Download Elementary Differential Equations and Boundary Value Problems

September 11, 2008 11:18
boyce-9e-bvp
Sheet number 108 Page number 88
88
cyan black
Chapter 2. First Order Differential Equations
place to permit successful breeding, and the population rapidly declined to extinction. The last survivor died in 1914. The precipitous decline in the passenger pigeon
population from huge numbers to extinction in a few decades was one of the early
factors contributing to a concern for conservation in this country.
PROBLEMS
Problems 1 through 6 involve equations of the form dy/dt = f (y). In each problem sketch
the graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one as
asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions
in the ty-plane.
1. dy/dt = ay + by2 ,
a > 0, b > 0, y0 ≥ 0
2. dy/dt = ay + by2 ,
a > 0, b > 0, −∞ < y0 < ∞
3. dy/dt = y(y − 1)(y − 2),
y0 ≥ 0
4. dy/dt = ey − 1,
−∞ < y0 < ∞
5. dy/dt = e−y − 1,
−∞ < y0 < ∞
6. dy/dt = −2(arctan y)/(1 + y2 ),
−∞ < y0 < ∞
7. Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the
property that solutions lying on one side of the equilibrium solution tend to approach it,
whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the
equilibrium solution is said to be semistable.
y
y
φ (t) = k
k
k
φ (t) = k
(a)
t
(b)
t
FIGURE 2.5.9 In both cases the equilibrium solution φ(t) = k is semistable.
(a) dy/dt ≤ 0; (b) dy/dt ≥ 0.
(a) Consider the equation
dy/dt = k(1 − y)2 ,
(i)
where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution φ(t) = 1.
(b) Sketch f (y) versus y. Show that y is increasing as a function of t for y < 1 and also
for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus
solutions below the equilibrium solution approach it, and those above it grow farther away.
Therefore φ(t) = 1 is semistable.
(c) Solve Eq. (i) subject to the initial condition y(0) = y0 and confirm the conclusions
reached in part (b).
Problems 8 through 13 involve equations of the form dy/dt = f (y). In each problem sketch
the graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one
September 11, 2008 11:18
2.5
boyce-9e-bvp
Sheet number 109 Page number 89
Autonomous Equations and Population Dynamics
cyan black
89
as asymptotically stable, unstable, or semistable (see Problem 7). Draw the phase line, and
sketch several graphs of solutions in the ty-plane.
8. dy/dt = −k(y − 1)2 ,
k > 0, −∞ < y0 < ∞
9. dy/dt = y2 (y2 − 1),
−∞ < y0 < ∞
10. dy/dt = y(1 − y2 ),
−∞ < y0 < ∞
√
11. dy/dt = ay − b y,
a > 0, b > 0, y0 ≥ 0
2
2
12. dy/dt = y (4 − y ),
−∞ < y0 < ∞
13. dy/dt = y2 (1 − y)2 ,
−∞ < y0 < ∞
14. Consider the equation dy/dt = f (y) and suppose that y1 is a critical point, that is, f (y1 ) = 0.
Show that the constant equilibrium solution φ(t) = y1 is asymptotically stable if f % (y1 ) < 0
and unstable if f % (y1 ) > 0.
15. Suppose that a certain population obeys the logistic equation dy/dt = ry[1 − (y/K)].
(a) If y0 = K/3, find the time τ at which the initial population has doubled. Find the value
of τ corresponding to r = 0.025 per year.
(b) If y0 /K = α, find the time T at which y(T)/K = β, where 0 < α, β < 1. Observe that
T → ∞ as α → 0 or as β → 1. Find the value of T for r = 0.025 per year, α = 0.1, and
β = 0.9.
16. Another equation that has been used to model population growth is the Gompertz14
equation
dy/dt = ry ln(K/y),
where r and K are positive constants.
(a) Sketch the graph of f (y) versus y, find the critical points, and determine whether each
is asymptotically stable or unstable.
(b) For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is
concave down.
(c) For each y in 0 < y ≤ K, show that dy/dt as given by the Gompertz equation is never
less than dy/dt as given by the logistic equation.
17. (a) Solve the Gompertz equation
dy/dt = ry ln(K/y),
subject to the initial condition y(0) = y0 .
Hint: You may wish to let u = ln(y/K).
(b) For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 × 106 kg,
y0 /K = 0.25), use the Gompertz model to find the predicted value of y(2).
(c) For the same data as in part (b), use the Gompertz model to find the time τ at which
y(τ ) = 0.75K.
18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose
that water flows in at a constant rate k and is lost through evaporation at a rate proportional
to the surface area.
(a) Show that the volume V(t) of water in the pond at time t satisfies the differential
equation
dV/dt = k − απ(3a/π h)2/3 V 2/3 ,
where α is the coefficient of evaporation.
14 Benjamin
Gompertz (1779–1865) was an English actuary. He developed his model for population
growth, published in 1825, in the course of constructing mortality tables for his insurance company.
September 11, 2008 11:18
2.6
boyce-9e-bvp
Sheet number 119 Page number 99
cyan black
Exact Equations and Integrating Factors
99
Thus, if (µM)y is to equal (µN)x , it is necessary that
My − Nx
dµ
=
µ.
dx
N
(27)
If (My − Nx )/N is a function of x only, then there is an integrating factor µ that also
depends only on x; further, µ(x) can be found by solving Eq. (27), which is both linear
and separable.
A similar procedure can be used to determine a condition under which Eq. (23)
has an integrating factor depending only on y; see Problem 23.
Find an integrating factor for the equation
EXAMPLE
4
(3xy + y2 ) + (x2 + xy)y" = 0
(19)
and then solve the equation.
In Example 3 we showed that this equation is not exact. Let us determine whether it has an
integrating factor that depends on x only. On computing the quantity (My − Nx )/N, we find
that
My (x, y) − Nx (x, y)
3x + 2y − (2x + y)
1
=
= .
(28)
N(x, y)
x2 + xy
x
Thus there is an integrating factor µ that is a function of x only, and it satisfies the differential
equation
dµ
µ
= .
(29)
dx
x
Hence
µ(x) = x.
(30)
Multiplying Eq. (19) by this integrating factor, we obtain
(3x2 y + xy2 ) + (x3 + x2 y)y" = 0.
(31)
The latter equation is exact, and its solutions are given implicitly by
x3 y + 21 x2 y2 = c.
(32)
Solutions may also be found in explicit form since Eq. (32) is quadratic in y.
You may also verify that a second integrating factor for Eq. (19) is
µ(x, y) =
1
,
xy(2x + y)
and that the same solution is obtained, though with much greater difficulty, if this integrating
factor is used (see Problem 32).
PROBLEMS
Determine whether each of the equations in Problems 1 through 12 is exact. If it is exact, find
the solution.
1. (2x + 3) + (2y − 2)y" = 0
2. (2x + 4y) + (2x − 2y)y" = 0
2
2
2
3. (3x − 2xy + 2) dx + (6y − x + 3) dy = 0
4. (2xy2 + 2y) + (2x2 y + 2x)y" = 0
September 11, 2008 11:18
boyce-9e-bvp
100
Sheet number 120 Page number 100
cyan black
Chapter 2. First Order Differential Equations
5.
7.
8.
9.
10.
11.
12.
dy
ax + by
dy
ax − by
=−
6.
=−
dx
bx + cy
dx
bx − cy
(ex sin y − 2y sin x) dx + (ex cos y + 2 cos x) dy = 0
(ex sin y + 3y) dx − (3x − ex sin y) dy = 0
(yexy cos 2x − 2exy sin 2x + 2x) dx + (xexy cos 2x − 3) dy = 0
(y/x + 6x) dx + (ln x − 2) dy = 0,
x>0
(x ln y + xy) dx + (y ln x + xy) dy = 0;
x > 0, y > 0
x dx
y dy
+ 2
=0
(x2 + y2 )3/2
(x + y2 )3/2
In each of Problems 13 and 14 solve the given initial value problem and determine at least
approximately where the solution is valid.
13. (2x − y) dx + (2y − x) dy = 0,
y(1) = 3
14. (9x2 + y − 1) dx − (4y − x) dy = 0,
y(1) = 0
In each of Problems 15 and 16 find the value of b for which the given equation is exact, and
then solve it using that value of b.
15. (xy2 + bx2 y) dx + (x + y)x2 dy = 0
16. (ye2xy + x) dx + bxe2xy dy = 0
17. Assume that Eq. (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is
therefore exact. Show that a possible function ψ(x, y) is
! x
! y
ψ(x, y) =
M(s, y0 ) ds +
N(x, t) dt,
x0
y0
where (x0 , y0 ) is a point in R.
18. Show that any separable equation
is also exact.
M(x) + N(y)y" = 0
In each of Problems 19 through 22 show that the given equation is not exact but becomes exact
when multiplied by the given integrating factor. Then solve the equation.
19. x2 y3 + x(1 + y2 )y" = 0,
µ(x, y) = 1/xy3
#
"
#
"
sin y
cos y + 2e−x cos x
20.
− 2e−x sin x dx +
dy = 0,
µ(x, y) = yex
y
y
21. y dx + (2x − yey ) dy = 0,
µ(x, y) = y
22. (x + 2) sin y dx + x cos y dy = 0,
µ(x, y) = xex
23. Show that if (Nx − My )/M = Q, where Q is a function of y only, then the differential
equation
M + Ny" = 0
has an integrating factor of the form
µ(y) = exp
!
Q(y) dy.
24. Show that if (Nx − My )/(xM − yN) = R, where R depends on the quantity xy only, then
the differential equation
M + Ny" = 0
has an integrating factor of the form µ(xy). Find a general formula for this integrating
factor.
September 11, 2008 11:18
boyce-9e-bvp
Sheet number 164 Page number 144
144
PROBLEMS
cyan black
Chapter 3. Second Order Linear Equations
In each of Problems 1 through 8 find the general solution of the given differential equation.
1. y!! + 2y! − 3y = 0
2. y!! + 3y! + 2y = 0
3. 6y!! − y! − y = 0
4. 2y!! − 3y! + y = 0
!!
!
5. y + 5y = 0
6. 4y!! − 9y = 0
!!
!
7. y − 9y + 9y = 0
8. y!! − 2y! − 2y = 0
In each of Problems 9 through 16 find the solution of the given initial value problem. Sketch
the graph of the solution and describe its behavior as t increases.
9. y!! + y! − 2y = 0,
y(0) = 1, y! (0) = 1
!!
!
10. y + 4y + 3y = 0,
y(0) = 2, y! (0) = −1
11. 6y!! − 5y! + y = 0,
y(0) = 4, y! (0) = 0
!!
!
12. y + 3y = 0,
y(0) = −2, y! (0) = 3
!!
!
13. y + 5y + 3y = 0,
y(0) = 1, y! (0) = 0
!!
!
14. 2y + y − 4y = 0,
y(0) = 0, y! (0) = 1
15. y!! + 8y! − 9y = 0,
y(1) = 1, y! (1) = 0
!!
16. 4y − y = 0,
y(−2) = 1, y! (−2) = −1
17. Find a differential equation whose general solution is y = c1 e2t + c2 e−3t .
18. Find a differential equation whose general solution is y = c1 e−t/2 + c2 e−2t .
19. Find the solution of the initial value problem
y!! − y = 0,
y(0) = 45 ,
y! (0) = − 43 .
Plot the solution for 0 ≤ t ≤ 2 and determine its minimum value.
20. Find the solution of the initial value problem
2y!! − 3y! + y = 0,
y(0) = 2,
y! (0) = 21 .
Then determine the maximum value of the solution and also find the point where the
solution is zero.
21. Solve the initial value problem y!! − y! − 2y = 0, y(0) = α, y! (0) = 2. Then find α so that
the solution approaches zero as t → ∞.
22. Solve the initial value problem 4y!! − y = 0, y(0) = 2, y! (0) = β. Then find β so that the
solution approaches zero as t → ∞.
In each of Problems 23 and 24 determine the values of α, if any, for which all solutions tend to
zero as t → ∞; also determine the values of α, if any, for which all (nonzero) solutions become
unbounded as t → ∞.
23. y!! − (2α − 1)y! + α(α − 1)y = 0
24. y!! + (3 − α)y! − 2(α − 1)y = 0
25. Consider the initial value problem
2y!! + 3y! − 2y = 0,
y(0) = 1,
y! (0) = −β,
where β > 0.
(a) Solve the initial value problem.
(b) Plot the solution when β = 1. Find the coordinates (t0 , y0 ) of the minimum point of
the solution in this case.
(c) Find the smallest value of β for which the solution has no minimum point.