for each of the models in part (b)? k 0.8, w t dw dt
k1200 w dV dt kV23 . S kV23. V S. 12.5 dv dt 43.2
1.75v 12.5 dv dt 43.2 1.25v t 0. v 0 44. HOW DO
YOU SEE IT? The differential equation represents
weight stretches another spring 9 in. If both weights
are simultaneously pulled down 1 in. below their
respective equilibrium positions and then released,
find the first time after when their velociti
highest point reached by the weight. Assume the
springmass system resides in a medium offering a
resistance of lb times the instantaneous velocity in
feet per second. 10> 1g > > > y0 t = 0 > > 212g +
x y x 1 y dy dx 1 y ex e y 1 1 y dy dt 3t 2 1 y xy 0 y y
5 1 y dy dx x 4x 0 2 4y dy dx 0 dy dx x2 3y y 2 dy dx
1 dy dx x2 3 y2 dy dx x 1 y3 dr ds 0.05s dr ds 0.05r
dy dx 1 x dy dx 2x x dy dx 1 y dy dx
Equations F5 Exercises F.1 Verifying Solutions In
Exercises 112, verify that the function is a solution
of the differential equation. See Example 1. Solution
Differential Equation 1. 2. 3. 4. 5. 6. 7.
without solving the differential equation. Explain
your reasoning. Differential Equation Solution 23. (a)
24. (b) 25. (c) 26. (d) Finding a Particular Solution In
Exercises 2734, find the particular s
function of time. 21. A 16-lb weight stretches a
spring 4 ft. This springmass system is in a medium
with a damping constant of 4.5 lb-sec ft, and an
external force given by (in pounds) is being applie
resistor to 4 times the instantaneous change in the
charge, across the capacitor to 10 times the charge,
and across the inductor to 2 times the
instantaneous change in the current. Find the
charge on
ln10 x, 110 x, 10 x 9781133108490_App_F2.qxp
12/6/11 8:30 AM Page F9 F10 Appendix F
Differential Equations Application Example 6
Corporate Investing A corporation invests part of its
receipts at a ra
x 0 x 10 dx dt k10 x, y 2 x 0. ex yy 0 y2 ex2 2. 1 1 C,
C 2. 12 e02 C. y2 ex2 C y2 2 1 2 ex2 C1 y dy xex2 dx y
dy xex2 dx xex2 y dy dx xex2 xex2 yy 0 x 0. xe y 1 x2
yy 0 STUDY TIP In Example 5, the co
2 3 y x General solution: y = Cx2 (1, 3) Solution
Curves for FIGURE F.1 xy 2y 0
9781133108490_App_F1.qxp 12/6/11 8:29 AM
Page F2 Appendix F.1 Solutions of Differential
Equations F3 Example 3 Finding a
bxy + cy = 0 P(x) = ax Q(x) = bx, R(x) = c 2 , P(x)y(x)
+ Q(x)y(x) + R(x)y(x) = 0 17.4 To solve Equation (1),
we first make the change of variables and . We next
use the chain rule to find the derivat
notice that the coefficients with even indices are
interrelated and the coefficients with odd indices
are also interrelated. Even indices: Here so the
power is From the last line in the table, we have
gy dy fx dx C. dy dx fxgy f g TECH TUTOR You can use
a symbolic integration utility to solve a differential
equation that has separable variables. Use a
symbolic integration utility to solve the diffe
or Thus, Writing the power series by grouping its
even and odd powers together and substituting for
the coefficients yields . From Table 9.1 in Section
9.10, we see that the first series on the right-
differential equation is When the population of the
herd is 100. After 2 years, the population has grown
to 160. (a) Write the population as a function of (b)
What is the population of the herd after
you should first convert to standard form so that
you can identify the functions and Example 2
Solving a Linear Differential Equation Find the
general solution of Assume SOLUTION Begin by
writing the
SOLUTION The particular solutions represented by
and are shown in Figure F.3. Graphs of Five
Particular Solutions FIGURE F.3 Checkpoint 4 Given
that is the general solution of sketch the particular
so
second. The mass is then pulled down 2 m below its
equilibrium position and released with a downward
velocity of 3 m sec. At this same instant an external
force given by cos t (in newtons) is applied
Equations F21 Earlier in the text, you studied two
models for population growth: exponential growth,
which assumes that the rate of change of is
proportional to and logistic growth, which assumes
that
function and its derivatives. For instance, in Figure
F.1, suppose you want to find the particular solution
whose graph passes through the point This initial
condition can be written as when Initial c
be written as Checkpoint 6 Use the result of
Example 6 to find when and t 25 years. A P
$550,000, r 5.9%, A P r ert 1. A t C P r 0 Cer0 P r A 0
t 0, C. A Cert P r A C3ert P r rA P ertC2 lnrA P rt C rA
coefficient is zero when Thus, we assume the
solution interval Substitution of the series form and
its derivatives gives us Next, we equate the
coefficients of each power of x to zero as
summarized in
1 y 4 0 y x xy y x3x 4 2 2x C x y Cx xy 3x 2y 0 2
3x y 2 x y 4x y 0 2 y 3 x y 2x y 0 3 y e y 2y 0 2x y Ce
y 4y 4x The following warm-up exercises involve
skills that were covered in earlier sections.
2x 0 2x2 y y 2 Cx3 2 C 42. HOW DO YOU SEE IT? The
graph shows a solution of one of the following
differential equations. Determine the correct
equation. Explain your reasoning. (i) (ii) (iii) (iv) y 4
general solution of Example 3 Solving a
Differential Equation Find the general solution of
Use a graphing utility to graph several solutions.
SOLUTION Begin by separating variables, then
integrate eac
form. EXAMPLE 1 Solve the equation by the powerseries method. Solution We assume the series
solution takes the form of and calculate the
derivatives and Substitution of these forms into the
second-ord
6! x6 - b cn+2 = 2n - 1 (n + 2)(n + 1) cn, 17.5
Power-Series Solutions 17-31 EXERCISES 17.5 In
Exercises 118, use power series to find the general
solution of the differential equation. 1. 2. 3. 4. 5.
F11 Exercises F.2 Separation of Variables In Exercises
16, decide whether the variables in the differential
equation can be separated. 1. 2. 3. 4. 5. 6. Solving a
Differential Equation In Exercises 72