September 11, 2008 11:18
boyce-9e-bvp
Sheet number 36 Page number 16
16
cyan black
Chapter 1. Introduction
(b) dy/dt = −2y + 5,
(c) dy/dt = −2y + 10,
y(0) = y0
y(0) = y0
2. Follow the instructions for Problem 1 for the following initial value problems:
(a) dy/dt = y − 5,
y(0) = y0
(b) dy/dt = 2y − 5,
y(0) = y0
(c) dy/dt = 2y − 10,
y(0) = y0
3. Consider the differential equation
dy/dt = −ay + b,
where both a and b are positive numbers.
(a) Solve the differential equation.
(b) Sketch the solution for several different initial conditions.
(c) Describe how the solutions change under each of the following conditions:
i. a increases.
ii. b increases.
iii. Both a and b increase, but the ratio b/a remains the same.
4. Consider the differential equation dy/dt = ay − b.
(a) Find the equilibrium solution ye .
(b) Let Y(t) = y − ye ; thus Y(t) is the deviation from the equilibrium solution. Find the
differential equation satisfied by Y(t).
5. Undetermined Coefficients.
Here is an alternative way to solve the equation
dy/dt = ay − b.
(i)
dy/dt = ay.
(ii)
(a) Solve the simpler equation
Call the solution y1 (t).
(b) Observe that the only difference between Eqs. (i) and (ii) is the constant −b in Eq. (i).
Therefore it may seem reasonable to assume that the solutions of these two equations
also differ only by a constant. Test this assumption by trying to find a constant k such that
y = y1 (t) + k is a solution of Eq. (i).
(c) Compare your solution from part (b) with the solution given in the text in Eq. (17).
Note: This method can also be used in some cases in which the constant b is replaced by a
function g(t). It depends on whether you can guess the general form that the solution is
likely to take. This method is described in detail in Section 3.5 in connection with second
order equations.
6. Use the method of Problem 5 to solve the equation
dy/dt = −ay + b.
7. The field mouse population in Example 1 satisfies the differential equation
dp/dt = 0.5p − 450.
(a) Find the time at which the population becomes extinct if p(0) = 850.
(b) Find the time of extinction if p(0) = p0 , where 0 < p0 < 900.
(c) Find the initial population p0 if the population is to become extinct in 1 year.
September 11, 2008 11:18
2.2
boyce-9e-bvp
Sheet number 67 Page number 47
cyan black
Separable Equations
47
we see that the interval ends when we reach points where the tangent line is vertical. It follows
from the differential equation (22) that these are points where 4 + y3 = 0, or
y = (−4)1/3 ∼
= ±3.3488. These
= −1.5874. From Eq. (24) the corresponding values of x are x ∼
points are marked on the graph in Figure 2.2.3.
Note 1: Sometimes an equation of the form (2)
dy
= f (x, y)
dx
has a constant solution y = y0 . Such a solution is usually easy to find because if
f (x, y0 ) = 0 for some value y0 and for all x, then the constant function y = y0 is a
solution of the differential equation (2). For example, the equation
dy
(y − 3) cos x
=
dx
1 + 2y2
(25)
has the constant solution y = 3. Other solutions of this equation can be found by
separating the variables and integrating.
Note 2: The investigation of a first order nonlinear equation can sometimes be
facilitated by regarding both x and y as functions of a third variable t. Then
dy
dy/dt
=
.
dx
dx/dt
(26)
dy
F(x, y)
=
,
dx
G(x, y)
(27)
If the differential equation is
then, by comparing numerators and denominators in Eqs. (26) and (27), we obtain
the system
dx/dt = G(x, y),
dy/dt = F(x, y).
(28)
At first sight it may seem unlikely that a problem will be simplified by replacing a
single equation by a pair of equations, but, in fact, the system (28) may well be more
amenable to investigation than the single equation (27). Chapter 9 is devoted to
nonlinear systems of the form (28).
Note 3: In Example 2 it was not difficult to solve explicitly for y as a function
of x. However, this situation is exceptional, and often it will be better to leave the
solution in implicit form, as in Examples 1 and 3. Thus, in the problems below and
in other sections where nonlinear equations appear, the words “solve the following
differential equation” mean to find the solution explicitly if it is convenient to do so,
but otherwise to find an equation defining the solution implicitly.
PROBLEMS
In each of Problems 1 through 8 solve the given differential equation.
1. y = x2 /y
2. y = x2 /y(1 + x3 )
3. y + y2 sin x = 0
4. y = (3x2 − 1)/(3 + 2y)
2
2
5. y = (cos x)(cos 2y)
6. xy = (1 − y2 )1/2
September 11, 2008 11:18
boyce-9e-bvp
48
Sheet number 68 Page number 48
cyan black
Chapter 2. First Order Differential Equations
7.
dy
x − e−x
=
dx
y + ey
8.
dy
x2
=
dx
1 + y2
In each of Problems 9 through 20:
(a) Find the solution of the given initial value problem in explicit form.
(b) Plot the graph of the solution.
(c) Determine (at least approximately) the interval in which the solution is defined.
9.
11.
13.
15.
17.
18.
19.
20.
y = (1 − 2x)y2 ,
y(0) = −1/6
10.
−x
x dx + ye dy = 0,
y(0) = 1
12.
y = 2x/(y + x2 y),
y(0) = −2
14.
y = 2x/(1 + 2y),
y(2) = 0
16.
y = (3x2 − ex )/(2y − 5),
y(0) = 1
y = (e−x − ex )/(3 + 4y),
y(0) = 1
sin 2x dx + cos 3y dy = 0,
y(π/2) = π/3
y2 (1 − x2 )1/2 dy = arcsin x dx,
y(0) = 1
y = (1 − 2x)/y,
y(1) = −2
2
dr/dθ = r /θ,
r(1) = 2
y = xy3 (1 + x2 )−1/2 ,
y(0) = 1
√
y = x(x2 + 1)/4y3 ,
y(0) = −1/ 2
Some of the results requested in Problems 21 through 28 can be obtained either by solving
the given equations analytically or by plotting numerically generated approximations to the
solutions. Try to form an opinion as to the advantages and disadvantages of each approach.
21. Solve the initial value problem
y = (1 + 3x2 )/(3y2 − 6y),
y(0) = 1
and determine the interval in which the solution is valid.
Hint: To find the interval of definition, look for points where the integral curve has a
vertical tangent.
22. Solve the initial value problem
y = 3x2 /(3y2 − 4),
y(1) = 0
and determine the interval in which the solution is valid.
Hint: To find the interval of definition, look for points where the integral curve has a
vertical tangent.
23. Solve the initial value problem
y = 2y2 + xy2 ,
y(0) = 1
and determine where the solution attains its minimum value.
24. Solve the initial value problem
y = (2 − ex )/(3 + 2y),
y(0) = 0
and determine where the solution attains its maximum value.
25. Solve the initial value problem
y = 2 cos 2x/(3 + 2y),
y(0) = −1
and determine where the solution attains its maximum value.
26. Solve the initial value problem
y = 2(1 + x)(1 + y2 ),
y(0) = 0
and determine where the solution attains its minimum value.
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
M(x) + N(y)y = 0
is also exact.
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
− 2e−x sin x dx +
dy = 0,
μ(x, y) = yex
20.
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
3.2
boyce-9e-bvp
Sheet number 175 Page number 155
cyan black
Solutions of Linear Homogeneous Equations; the Wronskian
PROBLEMS
155
In each of Problems 1 through 6 find the Wronskian of the given pair of functions.
1. e2t , e−3t/2
2. cos t, sin t
3. e−2t , te−2t
4. x, xex
t
t
5. e sin t, e cos t
6. cos2 θ, 1 + cos 2θ
In each of Problems 7 through 12 determine the longest interval in which the given initial value
problem is certain to have a unique twice differentiable solution. Do not attempt to find the
solution.
7. ty + 3y = t,
y(1) = 1, y (1) = 2
8. (t − 1)y − 3ty + 4y = sin t,
y(−2) = 2, y (−2) = 1
9. t(t − 4)y + 3ty + 4y = 2,
y(3) = 0, y (3) = −1
10. y + (cos t)y + 3(ln |t|)y = 0,
y(2) = 3, y (2) = 1
11. (x − 3)y + xy + (ln |x|)y = 0,
y(1) = 0, y (1) = 1
12. (x − 2)y + y + (x − 2)(tan x)y = 0,
y(3) = 1, y (3) = 2
13. Verify that y1 (t) = t 2 and y2 (t) = t −1 are two solutions of the differential equation
t 2 y − 2y = 0 for t > 0. Then show that y = c1 t 2 + c2 t −1 is also a solution of this equation for any c1 and c2 .
14. Verify that y1 (t) = 1 and y2 (t) = t 1/2 are solutions of the differential equation
yy + (y )2 = 0 for t > 0. Then show that y = c1 + c2 t 1/2 is not, in general, a solution of
this equation. Explain why this result does not contradict Theorem 3.2.2.
15. Show that if y = φ(t) is a solution of the differential equation y + p(t)y + q(t)y = g(t),
where g(t) is not always zero, then y = cφ(t), where c is any constant other than 1, is not a
solution. Explain why this result does not contradict the remark following Theorem 3.2.2.
16. Can y = sin(t 2 ) be a solution on an interval containing t = 0 of an equation
y + p(t)y + q(t)y = 0 with continuous coefficients? Explain your answer.
17. If the Wronskian W of f and g is 3e4t , and if f (t) = e2t , find g(t).
18. If the Wronskian W of f and g is t 2 et , and if f (t) = t, find g(t).
19. If W(f , g) is the Wronskian of f and g, and if u = 2f − g, v = f + 2g, find the Wronskian
W(u, v) of u and v in terms of W(f , g).
20. If the Wronskian of f and g is t cos t − sin t, and if u = f + 3g, v = f − g, find the Wronskian
of u and v.
21. Assume that y1 and y2 are a fundamental set of solutions of y + p(t)y + q(t)y = 0 and
let y3 = a1 y1 + a2 y2 and y4 = b1 y1 + b2 y2 , where a1 , a2 , b1 , and b2 are any constants. Show
that
W(y3 , y4 ) = (a1 b2 − a2 b1 )W(y1 , y2 ).
Are y3 and y4 also a fundamental set of solutions? Why or why not?
In each of Problems 22 and 23 find the fundamental set of solutions specified by Theorem 3.2.5
for the given differential equation and initial point.
22. y + y − 2y = 0,
t0 = 0
23. y + 4y + 3y = 0,
t0 = 1
In each of Problems 24 through 27 verify that the functions y1 and y2 are solutions of the given
differential equation. Do they constitute a fundamental set of solutions?
24. y + 4y = 0;
y1 (t) = cos 2t, y2 (t) = sin 2t
25. y − 2y + y = 0;
y1 (t) = et , y2 (t) = tet
September 11, 2008 11:18
boyce-9e-bvp
Sheet number 210 Page number 190
190
cyan black
Chapter 3. Second Order Linear Equations
21. Show that the solution of the initial value problem
L[y] = y + p(t)y + q(t)y = g(t),
y (t0 ) = y0
y(t0 ) = y0 ,
(i)
can be written as y = u(t) + v(t), where u and v are solutions of the two initial value
problems
L[u] = 0,
u (t0 ) = y0 ,
u(t0 ) = y0 ,
L[v] = g(t),
(ii)
v(t0 ) = 0,
v (t0 ) = 0,
(iii)
respectively. In other words, the nonhomogeneities in the differential equation and in
the initial conditions can be dealt with separately. Observe that u is easy to find if a
fundamental set of solutions of L[u] = 0 is known.
22. By choosing the lower limit of integration in Eq. (28) in the text as the initial point t0 ,
show that Y(t) becomes
Y(t) =
t
t0
y1 (s)y2 (t) − y1 (t)y2 (s)
g(s) ds.
y1 (s)y2 (s) − y1 (s)y2 (s)
Show that Y(t) is a solution of the initial value problem
L[y] = g(t),
y (t0 ) = 0.
y(t0 ) = 0,
Thus Y can be identified with v in Problem 21.
23. (a) Use the result of Problem 22 to show that the solution of the initial value problem
y + y = g(t),
is
y=
y(t0 ) = 0,
t
y (t0 ) = 0
(i)
sin(t − s)g(s) ds.
(ii)
t0
(b) Use the result of Problem 21 to find the solution of the initial value problem
y + y = g(t),
y(0) = y0 ,
y (0) = y0 .
24. Use the result of Problem 22 to find the solution of the initial value problem
y(t0 ) = 0,
L[y] = (D − a)(D − b)y = g(t),
y (t0 ) = 0,
where a and b are real numbers with a = b.
25. Use the result of Problem 22 to find the solution of the initial value problem
L[y] = [D2 − 2λD + (λ2 + μ2 )]y = g(t),
y(t0 ) = 0,
y (t0 ) = 0.
Note that the roots of the characteristic equation are λ ± iμ.
26. Use the result of Problem 22 to find the solution of the initial value problem
L[y] = (D − a)2 y = g(t),
where a is any real number.
y(t0 ) = 0,
y (t0 ) = 0,
September 11, 2008 11:18
4.3
boyce-9e-bvp
Sheet number 257 Page number 237
cyan black
The Method of Undetermined Coefficients
237
The method of undetermined coefficients can be used whenever it is possible to
guess the correct form for Y(t). However, this is usually impossible for differential
equations not having constant coefficients, or for nonhomogeneous terms other than
the type described previously. For more complicated problems we can use the method
of variation of parameters, which is discussed in the next section.
PROBLEMS
In each of Problems 1 through 8 determine the general solution of the given differential
equation.
1. y − y − y + y = 2e−t + 3
2. y(4) − y = 3t + cos t
−t
3. y + y + y + y = e + 4t
4. y − y = 2 sin t
5. y(4) − 4y = t 2 + et
6. y(4) + 2y + y = 3 + cos 2t
(6)
7. y + y = t
8. y(4) + y = sin 2t
In each of Problems 9 through 12 find the solution of the given initial value problem. Then
plot a graph of the solution.
9. y + 4y = t;
y(0) = y (0) = 0, y (0) = 1
(4)
10. y + 2y + y = 3t + 4;
y(0) = y (0) = 0, y (0) = y (0) = 1
11. y − 3y + 2y = t + et ;
y(0) = 1, y (0) = − 41 , y (0) = − 23
(4)
12. y + 2y + y + 8y − 12y = 12 sin t − e−t ;
y(0) = 3, y (0) = 0,
y (0) = −1, y (0) = 2
In each of Problems 13 through 18 determine a suitable form for Y(t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
13. y − 2y + y = t 3 + 2et
14. y − y = te−t + 2 cos t
(4)
t
15. y − 2y + y = e + sin t
16. y(4) + 4y = sin 2t + tet + 4
(4)
2
17. y − y − y + y = t + 4 + t sin t
18. y(4) + 2y + 2y = 3et + 2te−t + e−t sin t
19. Consider the nonhomogeneous nth order linear differential equation
a0 y(n) + a1 y(n−1) + · · · + an y = g(t),
(i)
where a0 , . . . , an are constants. Verify that if g(t) is of the form
eαt (b0 t m + · · · + bm ),
then the substitution y = eαt u(t) reduces Eq. (i) to the form
k0 u(n) + k1 u(n−1) + · · · + kn u = b0 t m + · · · + bm ,
(ii)
where k0 , . . . , kn are constants. Determine k0 and kn in terms of the a’s and α. Thus
the problem of determining a particular solution of the original equation is reduced to
the simpler problem of determining a particular solution of an equation with constant
coefficients and a polynomial for the nonhomogeneous term.
Method of Annihilators. In Problems 20 through 22 we consider another way of arriving at the
proper form of Y(t) for use in the method of undetermined coefficients. The procedure is based
on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of
such terms) can be viewed as solutions of certain linear homogeneous differential equations
with constant coefficients. It is convenient to use the symbol D for d/dt. Then, for example, e−t
is a solution of (D + 1)y = 0; the differential operator D + 1 is said to annihilate, or to be an
annihilator of, e−t . Similarly, D2 + 4 is an annihilator of sin 2t or cos 2t, (D − 3)2 = D2 − 6D + 9
is an annihilator of e3t or te3t , and so forth.