an RL circuit, R 3750 and L 0.150 H. It is connected to a dc source
of 250 V at t 0. If i is zero at t 0, show that
i 66.71 e25,000t mA
3. The current in an RLC circuit satisfies the equation
0.55i 482i 7350i 120
If i and i are zero at t 0, show that
i 16.41 e16t mA
4. In an RLC circuit, R 2950 , L 0.120 H, and C 1.55 mF. It is connected
to a dc source of 115 V at t 0. If i is zero at t 0, show that
i 39.7e221t mA
5. The current in a certain RLC circuit satisfies the equation
3.15i 0.0223i 1680
i dt 1
t
0
If i is zero at t 0, show that
i 169e70.5t sin 265t mA
6. In an RL circuit, R 4150 and L 0.127 H. It is connected to a dc source
of 28.4 V at t 0. If i is zero at t 0, show that
i 6.841 e32,700t mA
7. In an RL circuit, R 8.37 and L 0.250 H. It is connected to a dc source
of 50.5 V at t 0. If i is zero at t 0, show that
i 6.031 e33.5t A
5 Numerical Solution of First-Order Differential Equations
In Sec. 28–2, we showed how to use Euler’s method to graphically and numerically
solve a differential equation, but we mentioned there that it is not the most accurate
method available. It will, however, serve as a good introduction to other methods. We
will now show both the modified Euler’s method that includes predictor-corrector
steps and the Runge-Kutta method. We will apply these methods to both first- and
second-order equations.
Modified Euler’s Method with Predictor-Corrector Steps
The various predictor-corrector methods all have two main steps. First, the predictor step tries to predict the next point on the curve from preceding values, and then
the corrector step tries to improve the predicted values.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 21
Section 5
◆
21
Numerical Solution of First-Order Differential Equations
Here we modify Euler’s method of Sec. 28–2 as the predictor, using the coordinates and slope at the present point P to estimate the next point Q. The corrector
step then recomputes Q using the average of the slopes at P and Q. Our iteration
formula is then the following:
xq xp x
Modified
Euler’s
Method
9
mp mq
yq yp x
2
Example 23: Find an approximate solution to y x2/y, with the boundary
condition that y 2 when x 3. Calculate y for x 3 to 10 in steps of 1.
◆◆◆
Solution: Using data from before, the slope m1 at (3, 2) was 4.5, the second point
was predicted to be (4, 6.5), and the slope m2 at (4, 6.5) was 2.462. The recomputed
ordinate at the second point is then, by Eq. 9,
4.5 2.462
y2 2 (1)
2
5.481
so our corrected second point is (4, 5.481). The slope at this point is
m2 y(4, 5.481) 2.919
We use this to predict y3.
y3 (predicted) y2 m2 x
5.481 2.919(1)
8.400
The slope at (5, 8.400) is
m3 y(5, 8.400) 2.976
The corrected y3 is then
2.919 2.976
y3 (corrected) 5.481 (1) 8.429
2
The remaining values are given in Table 2.
TABLE 2
x
Approximate y
Exact y
Error
3
4
5
6
7
8
9
10
2.00000
5.48077
8.42850
11.49126
14.73299
18.16790
21.79642
25.61434
2.00000
5.35413
8.32666
11.40176
14.65151
18.09236
21.72556
25.54734
0.00000
0.12664
0.10184
0.08950
0.08148
0.07555
0.07086
0.06700
Note that the error at x 10 is about 0.26%, compared with 0.6% before. We
have gained in accuracy at the cost of added computation.
◆◆◆
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22
Solving Differential Equations by the Laplace Transform and by Numerical Methods
Runge-Kutta Method
The Runge-Kutta method, like the modified Euler’s, is a predictor-corrector method.
Prediction: Starting at P (Fig. 9), we use
This method is named after two German
mathematicians, Carl Runge
(1856–1927) and Wilhelm Kutta
(1867–1944).
the slope mp to predict R,
the slope mr to predict S, and
the slope ms to predict Q.
y
Q
yq
ms
S
ys
mr
R
P
yp
mp
∆x
2
xp
0
∆x
2
xr
xs
xq
x
FIGURE 9 Runge-Kutta method.
Correction: We then take a weighted average of the slopes at P, Q, R, and S,
giving twice the weight to the slopes at the midpoints R and S as at the endpoints P
and Q.
1
mavg (mp 2mr 2ms mq)
6
We use the average slope and the coordinates of P to find Q.
yq yp mavg x
xq xp x
yq yp mavg x
where
RungeKutta
Method
and
1
mavg (mp 2mr 2ms mq)
6
mp f(xp, yp)
x
x
mr f xp , yp mp 2
2
m fx
x
m 2
10
x
, yp
r
2
mq f(xp x, yp ms x)
s
◆◆◆
p
Example 24: Repeat Example 23 using the Runge-Kutta method and a step of 1.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 23
Section 5
◆
Numerical Solution of First-Order Differential Equations
Solution: Starting at P(3, 2), with mp 4.5,
yr 2 4.5(0.5) 4.25
(3.5)2
mr f(3.5, 4.25) 2.882
4.25
ys 2 2.882(0.5) 3.441
(3.5)2
ms f(3.5, 3.441) 3.560
3.441
yq 2 3.560(1) 5.560
42
mq f(4, 5.560) 2.878
5.560
4.5 2(2.882) 2(3.560) 2.878
mavg 3.377
6
yq (corrected) yp mavg x 2 3.377(1) 5.377
So our second point is (3, 5.377). The rest of the calculation is given in Table 3.
TABLE 3
x
Approximate y
Exact y
Error
3
4
5
6
7
8
9
10
2.00000
5.37703
8.34141
11.41255
14.65992
18.09918
21.73125
25.55219
2.000000
5.354126
8.326664
11.401760
14.651510
18.092360
21.725560
25.547340
0.00000
0.02290
0.01475
0.01079
0.00841
0.00682
0.00569
0.00484
The error here at x 10 with the Runge-Kutta method is about 0.02%,
compared with 0.26% for the modified Euler method.
◆◆◆
Comparison of the Three Methods
Table 4 shows the errors obtained when solving the same differential equation by
our three different methods, each with a step size of 1.
TABLE 4
x
Euler
Modified Euler
Runge-Kutta
3
4
5
6
7
8
9
10
0.00000
1.14587
0.63487
0.34948
0.16324
0.02991
0.07173
0.15283
0.00000
0.12664
0.10184
0.08950
0.08148
0.07555
0.07086
0.06700
0.00000
0.02290
0.01475
0.01079
0.00841
0.00682
0.00569
0.00484
Divergence
As with most numerical methods, the computation can diverge, with our answers
getting further and further from the true values. The computations shown here can
23
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24
Solving Differential Equations by the Laplace Transform and by Numerical Methods
diverge if either the step size x is too large or the range of x over which we compute
is too large. Numerical methods can also diverge if the points used in the computation are near a discontinuity, a maximum or minimum point, or a point of inflection.
When doing the computations on the computer (the only practical way), it is
easy to change the step size. A good practice is to reduce the step size by a factor
of 10 each time (1, 0.1, 0.01, . . .) until the answers do not change in a given decimal place. Further reduction of the step size will eventually result in numbers too
small for the computer to handle, and unreliable results will occur.
Exercise 5 ◆ Numerical Solution of
First-Order Differential Equations
Solve each differential equation and find the value of y asked for. Use either the
modified Euler method or the Runge-Kutta method, with a step size of 0.1.
Your answers may differ from those
given for this exercise, depending on
the method used and roundoff errors.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
y xy 3; y(4) 2. Find y(5).
y 5y2 3x; y(0) 3. Find y(1).
x4y yex xy3; y(2) 2. Find y(3).
xy y2 3 ln x; y(1) 3. Find y(2).
1.94x2y 8.23x sin y 2.99y 0; y(1) 3. Find y(2).
y ln x 3.99(x y)2 4.28y; y(2) 1.76. Find y(3).
(x y)y 2.76x 2y2 5.33y; y(0) 1. Find y(1).
y 2.97xex 1.92x cos y 0; y(2) 1. Find y(3).
xy xy2 8y lnhxyh; y(1) 3. Find y(2).
x2y 8.85x2 ln y 3.27x 0; y(1) 23.9. Find y(2).
Computer
11. Write a program or use a spreadsheet to solve a first-order differential equation
by Euler’s method. Try it on any of the equations above.
12. Write a program or use a spreadsheet to solve a first-order differential equation
by the modified Euler’s method. Try it on any of the equations above.
13. Write a program or use a spreadsheet to solve a first-order differential equation
by the Runge-Kutta method. Try it on any of the equations above.
14. If you have done one of the calculations above using a spreadsheet that has a
graphing utility, use that utility to make a graph of x versus y over the given
range.
15. Some computer algebra systems have built-in functions for numerically solving
a differential equation. In Derive, the command Euler will invoke Euler’s
method, and the command RK will call for the Runge-Kutta method.
In Maple, the commands firsteuler, impeuler, and rungekutta are used for
solutions by Euler’s method, the modified Euler’s method, and the Runge-Kutta
method. Mathematica uses NDSolve for the numerical solution of a differential
equation.
Consult your manual for specific instructions on how to use these commands, and use one or more to solve any of the equations in this exercise set.
6 Numerical Solution of Second-Order
Differential Equations
To solve a second-order differential equation numerically, we first transform the
given equation into two first-order equations by substituting another variable, m,
for y.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 25
Section 6
◆
25
Numerical Solution of Second-Order Differential Equations
◆◆◆ Example 25: By making the substitution y m, we can transform the secondorder equation
y 3y 4xy 0
into the two first-order equations
y m
and
m 3m 4xy
f(x, y, m)
◆◆◆
As before, we start with the given point P(xp, yp), at which we must also know
the slope mp. We proceed from P to the next point Q(xq, yq) as follows:
1. Compute the average first derivative, mavg, over the interval x.
2. Compute the average second derivative mavg over that interval.
3. Find the coordinates xq and yq at Q by using the equations
xq xp x
and
yq yp mavg x
4. Find the first derivative mq at Q by using
mq mp mavg x
5. Return to Step 1 and repeat the sequence for the next interval.
We can find mavg and mavg in Steps 1 and 2 either by the modified Euler’s
method or by the Runge-Kutta method. The latter method, as before, requires us to
find values at the intermediate points R and S. The equations for second-order equations are similar to those we had for first-order equations.
At P: xp, yp, and mp are given:
At R:
x
xr xp 2
x
mr mp mp 2
At S:
At Q:
◆◆◆
mp f(xp, yp, mp).
x
yr yp mp 2
mr f(xr, yr, mr)
x
xs xp 2
x
ms mp mr 2
xq xp x
mq mp ms x
x
ys yp mr 2
ms f(xs, ys, ms)
yq yp ms x
mq f(xq, yq, mq)
Example 26: Find an approximate solution to the equation
y 2y y 2ex
with boundary conditions y 2e at (1, 0). Find points from 1 to 3 in steps of 0.1.
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Solving Differential Equations by the Laplace Transform and by Numerical Methods
Solution: Replacing y by m and solving for m gives
m f(x, y, m)
2ex y 2m
At P:
The boundary values are xp 1, yp 0, and mp 5.4366. Then
mp f(xp, yp, mp) 2e1 0 2(5.4366) 16.3097
At R:
x
xr xp 1 0.05 1.05
2
x
yr yp mp 0 5.4366(0.05) 0.2718
2
x
mr mp mp 5.4366 16.3097(0.05) 6.2521
2
1.05
mr 2e
0.2718 2(6.2521) 17.9477
xs 1.05
At S:
x
ys yp mr 0 6.2521(0.05) 0.3126
2
x
ms mp mr 5.4366 17.9477(0.05) 6.3360
2
1.05
ms 2e
0.3126 2(6.3360) 18.0707
At Q:
xq
yq
mq
mq
xp x 1.1
yp ms x 0 6.3360(0.1) 0.6336
mp ms x 5.4366 18.0707(0.1) 7.2437
2e1.1 0.6336 2(7.2437) 19.8621
Then
mp 2mr 2ms mq
mavg 6
5.4366 2(6.2521) 2(6.3360) 7.2437
6.3094
6
and
mp 2mr 2ms mq
mavg 6
16.3097 2(17.9477) 2(18.0707) 19.8621
18.0348
6
So
yq yp mavg x 0 6.3094(0.1) 0.63094
and
mq mp mavg x 5.4366 18.0348(0.1) 7.2401
The computation is then repeated. The remaining values are given in Table 5,
along with the exact y and exact slope, found by analytical solution of the given
equation.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 27
27
Review Problems
TABLE 5
x
y
Exact y
Slope
Exact Slope
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0.000
0.631
1.461
2.532
3.893
5.602
7.727
10.346
13.551
17.450
22.167
27.846
34.656
42.789
52.470
63.958
77.551
93.593
112.480
134.669
160.683
0.000
0.631
1.461
2.532
3.893
5.602
7.727
10.346
13.551
17.450
22.167
27.847
34.656
42.789
52.470
63.958
77.551
93.593
112.481
134.670
160.684
5.437
7.240
9.429
12.072
15.248
19.047
23.576
28.957
35.330
42.856
51.723
62.144
74.366
88.670
105.381
124.870
147.562
173.943
204.570
240.079
281.196
5.437
7.240
9.429
12.072
15.248
19.047
23.576
28.957
35.330
42.857
51.723
62.145
74.366
88.670
105.382
124.870
147.562
173.944
204.571
240.080
281.197
◆◆◆
Exercise 6 ◆ Numerical Solution of
Second-Order Differential Equations
Solve each equation by the Runge-Kutta method, and find the value of y at x 2.
Take a step size of 0.1 unit.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
y y xy 3, y(1, 1) 2
y xy 3xy2 2y, y(1, 1) 2
3y y yex xy3, y(1, 2) 2
2y xy y2 5 ln x, y(1, 0) 1
xy 2.4x2y 5.3x sin y 7.4y 0, y(1, 1) 0.1
y y ln x 2.69(x y)2 6.26y, y(1, 3.27) 8.26
x2y (x y)y 8.26x2y2 1.83y, y(1, 1) 2
y y 0.533xyex 0.326x cos y 0, y(1, 73.4) 273
y xy 59.2xy 74.1y lnhxyh, y(1, 1) 2
xy x2y 11.4x2 ln y 62.2x 0, y(1, 8) 52.1
Computer
11. Write a program or use a spreadsheet to solve a second-order differential equation by the Runge-Kutta method. Try your program on any of the equations
above. If your spreadsheet has a graphing utility, use it to graph x versus y over
the given range.
◆◆◆
REVIEW PROBLEMS
◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
Find the Laplace transform by direct integration.
1. f(t) 2t
2. f(t) 3t2
3. f(t) cos 3t
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 28
28
Solving Differential Equations by the Laplace Transform and by Numerical Methods
Find the Laplace transform by using Table 1.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
f(t) 2t3 3t
f(t) 3te2t
f(t) 2t 3e2t
f(t) 3et 2t2
f(t) 3t3 5t2 4et
y 3y, y(0) 1
y 6y, y(0) 1
3y 2y, y(0) 0
2y y 3y, y(0) 1 and y(0) 3
y 3y 4y, y(0) 1 and y(0) 3
Find the inverse transform.
6
14. F(s) 2
4s
s
15. F(s) 2
(s 3)
s6
16. F(s) 2
(s 2)
2s
17. F(s) (s2 5)2
3
18. F(s) (s 4)2 5
s1
19. F(s) s2 6s 8
4s2 2
20. F(s) s(s2 3)
3s
21. F(s) s2 2s 1
3
22. F(s) s2 s 4
4s
23. F(s) s2 4s 4
Solve each differential equation by the Laplace transform.
24. y 2y 0, y(0) 1
25. y 3y t2, y(0) 2
26. 3y y 2t, y(0) 1
27. y 2y y 2, y(0) 0 and y(0) 0
28. y 3y 2y t, y(0) 1 and y(0) 2
29. y 2y 4t, y(0) 3 and y(0) 0
30. y 3y 2y e2t, y(0) 2 and y(0) 1
31. A 21.5-lb ball is dropped from rest through air whose resisting force is equal
to 2.75 times the ball’s speed in ft/s. Write an expression for the speed of
the ball.
32. A weight that hangs from a spring is pulled down 0.500 cm at t 0 and
released from rest. The differential equation of motion is x 3.22x 18.5 0. Write the equation for x as a function of time.
36CH_PHCalter_TechMath_950299 23/2/2007 2:28 PM Page 29
29
Review Problems
33. The current in a certain RC circuit satisfies the equation 8.24i 149i 100.
If i is zero at t 0, write an equation for i.
34. The current in an RLC circuit satisfies the equation 5.45i 2820i 9730i 10.
If i and i are zero at t 0, write an equation for i.
35. In an RLC circuit, R 3750 , L 0.150 H, and C 1.25 µF. The circuit
is connected to a dc source of 150 V at t 0. If i is zero at t 0, write an
equation for the instantaneous current.
36. In an RL circuit, R 3750 and L 0.150 H. The circuit is connected to an
ac source of 28.4 sin 84t V at t 0. If i is zero at t 0, write an equation for
the instantaneous current.
Solve each differential equation. Use any numerical method with a step size of
0.1 unit.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
y xy 4y, y(2) 2. Find y(3).
xy 3y2 5xy, y(1) 5. Find y(2).
3y y xy2 2 ln y, y(1, 2) 8. Find y(2).
yy 7.2yy 2.8x sin x 2.2y 0, y(1, 5) 8.2. Find y(2).
y y ln y 5.27(x 2y)2 2.84x, y(1, 1) 3. Find y(2).
x2y (1 y)y 3.67x2 5.28x, y(1, 2) 3. Find y(2).
y xy 6ex 126 cos y 0, y(1, 2) 5. Find y(2).
y xy2 8 ln y, y(1) 7. Find y(2).
7.35y 2.85x sin y 7.34x 0, y(2) 4.3. Find y(3).
yy y 77.2y2 28.4x lnhxyh, y(3, 5) 3. Find y(4).
Writing
47. Suppose that your company gets a new project requiring you to solve many differential equations. You see this as a chance to get that computer you’ve always
wanted. Write a memo to your boss explaining how the computer can be used
to solve those differential equations.
Team Project
48. Given
y y e2t 0
where y(0) 2, solve this DE by all of the methods at your disposal:
(a)
(b)
(c)
(d)
(e)
analytically
graphically
numerically
by the Laplace transform
by CAS
Find y when t 2.