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Part A. ORDINARY DIFFERENTIAL
EQUATIONS (ODEs)
CHAPTER 1
First-Order ODEs
Major Changes
There is more material on modeling in the text as well as in the problem set.
Some additions on population dynamics appear in Sec. 1.5.
Electric circuits are shifted to Chap. 2, where second-order ODEs will be available.
This avoids repetitions that are unnecessary and practically irrelevant.
Team Projects, CAS Projects, and CAS Experiments are included in most problem sets.
SECTION 1.1. Basic Concepts. Modeling, page 2
Purpose. To give the students a first impression what an ODE is and what we mean by
solving it.
Background Material. For the whole chapter we need integration formulas and
techniques, which the student should review.
General Comments
This section should be covered relatively rapidly to get quickly to the actual solution
methods in the next sections.
Equations (1)–(3) are just examples, not for solution, but the student will see that
solutions of (1) and (2) can be found by calculus, and a solution y e x of (3) by inspection.
Problem Set 1.1 will help the student with the tasks of
Solving y ƒ(x) by calculus
Finding particular solutions from given general solutions
Setting up an ODE for a given function as solution
Gaining a first experience in modeling, by doing one or two problems
Gaining a first impression of the importance of ODEs
without wasting time on matters that can be done much faster, once systematic methods
are available.
Comment on “General Solution” and “Singular Solution”
Usage of the term “general solution” is not uniform in the literature. Some books use the
term to mean a solution that includes all solutions, that is, both the particular and the
singular ones. We do not adopt this definition for two reasons. First, it is frequently quite
difficult to prove that a formula includes all solutions; hence, this definition of a general
solution is rather useless in practice. Second, linear differential equations (satisfying rather
general conditions on the coefficients) have no singular solutions (as mentioned in the
text), so that for these equations a general solution as defined does include all solutions.
For the latter reason, some books use the term “general solution” for linear equations only;
but this seems very unfortunate.
1
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SOLUTIONS TO PROBLEM SET 1.1, page 8
2.
6.
10.
12.
14.
16.
y e3x/3 c 4. y (sinh 4x)/4 c
Second order. 8. First order.
y ce0.5x, y(2) ce 2, c 2/e, y (2/e)e0.5x 0.736e0.5x
y ce x x 1, y(0) c 1 3, c 2, y 2e x x 1
y c sec x, y(0) c/cos 0 c _12, y _12 sec x
Substitution of y cx c 2 into the ODE gives
y2 xy y c 2 xc (cx c 2) 0.
Similarly,
y _14x 2,
y _12x,
_1 x 2 x(_1 x) _1 x 2 0.
4
2
4
thus
18. In Prob. 17 the constants of integration were set to zero. Here, by two integrations,
y g,
v y gt c1,
y _21gt 2 c1t c2,
y(0) c2 y0,
and, furthermore,
v(0) c1 v0,
hence
y _12gt 2 v0 t y0,
as claimed. Times of fall are 4.5 and 6.4 sec, from t 100/4.
9 and 200/4.
9.
20. y ky. Solution y y0 ekx, where y0 is the pressure at sea level x 0. Now
y(18000) y0 ek18000 _12y0 (given). From this,
ek18000 _1 , y(36000) y ek218000 y (ek18000)2 y (_1 )2 _1 y .
0
2
0
0 2
4 0
22. For 1 year and annual, daily, and continuous compounding we obtain the values
ya(1) 1060.00,
yd(1) 1000(1 0.06/365)365 1061.83,
yc(1) 1000e0.06 1061.84,
respectively. Similarly for 5 years,
ya(5) 1000 1.065 1338.23,
yd(5) 1000(1 0.06/365)3655 1349.83,
yc(5) 1000e0.065 1349.86.
We see that the difference between daily compounding and continuous compounding
is very small.
The ODE for continuous compounding is yc ryc.
SECTION 1.2. Geometric Meaning of y ƒ(x, y). Direction Fields, page 9
Purpose. To give the student a feel for the nature of ODEs and the general behavior of
fields of solutions. This amounts to a conceptual clarification before entering into formal
manipulations of solution methods, the latter being restricted to relatively small—albeit
important—classes of ODEs. This approach is becoming increasingly important, especially
because of the graphical power of computer software. It is the analog of conceptual
studies of the derivative and integral in calculus as opposed to formal techniques of
differentiation and integration.
Comment on Isoclines
These could be omitted because students sometimes confuse them with solutions. In the
computer approach to direction fields they no longer play a role.
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Comment on Order of Sections
This section could equally well be presented later in Chap. 1, perhaps after one or two
formal methods of solution have been studied.
SOLUTIONS TO PROBLEM SET 1.2, page 11
2. Semi-ellipse x 2/4 y 2/9 13/9, y 0. To graph it, choose the y-interval large
enough, at least 0 y 4.
4. Logistic equation (Verhulst equation; Sec. 1.5). Constant solutions y 0 and y _12.
For these, y 0. Increasing solutions for 0 y(0) _12, decreasing for y(0) _12.
6. The solution (not of interest for doing the problem) is obtained by using
dy/dx 1/(dx/dy) and solving
x c 2/(tan _12 y 1);
dx/dy 1/(1 sin y) by integration,
thus y 2 arctan ((x 2 c)/(x c)).
8. Linear ODE. The solution involves the error function.
12. By integration, y c 1/x.
16. The solution (not needed for doing the problem) of y 1/y can be obtained by
separating variables and using the initial condition; y 2/2 t c, y 2t
.
1
18. The solution of this initial value problem involving the linear ODE y y t 2 is
y 4et t 2 2t 2.
20. CAS Project. (a) Verify by substitution that the general solution is y 1 cex.
Limit y 1 (y(x) 1 for all x), increasing for y(0)
1, decreasing for
y(0) 1.
(b) Verify by substitution that the general solution is x 4 y 4 c. More “squareshaped,” isoclines y kx. Without the minus on the right you get “hyperbola-like”
curves y 4 x 4 const as solutions (verify!). The direction fields should turn out in
perfect shape.
(c) The computer may be better if the isoclines are complicated; but the computer
may give you nonsense even in simpler cases, for instance when y(x) becomes
imaginary. Much will depend on the choice of x- and y-intervals, a method of trial
and error. Isoclines may be preferable if the explicit form of the ODE contains roots
on the right.
SECTION 1.3. Separable ODEs. Modeling, page 12
Purpose. To familiarize the student with the first “big” method of solving ODEs, the
separation of variables, and an extension of it, the reduction to separable form by a
transformation of the ODE, namely, by introducing a new unknown function.
The section includes standard applications that lead to separable ODEs, namely,
1. the ODE giving tan x as solution
2. the ODE of the exponential function, having various applications, such as in
radiocarbon dating
3. a mixing problem for a single tank
4. Newton’s law of cooling
5. Torricelli’s law of outflow.
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In reducing to separability we consider
6. the transformation u y/x, giving perhaps the most important reducible class of
ODEs.
Ince’s classical book [A11] contains many further reductions as well as a systematic
theory of reduction for certain classes of ODEs.
Comment on Problem 5
From the implicit solution we can get two explicit solutions
y c (6x)
2
representing semi-ellipses in the upper half-plane, and
y c (6x)
2
representing semi-ellipses in the lower half-plane. [Similarly, we can get two explicit
solutions x(y) representing semi-ellipses in the left and right half-planes, respectively.]
On the x-axis, the tangents to the ellipses are vertical, so that y(x) does not exist. Similarly
for x(y) on the y-axis.
This also illustrates that it is natural to consider solutions of ODEs on open rather than
on closed intervals.
Comment on Separability
An analytic function ƒ(x, y) in a domain D of the xy-plane can be factored in D,
ƒ(x, y) g(x)h(y), if and only if in D,
ƒxyƒ ƒx ƒy
[D. Scott, American Math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide,
but this may save time in cases of more complicated ODEs, some of which may perhaps
be of practical interest. You may perhaps ask your students to derive such a criterion.
Comments on Application
Each of those examples can be modified in various ways, for example, by changing the
application or by taking another form of the tank, so that each example characterizes a
whole class of applications.
The many ODEs in the problem set, much more than one would ordinarily be willing
and have the time to consider, should serve to convince the student of the practical
importance of ODEs; so these are ODEs to choose from, depending on the students’
interest and background.
Comment on Footnote 3
Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22.
Philosophiae Naturalis Principia Mathematica was his most influential work.
Leibniz invented calculus independently in 1675 and introduced notations that were
essential to the rapid development in this field. His first publication on differential calculus
appeared in 1684.
SOLUTIONS TO PROBLEM SET 1.3, page 18
2. dy/y 2 (x 2)dx. The variables are now separated. Integration on both sides gives
1
_12x 2 2x c*.
y
Hence
y
2
.
x 2 4x c
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4. Set y 9x v. Then y v 9x. By substitution into the given ODE you obtain
y v 9 v2.
dv
dx.
v 9
By separation,
2
Integration gives
1
v
arctan
x c*,
3
3
arctan
v
3x c
3
and from this and substitution of y v 9x,
v 3 tan (3x c),
y 3 tan (3x c) 9x.
6. Set u y/x. Then y xu, y u xu. Substitution into the ODE and subtraction
of u on both sides gives
y 4x
y
4
u xu u,
y
x
u
xu 4
.
u
Separation of variables and replacement of u with y/x yields
2u du 8
dx,
x
u2 8 ln x c,
y 2 x 2 (8 ln x c).
8. u y/x, y xu, y u xu. Substitute u into the ODE, drop xu on both sides,
and divide by x 2 to get
xy xu x 2u _12x 2u2 xu,
u _12u2.
Separate variables, integrate, and solve algebraically for u:
du
_12 dx,
u2
1
_12(x c*),
u
Hence
y xu u
2
.
cx
2x
.
cx
10. By separation, y dy 4x dx. By integration, y 2 4x 2 c. The initial condition
y(0) 3, applied to the last equation, gives 9 0 c. Hence y 2 4x 2 9.
12. Set u y/x. Then y u xu. Divide the given ODE by x 2 and substitute u and
u into the resulting equation. This gives
2u(u xu) 3u2 1.
Subtract 2u2 on both sides and separate the variables. This gives
dx
2u du
.
2
x
u 1
2xuu u2 1,
Integrate, take exponents, and then take the square root:
ln (u2 1) ln x c*,
u2 1 cx,
u
cx
.
1
Hence
y xu xcx
.
1
From this and the initial condition, y(1) c
1 2, c 5. This gives the answer
.
y x5x
1
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14. Set u y/x. Then y xu, y u xu. Substitute this into the ODE, subtract u on
both sides, simplify algebraically, and integrate:
xu 2x 2
cos (x 2)
u
uu 2x cos (x 2),
u2/2 sin (x 2) c.
Hence y 2 2x 2(sin (x 2) c). By the initial condition, (sin _12 c), c 0,
y xu x
2 sin (
x 2).
y
6
5
4
3
2
1
–4
–3
–2
–1
0
1
2
3
4
x
–1
–2
–3
–4
–5
–6
Problem Set 1.3. Problem 14. First five real branches of the solution
16. u y/x, y xu, y u xu u 4x 4 cos2u. Simplify, separate variables, and
integrate:
u 4x 3 cos2u,
du/cos2u 4x 3 dx,
tan u x 4 c.
Hence
y xu x arctan (x 4 c).
From the initial condition, y(2) 2 arctan (16 c) 0, c 16. Answer:
y x arctan (x 4 16).
18. Order terms:
dr
(1 b cos ) br sin .
d
Separate variables and integrate:
dr
b sin
r
1 b cos
d ,
ln r ln (1 b cos ) c*.
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Take exponents and use the initial condition:
r c(1 b cos ),
r(
) c(1 b 0) ,
2
c .
Hence the answer is r (1 b cos ).
20. On the left, integrate g(w) over w from y0 to y. On the right, integrate ƒ(t) over t from
x0 to x. In Prob. 19,
we
y
(t 1) dt.
x
w2
dw 1
0
22. Consider any straight line y ax through the origin. Its slope is y/x a. The slope
of a solution curve at a point of intersection (x, ax) is y g(y/x) g(a) const,
independent of the point (x, y) on the straight line considered.
24. Let kB and kD be the constants of proportionality for the birth rate and death rate,
respectively. Then y kB y kD y, where y(t) is the population at time t. By separating
variables, integrating, and taking exponents,
dy/y (kB kD) dt,
ln y (kB kD)t c*,
y ce(kB
k )t
D
.
26. The model is y Ay ln y with A 0. Constant solutions are obtained from
y 0 when y 0 and 1. Between 0 and 1 the right side is positive (since ln y 0),
so that the solutions grow. For y 1 we have ln y 0; hence the right side is negative,
so that the solutions decrease with increasing t. It follows that y 1 is stable. The
general solution is obtained by separation of variables, integration, and two subsequent
exponentiations:
dy/(y ln y) A dt,
At
ln y ce
,
ln (ln y) At c*,
y exp (ceAt).
28. The temperature of the water is decreasing exponentially according to Newton’s law
of cooling. The decrease during the first 30 min, call it d1, is greater than that, d2,
during the next 30 min. Thus d1 d2 190 110 80 as measured. Hence the
temperature at the beginning of parking, if it had been 30 min earlier, before the arrest,
would have been greater than 190 80 270, which is impossible. Therefore Jack
has no alibi.
30. The cross-sectional area A of the hole is multiplied by 4. In the particular solution,
15.00 0.000332t is changed to 15.00 4 0.000332t because the second term
contains A/B. This changes the time t 15.00/0.000332 when the tank is empty, to
t 15.00/(4 0.000332), that is, to t 12.6/4 3.1 hr, which is 1/4 of the original
time.
32. According to the physical information given, you have
S 0.15S .
Now let * 0. This gives the ODE dS/d 0.15S. Separation of variables yields
the general solution S S0 e0.15 with the arbitrary constant denoted by S0. The
angle should be so large that S equals 1000 times S0. Hence e0.15 1000,
(ln 1000)/0.15 46 7.3 2, that is, eight times, which is surprisingly little.
Equally remarkable is that here we see another application of the ODE y ky and
a derivation of it by a general principle, namely, by working with small quantities
and then taking limits to zero.
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36. B now depends on h, namely, by the Pythagorean theorem,
B(h) r 2 (R2 (R h)2 ) (2Rh h2).
Hence you can use the ODE
h 26.56(A/B)h
in the text, with constant A as before and the new B. The latter makes the further
calculations different from those in Example 5.
From the given outlet size A 5 cm2 and B(h) we obtain
dh
5
26.56 h
.
dt
(2Rh h2)
Now 26.56 5/ 42.27, so that separation of variables gives
(2Rh1/2 h3/2 ) dh 42.27 dt.
By integration,
_4 Rh3/2 _2h5/2 42.27t c.
3
5
From this and the initial condition h(0) R we obtain
_4 R5/2 _2R5/2 0.9333R5/2 c.
3
5
Hence the particular solution (in implicit form) is
_4 Rh3/2 _2h5/2 42.27t 0.9333R5/2.
3
5
The tank is empty (h 0) for t such that
0 42.27t 0.9333R5/2;
hence
t
0.9333 5/2
R 0.0221R5/2.
42.27
For R 1 m 100 cm this gives
t 0.0221 1005/2 2210 [sec] 37 [min].
The tank has water level R/2 for t in the particular solution such that
4
2 R5/2
R3/2
R 3/2 0.9333R5/2 42.27t.
3
2
5 25/2
The left side equals 0.4007R5/2. This gives
t
0.4007 0.9333 5/2
R 0.01260R5/2.
42.27
For R 100 this yields t 1260 sec 21 min. This is slightly more than half the
time needed to empty the tank. This seems physically reasonable because if the water
level is R/2, this means that 11/16 of the total water volume has flown out, and 5/16
is left—take into account that the velocity decreases monotone according to
Torricelli’s law.
R
R=h
r
h
Problem Set 1.3. Tank in Problem 36
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SECTION 1.4. Exact ODEs. Integrating Factors, page 19
Purpose. This is the second “big” method in this chapter, after separation of variables, and
also applies to equations that are not separable. The criterion (5) is basic. Simpler cases
are solved by inspection, more involved cases by integration, as explained in the text.
Comment on Condition (5)
Condition (5) is equivalent to (6) in Sec. 10.2, which is equivalent to (6) in the case of two
variables x, y. Simple connectedness of D follows from our assumptions in Sec. 1.4. Hence
the differential form is exact by Theorem 3, Sec. 10.2, part (b) and part (a), in that order.
Method of Integrating Factors
This greatly increases the usefulness of solving exact equations. It is important in itself
as well as in connection with linear ODEs in the next section. Problem Set 1.4 will help
the student gain skill needed in finding integrating factors. Although the method has
somewhat the flavor of tricks, Theorems 1 and 2 show that at least in some cases one can
proceed systematically—and one of them is precisely the case needed in the next section
for linear ODEs.
SOLUTIONS TO PROBLEM SET 1.4, page 25
2. (x y) dx (y x) dy 0. Exact; the test gives 1 on both sides. Integrate
x y over x:
u _12x 2 xy k(y).
Differentiate this with respect to y and compare with N:
uy x k y x.
Thus
k y,
k _12 y 2 c*.
Answer: _12x 2 xy _12 y 2 _12(x y)2 c; thus y x c.
4. Exact; the test gives ey e x on both sides. Integrate M with respect to x to get
u xey ye x k(y).
Differentiate this with respect to y and equate the result to N:
uy xey e x k N xey e x.
Hence k 0, k const. Answer: xey ye x c.
6. Exact; the test gives e x sin y on both sides. Integrate M with respect to x:
u e x cos y k(y).
uy e x sin y k.
Differentiate:
Equate this to N e x sin y. Hence k 0, k const. Answer: e x cos y c.
8. Exact; 1/x 2 1/y 2 on both sides of the equation. Integrate M with respect to x:
u x2 x
y
k(y).
y
x
Differentiate this with respect to y and equate the result to N:
uy x
1
k N,
2 y
x
k 2y,
Answer:
x2 x
y
y 2 c.
y
x
k y 2.
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10. Exact; the test gives 2x sin (x 2) on both sides. Integrate N with respect to y to get
u y cos (x 2) l(x).
Differentiate this with respect to x and equate the result to M:
ux 2xy sin (x 2) l M 2xy sin (x 2),
l 0.
Answer: y cos (x 2) c.
12. Not exact. Try Theorem 1. In R you have
Py Qx exy 1 exy(x 1) xexy 1 Q
so that R 1, F ex, and the exact ODE is
(ey yex) dx (xey ex) dy 0.
The test gives ey ex on both sides of the equation. Integration of M FP with
respect to x gives
u xey yex k(y).
Differentiate this with respect to y and equate it to N FQ:
uy xey ex k N xey ex.
Hence k 0. Answer: xey yex c.
14. Not exact; 2y y. Try Theorem 1; namely,
R (Py Qx)/Q (2y y)/(xy) 3/x.
Hence
F 1/x 3.
The exact ODE is
(x y2
y
) dx 2 dy 0.
x3
x
The test gives 2y/x 3 on both sides of the equation. Obtain u by integrating N FQ
with respect to y:
u
y2
l(x).
2x 2
Thus
ux y2
y2
l
M
x
.
x3
x3
Hence l x, l x 2/2, y 2/2x 2 x 2/2 c*. Multiply by 2 and use the initial
condition y(2) 1:
y2
x 2 2 c 3.75
x
because inserting y(2) 1 into the last equation gives 4 0.25 3.75.
16. The given ODE is exact and can be written as d(cos xy) 0; hence cos xy c, or
you can solve it for y by the usual procedure. y(1) gives 1 c.
Answer: cos xy 1.
18. Try Theorem 2. You have
R* (Qx Py)/P [
1
x
cos xy x sin xy (x sin xy 2 )]
y
y
Hence F* y. This gives the exact ODE
(y cos xy x) dx (y x cos xy) dy 0.
P
1
.
y
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In the test, both sides of the equation are cos xy xy sin xy. Integrate M with respect
to x:
u sin xy _1 x 2 k(y).
Hence
u x cos xy k(y).
y
2
Equate the last equation to N y x cos xy. This shows that k y; hence
k y 2/2. Answer: sin xy _12x 2 _12y 2 c.
20. Not exact; try Theorem 2:
R* (Qx Py)/P [1 (cos2 y sin2 y 2x cos y sin y)]/P
[2 sin2 y 2x cos y sin y]/P
2(sin y)(sin y x cos y)/(sin y cos y x cos2 y)
2(sin y)/cos y 2 tan y.
Integration with respect to y gives 2 ln (cos y) ln (1/cos2 y); hence F* 1/cos2 y.
The resulting exact equation is
x
(tan y x) dx dy 0.
cos2 y
The exactness test gives 1/cos2 y on both sides. Integration of M with respect to x
yields
x
u x tan y _12x 2 k(y).
From this,
uy k.
cos2 y
Equate this to N x/cos2 y to see that k 0, k const. Answer: x tan y _1x 2 c.
2
22. (a) Not exact. Theorem 2 applies and gives F* 1/y from
R* (Qx Py)/P (0 cos x)/(y cos x) 1
.
y
Integrating M in the resulting exact ODE
cos x dx 1
dy 0
y2
with respect to x gives
u sin x k(y).
From this,
uy k N 1
.
y2
Hence k 1/y. Answer: sin x 1/y c.
Note that the integrating factor 1/y could have been found by inspection and by the
fact that an ODE of the general form
ƒ(x) dx g(y) dy 0
is always exact, the test resulting in 0 on both sides.
(b) Yes. Separation of variables gives
dy/y 2 cos x dx.
By integration,
1/y sin x c*
in agreement with the solution in (a).
(d) seems better than (c). But this may depend on your CAS. In (d) the CAS may
draw vertical asymptotes that disturb the figure.
From the solution in (a) or (b) the student should conclude that for each nonzero
y(x0) y0 there is a unique particular solution because
sin x0 1/y0 c.
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24. (A) ey cosh x c.
(B) R* tan y, F 1/cos y. Separation:
dy/cos2 y (1 2x) dx,
tan y x x 2 c.
(C) R 2/x, F 1/x 2, x y 2/x c. v y/x, and separation:
2v dv/(1 v2) dx/x,
x 2 y 2 cx;
divide by x.
(D) Separation is simplest. y cx3/4. R 9/(4x), F(x) x9/4, x 3y 4 c.
R* 3/y, F*(y) y 3.
SECTION 1.5. Linear ODEs. Bernoulli Equation. Population Dynamics,
page 26
Purpose. Linear ODEs are of great practical importance, as Problem Set 1.5 illustrates
(and even more so are second-order linear ODEs in Chap. 2). We show that the
homogeneous ODE of the first order is easily separated and the nonhomogeneous ODE
is solved, once and for all, in the form of an integral (4) by the method of integrating
factors. Of course, in simpler cases one does not need (4), as our examples illustrate.
Comment on Notation
We write
y p(x)y r(x).
p(x) seems standard. r(x) suggests “right side.” The notation
y p(x)y q(x)
used in some calculus books (which are not concerned with higher order ODEs) would
be shortsighted here because later, in Chap. 2, we turn to second-order ODEs
y p(x)y q(x)y r(x),
where we need q(x) on the left, thus in a quite different role (and on the right we would
have to choose another letter different from that used in the first-order case).
Comment on Content
Bernoulli’s equation appears occasionally in practice, so the student should remember
how to handle it.
A special Bernoulli equation, the Verhulst equation, plays a central role in population
dynamics of humans, animals, plants, and so on, and we give a short introduction to this
interesting field, along with one reference in the text.
Riccati and Clairaut equations are less important than Bernoulli’s, so we have put
them in the problem set; they will not be needed in our further work.
Input and output have become common terms in various contexts, so we thought this
a good place to mention them.
Problems 37–42 express properties that make linearity important, notably in obtaining
new solutions from given ones. The counterparts of these properties will, of course,
reappear in Chap. 2.
Comment on Footnote 5
Eight members of the Bernoulli family became known as mathematicians; for more details,
see p. 220 in Ref. [GR2] listed in App. 1.
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SOLUTIONS TO PROBLEM SET 1.5, page 32
4. The standard form (1) is y 4y x, so that (4) gives
y e 4x [ e4xx dx c] ce 4x x/4 1/16.
1
3
y 3 . From this and (4) we obtain, with
x
x
c 2 from the initial condition,
6. The standard form (1) is y y x3 [ x 3x3 dx c] x3[x c] x2 2x3.
8. From (4) with p 2, h 2x, r 4 cos 2x we obtain
y e2x [ e 2x 4 cos 2x dx c] e2x[e 2x(cos 2x sin 2x) c].
It is perhaps worthwhile mentioning that integrals of this type can more easily be
evaluated by undetermined coefficients. Also, the student should verify the result by
differentiation, even if it was obtained by a CAS. From the initial condition we obtain
y(_14) ce /2 0 1 2;
hence
c e /2.
The answer can be written
y e/22x cos 2x sin 2x.
10. In (4) we have p 4x 2; hence h 4x 3/3, so that (4) gives
y e4x /3 [ e (4x
3
3
/3)x 2/ 2
(4x 2 x) dx c].
The integral can be evaluated by noting that the factor of the exponential function
under the integral sign is the derivative of the exponent of that function. We thus
obtain
3
3
2
3
2
y e4x /3 [e(4x /3)x /2 c] ce4x /3 ex /2.
12. y tan x 2(y 4). Separation of variables gives
dy
cos x
2
dx.
y4
sin x
By integration,
ln y 4 2 ln sin x c*.
Taking exponents on both sides gives
y 4 c sin2 x,
y c sin2 x 4.
The desired particular solution is obtained from the initial condition
y(_12) c 4 0,
c 4.
Answer: y 4 4 sin2 x.
14. In (4) we have p tan x, h ln (cos x), eh 1/cos x, so that (4) gives
y (cos x) [
cos x
cos
e
x
0.01x
dx c] [100 e0.01x c] cos x.
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The initial condition gives y(0) 100 c 0; hence c 100. The particular
solution is
y 100(1 e0.01x) cos x.
The factor 0.01, which we included in the exponent, has the effect that the graph of
y shows a long transition period. Indeed, it takes x 460 to let the exponential function
e0.01x decrease to 0.01. Choose the x-interval of the graph accordingly.
16. The standard form (1) is
3
1
y
.
2
cos x
cos2 x
y Hence h 3 tan x, and (4) gives the general solution
ecos
3 tan x
y e3 tan x [
2
x
dx c] .
To evaluate the integral, observe that the integrand is of the form
_1 (3 tan x) e 3 tan x;
3
that is,
_1 (e3 tan x).
3
Hence the integral has the value _13 e3 tan x. This gives the general solution
y e3 tan x [_1 e3 tan x c] _1 ce3 tan x.
3
3
The initial condition gives from this
y( _14) _13 ce3 _43;
The answer is y _1 e33 tan x.
hence
c e 3.
3
18. Bernoulli equation. First solution method: Transformation to linear form. Set
y 1/u. Then y y u/u2 1/u 1/u2. Multiplication by u2 gives the linear
ODE in standard form
u u 1.
General solution
u ce x 1.
Hence the given ODE has the general solution
y 1/(ce x 1).
From this and the initial condition y(0) 1 we obtain
y(0) 1/(c 1) 1,
c 2,
Answer:
y 1/(1 2e x).
Second solution method: Separation of variables and use of partial fractions.
1
1
dy
(
) dy dx.
y
y(y 1)
y1
Integration gives
ln y 1 ln y ln j
y1
j x c*.
y
Taking exponents on both sides, we obtain
y1
1
1
ce x,
y
y
We now continue as before.
1
1
ce x,
y
y
1
.
1 ce x
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20. Separate variables, integrate, and take exponents:
cot y dy dx/(x 2 1),
ln sin y arctan x c*
and
sin y cearctan x.
Now use the initial condition y(0) _12 :
1 ce 0,
c 1.
Answer: y arcsin (earctan x).
22. First solution method: by setting z cos 2y (linearization): From z we have
z (2 sin 2y)y.
_12 z xz 2x.
From the ODE,
This is a linear ODE. Its standard form is
z 2xz 4x.
p 2x,
In (4) this gives
h x 2.
Hence (4) gives the solution in terms of z in the form
z ex [ ex (4x) dx c] ex [2ex c] 2 cex .
2
2
2
2
2
From this we obtain the solution
2
y _12 arccos z _12 arccos (2 cex ).
Second solution method: Separation of variables. By algebra,
y sin 2y x(cos 2y 2).
Separation of variables now gives
sin 2y
dy x dx.
2 cos 2y
_1 ln 2 cos 2y _1 x 2 c*.
2
2
Integrate:
Multiply by 2 and take exponents:
2 cos 2y cex .
2
ln 2 cos 2y x 2 2c*,
Solve this for y:
cos 2y 2 cex ,
2
2
y _12 arccos (2 cex ).
24. Bernoulli ODE. Set u y 3 and note that u 3y 2y. Multiply the given ODE by
3y 2 to obtain
3
3y 2y 3x 2y 3 ex sinh x.
In terms of u this gives the linear ODE
u 3x 2u ex sinh x.
3
In (4) we thus have h x 3. The solution is
u ex [ ex ex sinh x dx c] ex [cosh x c]
3
and y u1/3.
3
3
3
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26. The salt content in the inflow is 50(1 cos t). Let y(t) be the salt content in the tank
to be determined. Then y(t)/1000 is the salt content per gallon. Hence (50/1000)y(t) is
the salt content in the outflow per minute. The rate of change y equals the balance,
y ln Out 50(1 cos t) 0.05y.
Thus y 0.05y 50(1 cos t). Hence p 0.05, h 0.05t, and (4) gives the
general solution
y e0.05t ( e0.05t 50(1 cos t) dt c)
e0.05t (e0.05t (1000 a cos t b sin t) c)
1000 a cos t b sin t ce0.05t
where a 2.5/(1 0.052) 2.494 and b 50/(1 0.052) 49.88, which we
obtained by evaluating the integral. From this and the initial condition y(0) 200 we
have
y(0) 1000 a c 200,
c 200 1000 a 802.5.
Hence the solution of our problem is
y(t) 1000 2.494 cos t 49.88 sin t 802.5e0.05t.
Figure 20 shows the solution y(t). The last term in y(t) is the only term that depends
on the initial condition (because c does). It decreases monotone. As a consequence,
y(t) increases but keeps oscillating about 1000 as the limit of the mean value.
This mean value is also shown in Fig. 20. It is obtained as the solution of the ODE
y 0.05y 50.
Its solution satisfying the initial condition is
y 1000 800e0.05t.
28. k1(T Ta) follows from Newton’s law of cooling. k2(T Tw) models the effect of
heating or cooling. T Tw calls for cooling; hence k2(T Tw) should be negative
in this case; this is true, since k2 is assumed to be negative in this formula. Similarly
for heating, when heat should be added, so that the temperature increases.
The given model is of the form
T kT K k1C cos (/12)t.
This can be seen by collecting terms and introducing suitable constants, k k1 k2
(because there are two terms involving T ), and K k1A k2 Tw P. The general
solution is
T cekt K/k L(k cos ( t/12) (/12) sin ( t/12)),
where L k1C/(k 2 2/144). The first term solves the homogeneous ODE
T kT and decreases to zero. The second term results from the constants A (in Ta),
Tw, and P. The third term is sinusoidal, of period 24 hours, and time-delayed against
the outside temperature, as is physically understandable.
30. y ky(1 y) ƒ(y), where k 0 and y is the proportion of infected persons.
Equilibrium solutions are y 0 and y 1. The first, y 0, is unstable because
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ƒ(y) 0 if 0
y
1 but ƒ(y)
0 for negative y. The solution y 1 is stable
because ƒ(y) 0 if 0 y 1 and ƒ(y) 0 if y 1. The general solution is
y
1
.
1 cekt
It approaches 1 as t * . This means that eventually everybody in the population
will be infected.
32. The model is
y Ay By 2 Hy Ky By 2 y(K By)
where K A H. Hence the general solution is given by (9) in Example 4 with A
replaced by K A H. The equilibrium solutions are obtained from y 0; hence
they are y1 0 and y2 K/B. The population y2 remains unchanged under harvesting,
and the fraction Hy2 of it can be harvested indefinitely—hence the name.
34. For the first 3 years you have the solution
y1 4/(5 3e0.8t)
from Prob. 32. The idea now is that, by continuity, the value y1(3) at the end of the
first period is the initial value for the solution y2 during the next period. That is,
y2(3) y1(3) 4/(5 3e2.4).
Now y2 is the solution of y y y 2 (no fishing!). Because of the initial condition
this gives
y2 4/(4 e3t 3e0.6t).
Check the continuity at t 3 by calculating
y2(3) 4/(4 e 0 3e2.4).
Similarly, for t from 6 to 9 you obtain
y3 4/(5 e4.80.8t e1.80.8t 3e0.60.8t).
This is a period of fishing. Check the continuity at t 6:
y3(6) 4/(5 e 0 e3 3e5.4).
This agrees with
y2(6) 4/(4 e3 3e5.4).
36. y1 1/u1,
u1(0) 1/y1(0) 0.5,
y 1 u 1/u12 0.8y1 y12 0.8/u1 1/u12,
u 1 0.8u1 1,
u1 1.25 c1e0.8t
1.25 0.75e0.8t 1/y1
for 0
t
3. u 2 u2 1, u2 1 c2 et. The continuity condition is
u2(3) 1 c2 e3 u1(3) 1.25 0.75e2.4.
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For c2 this gives
c2 e 3(1 1.25 0.75e2.4) 0.25e 3 0.75e0.6.
This gives for 3
t
6
u2 1 0.25e3t 0.75e0.6t 1/y2.
Finally, for 6
t
9 we have the ODE is u 3 0.8u3 1, whose general solution is
u3 1.25 c3 e0.8t.
c3 is determined by the continuity condition at t 6, namely,
u3(6) 1.25 c3 e4.8 u2(6) 1 0.25e3 0.75e5.4.
This gives
c3 e4.8(1.25 1 0.25e3 0.75e5.4)
0.25e4.8 0.25e1.8 0.75e0.6.
Substitution gives the solution for 6
u3 1.25 (0.25e
4.8
t
9:
0.25e1.8 0.75e0.6)e0.8t 1/y3.
38. Substitution gives the identity 0 0.
These problems are of importance because they show why linear ODEs are
preferable over nonlinear ones in the modeling process. Thus one favors a linear ODE
over a nonlinear one if the model is a faithful mathematical representation of the
problem. Furthermore, these problems illustrate the difference between homogeneous
and nonhomogeneous ODEs.
40. We obtain
(y1 y2) p(y1 y2) y 1 y 2 py1 py2
(y 1 py1) (y 2 py2)
rr
0.
42. The sum satisfies the ODE with r1 r2 on the right. This is important as the key to
the method of developing the right side into a series, then finding the solutions
corresponding to single terms, and finally, adding these solutions to get a solution of
the given ODE. For instance, this method is used in connection with Fourier series,
as we shall see in Sec. 11.5.
44. (a) y Y v reduces the Riccati equation to a Bernoulli equation by removing the
term h(x). The second transformation, v 1/u, is the usual one for transforming a
Bernoulli equation with y 2 on the right into a linear ODE.
Substitute y Y 1/u into the Riccati equation to get
Y u/u2 p(Y 1/u) g(Y 2 2Y/u 1/u2) h.
Since Y is a solution, Y pY gY 2 h. There remains
u/u2 p/u g(2Y/u 1/u2).
Multiplication by u2 gives u pu g(2Yu 1). Reshuffle terms to get
u (2Yg p)u g,
the linear ODE as claimed.
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(b) Substitute y Y x to get 1 2x 4 x x 4 x 4 x 1, which is true.
Now substitute y x 1/u. This gives
1 u/u2 (2x 3 1)(x 1/u) x 2(x 2 2x/u 1/u2) x 4 x 1.
Most of the terms cancel on both sides. There remains u/u2 1/u x 2/u2.
Multiplication by u2 finally gives u u x 2. The general solution is
u cex x 2 2x 2
and y x 1/u. Of course, instead performing this calculation we could have used
the general formula in (a), in which
2Yg p 2x(x 2) 2x 3 1 1
and
g x 2.
(c) Substitution of Y x 2 shows that this is a solution. In the ODE for u you find
2Yg p 2x 2 (sin x) (3 2x 2 sin x) 3.
Also, g sin x. Hence the ODE for u is u 3u sin x. Solution:
u ce3x 0.1 cos x 0.3 sin x
and
y x 2 1/u.
(e) y y xy y /y2 by the chain rule. Hence y (x 1/y2) 0.
(A) From y 0 we obtain by integrations y cx a. Substitution into the given
ODE gives cx a xc 1/c; hence a 1/c. This is a family of straight lines.
(B) The other factor is zero when x 1/y2. By integration, y 2x1/2 c*.
Substituting y and y x1/2 into the given equation y xy 1/y, we obtain
2x1/2 c* x x1/ 2 1/x1/ 2;
hence c* 0. This gives the singular solution y 2x, a curve, to which those
straight lines in (A) are tangent.
(f) By differentiation, 2yy y xy y 0, y (2y x) 0, (A) y 0,
y cx a. By substitution, c 2 xc cx a 0, a c 2, y cx c 2, a family
of straight lines. (B) y x/2, y x 2/4 c*. By substitution into the given ODE,
x 2/4 x 2/2 x 2/4 c* 0, c* 0, y x 2/4, the envelope of the family; see
Fig. 6 in Sec. 1.1.
SECTION 1.6. Orthogonal Trajectories. Optional, page 35
Purpose. To show that families of curves F(x, y, c) 0 can be described by ODEs
y 1/ƒ(x, y) produces as general solution the orthogonal
y ƒ(x, y) and the switch to trajectories. This is a nice application that may also help the student to gain more
self-confidence, skill, and a deeper understanding of the nature of ODEs.
We leave this section optional, for reasons of time. This will cause no gap.
The reason why ODEs can be applied in this fashion results from the fact that
general solutions of ODEs involve an arbitrary constant that serves as the parameter
of this one-parameter family of curves determined by the given ODE, and then another
general solution similarly determines the one-parameter family of the orthogonal
trajectories.
Curves and their orthogonal trajectories play a role in several physical applications (e.g.,
in connection with electrostatic fields, fluid flows, and so on).
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SOLUTIONS TO PROBLEM SET 1.6, page 36
2. xy c, and by differentiation, y xy 0; hence y y/x. The ODE of the
trajectories is y x/y. By separation and integration, y2/2 x 2/2 c*. Hyperbolas.
(So are the given curves.)
4. By differentiation, 2yy 4x; hence y 2x/y. Thus the ODE of the trajectories is
y y/2x. By separating, integrating, and taking exponents on both sides,
dy/y dx/2x,
ln y _12 ln x c**,
y c*/x.
6. ye 3x c. Differentiation gives
(y 3y)e 3x 0.
Hence the ODE of the given family is y 3y. For the trajectories we obtain
_1
2
_1
y 1/(3y),
y
y _13,
y _32
x c*.
2 y 3 x c**,
8. 2x 2yy 0, so that the ODE of the curves is y x/y.
Hence the ODE of the trajectories is y y/x. Separating variables, integrating,
and taking exponents gives hyperbolas as trajectories; namely,
y/y 1/x,
ln y ln x c**,
xy c*.
10. xy1/2 ĉ, or x 2y1 c. By differentiation,
2xy1 x 2y2y 0,
y 2y/x.
This is the ODE of the given family. Hence the orthogonal trajectories have the ODE
x
y .
2y
2y y x,
Thus
y2 _21x 2 c*
12. x 2 y 2 2cy 0. Solve algebraically for 2c:
x2 y2
x2
y 2c.
y
y
Differentiation gives
2x
x 2y
y 0.
y
y2
By algebra,
y(
Solve for y:
y 2x
y
x2
2x
1) .
y2
y
(
2xy
y2 x2
) 2
.
2
y x2
y
This is the ODE of the given family. Hence the ODE
y 2 x2
y
1 y ( 2xy
2 x
of the trajectories is
x
).
y
To solve this equation, set u y/x. Then
1
1
y xu u (u ) .
u
2
(ellipses).
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Subtract u on both sides to get
xu u2 1
.
2u
Now separate variables, integrate, and take exponents, obtaining
2u du
dx
,
x
u2 1
ln (u2 1) ln x c1,
u2 1 c2
.
x
Write u y/x and multiply by x 2 on both sides of the last equation. This gives
y 2 x 2 c2 x.
The answer is
(x c3)2 y 2 c32.
Note that the given circles all have their centers on the y-axis and pass through the
origin. The result shows that their orthogonal trajectories are circles, too, with centers
on the x-axis and passing through the origin.
14. By differentiation, gx dx gy dy 0. Hence y gx /gy. This implies that the
trajectories are obtained from
y gy /gx.
For Prob. 6 we obtain ye 3x c and by differentiation,
y e 3x
1
,
3ye 3x
3y
1
y
y ,
3
y2
x
c**
2
3
and so forth.
16. Differentiating xy c, we have y xy 0, so that the ODE of the given hyperbolas
is y y/x. The trajectories are thus obtained by solving y x/y. By separation
of variables and integration we obtain
y
y x
y 2 x 2 c*
and
(hyperbolas).
18. Setting y 0 gives from x 2 (y c)2 1 c 2 the equation x 2 c 2 1 c 2;
hence x 1 and x 1, which verifies that those circles all pass through 1 and
1, each of them simultaneously through both points. Subtracting c 2 on both sides of
the given equation, we obtain
x 2 y 2 2cy 1,
x 2 y 2 1 2cy,
x2 1
y 2c.
y
Emphasize to your class that the ODE for the given curves must always be free of c.
Having accomplished this, we can now differentiate. This gives
2x
x2 1
(
1) y 0.
y
y2
This is the ODE of the given curves. Replacing y with 1/y and y with y, we obtain
the ODE of the trajectories:
2x
x2 1
(
1)
y
y2
(y) 0.
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Multiplying this by y, we get
2xy
x2 1
1 0.
y
y2
Multiplying this by y2/x 2, we obtain
d
1
1
y2
2y
y
y2
1 2 2 ( ) 1 2 0.
x
dx
x
x
x
x
By integration,
1
y2
x
2c*.
x
x
y 2 x 2 1 2c*x.
Thus,
We see that these are the circles
y 2 (x c*)2 c*2 1
dashed in Fig. 25, as claimed.
20. (B) By differentiation,
2x
2yy
0,
a2
b2
y Hence the ODE of the orthogonal trajectories is
a2 y
y 2 .
By separation,
b x
2x/a 2
b 2x
.
2 a 2y
2y/b
a 2 dx
dy
.
b2 x
y
Integration and taking exponents gives
a2
ln y 2 ln x c**,
b
2 2
y c*xa /b .
This shows that the ratio a 2/b 2 has substantial influence on the form of the trajectories.
For a 2 b 2 the given curves are circles, and we obtain straight lines as trajectories.
a 2/b 2 2 gives quadratic parabolas. For higher integer values of a 2/b 2 we obtain
parabolas of higher order. Intuitively, the “flatter” the ellipses are, the more rapidly
the trajectories must increase to have orthogonality.
Note that our discussion also covers families of parabolas; simply interchange the
roles of the curves and their trajectories.
(C) For hyperbolas we have a minus sign in the formula of the given curves. This
produces a plus sign in the ODE for the curves and a minus sign in the ODE for the
trajectories:
a 2y
y 2 .
b x
Separation of variables and integration gives
y c*xa /b .
2
2
For a 2/b 2 1 we obtain the hyperbolas y c*x, and for higher values of a 2/b 2 we
obtain less familiar curves.
(D) The problem set of this section contains other families of curves whose trajectories
can be readily obtained by solving the corresponding ODEs.
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SECTION 1.7. Existence and Uniqueness of Solutions, page 37
Purpose. To give the student at least some impression of the theory that would occupy a
central position in a more theoretical course on a higher level.
Short Courses. This section can be omitted.
Comment on Iteration Methods
Iteration methods were used rather early in history, but it was Picard who made them
popular. Proofs of the theorems in this section (given in books of higher level, e.g., [A11])
are based on the Picard iteration (see CAS Project 10).
Iterations are well suited for the computer because of their modest storage demand and
usually short programs in which the same loop or loops are used many times, with different
data. Because integration is generally not difficult for a CAS, Picard’s method has gained
some popularity during the past few decades.
SOLUTIONS TO PROBLEM SET 1.7, page 41
2. The initial condition is given at the point x 1. The coefficient of y is 0 at that
point, so from the ODE we already see that something is likely to go wrong. Separating
variables, integrating, and taking exponents gives
dy
2 dx
,
y
x1
ln y 2 ln x 1 c*,
y c(x 1)2.
This last expression is the general solution. It shows that y(1) 0 for any c. Hence
the initial condition y(1) 1 cannot be satisfied. This does not contradict the theorems
because we first have to write the ODE in standard form:
y ƒ(x, y) 2y
.
x1
This shows that ƒ is not defined when x 1 (to which the initial condition refers).
4. For k 0 we still get no solution, violating the existence as in Prob. 2. For k 0
we obtain infinitely many solutions, because c remains unspecified. Thus in this case
the uniqueness is violated. Neither of the two theorems is violated in either case.
6. By separation and integration,
dy
2x 4
2
dx,
y
x 4x
ln y ln x 2 4x c*.
Taking exponents gives the general solution
y c(x 2 4x).
From this we can see the answers:
No solution if y(0) k 0 or y(4) k 0.
A unique solution if y(x0) equals any y0 and x0 0 or x0 4.
Infinitely many solutions if y(0) 0 or y(4) 0.
This does not contradict the theorems because
ƒ(x, y) is not defined when x 0 or 4.
2x 4
x 2 4x
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8. (A) The student should gain an understanding for the “intermediate” position of a
Lipschitz condition: it is more than continuity but less than partial differentiability.
(B) Here the student should realize that the linear ODE is basically simpler than a
nonlinear ODE. The calculation is straightforward because
ƒ(x, y) r(x) p(x)y
and implies that
ƒ(x, y2) ƒ(x, y1) p(x) y2 y1 My2 y1
where the boundedness p(x) M for x x0 a follows from the continuity of p
in this closed interval.
10. (B) yn x2
x3
xn1
•••
,
2!
3!
(n 1)!
y ex x 1
(C) y0 1, y1 1 2x, y2 1 2x 4x 2 8x 3
,•••
3
1
1 2x 4x 2 8x 3 • • •
1 2x
(D) y (x 1)2, y 0. It approximates y 0. General solution y (x c)2.
(E) y y would be a good candidate to begin with. Perhaps you write the initial
choice as y0 a; then a 0 corresponds to the choice in the text, and you see how
the expressions in a are involved in the approximations. The conjecture is true for
any choice of a constant (or even of a continuous function of x).
It was mentioned in footnote 9 that Picard used his iteration for proving his existence
and uniqueness theorems. Since the integrations involved in the method can be handled
on the computer quite efficiently, the method has gained in importance in numerics.
y(x) SOLUTIONS TO CHAP. 1 REVIEW QUESTIONS AND PROBLEMS, page 42
12. Linear ODE. Formula (4) in Sec. 1.5 gives, since p 3, h 3x,
y e 3x ( e3x(2x) dx c) e 3x(e3x(_32x _92 ) c) ce 3x _32x _92.
14. Separate variables. y dy 16x dx, _12y 2 8x 2 c*, y 2 16x 2 c. Hyperbolas.
16. Linear ODE. Standard form y xy x 3 x. Use (4), Sec. 1.5, with p x,
h x 2/2, obtaining the general solution
y ex /2 ( ex /2 (x 3 x) dx c) ex /2[ex /2(x 2 1) c]
2
2
2
2
2
cex /2 x 2 1.
18. Exact; the exactness test gives 3 sin x sinh 3y on both sides. Integrate the
coefficient function of dx with respect to x, obtaining
u
M dx cos x cosh 3y k(y).
Differentiate this with respect to y and equate the result to the coefficient function
of dy:
uy 3 cos x sinh 3y k(y) N.
Hence k 0.
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The implicit general solution is
cos x cosh 3y c.
20. Solvable (A) as a Bernoulli equation or (B) by separating variables.
(A) Set y 2 u since a 1; hence 1 a 2. Differentiate u y 2, substitute y
from the given ODE, and express the resulting equation in terms of u; that is,
u 2yy 2y(y 1/y) 2u 2.
This is a linear ODE with unknown u. Its standard form is u 2u 2. Solve it by
(4) in Sec. 1.5 or by noting that the homogeneous ODE has the general solution ce 2x,
and a particular solution of the nonhomogeneous ODE is 1. Hence u ce 2x 1,
and u y 2.
(B) y y 1/y (y 2 1)/y, y dy/(y 2 1) dx. Integrate and take exponents
on both sides:
_1 ln (y 2 1) x c*,
y 2 1 ce 2x.
2
22. The argument of the tangent suggests to set y/x u. Then y xu, and by differentiation
and use of the given ODE divided by x,
y u xu tan u u;
xu tan u.
hence
Separation of variables gives
cot u du dx/x,
ln sin u ln x c*,
sin u cx.
This yields the general solution
y xu x arcsin cx.
24. We set y 2x z as indicated. Then y z 2x, y z 2 and by substitution
into the given ODE,
xy xz 2x z 2 z 2x.
Subtraction of 2x on both sides gives xz z 2 z. By separation of variables and
integration we obtain
1
dx
dz
1
( ) dz ,
z
x
z2 z
z1
ln j
z
j ln x c*.
z1
We now take exponents and simplify algebraically. This yields
y 2x
z
cx,
y 2x 1
z1
y 2x cx(y 2x 1).
Solving for y, we finally have
(1 cx)y 2x(1 cx) cx,
y 2x 1 1/(1 cx).
26. The first term on the right suggests the substitution u y/x. Then y xu, and from
the ODE, xy x(u xu) u3 xu. Subtract xu on both sides to get x 2u u3.
Separate variables and integrate:
1
u3 du x2 dx,
_12u2 x1 c*;
hence
u2 .
c 2/x
This gives the general solution
y
xu x
cx
2/
.
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28. Logistic equation. y 1/u, y u/u2 3/u 12/u2. Multiplication by u2 gives
the linear ODE
u 3u 12.
u ce3x 4.
Solution:
Hence the general solution of the given ODE is
y 1/u 1/(ce3x 4).
From the given initial condition we obtain y(0) 1/(c 4) 2; hence c 3.5.
The answer is
1
y
.
3x
3.5e
4
30. Linear ODE. The corresponding homogeneous ODE has the general solution
y ce x. A solution of the nonhomogeneous equation can be found without
integration by parts and recursion if we substitute
y A cos x B sin x
y A sin x B cos x
and
and equate the result to the right side; that is,
y y (B A) cos x (A B) sin x 2b cos x.
This gives A B b/. The general solution is
y ce x b
(cos x sin x).
Thus
y(0) c b/ 0, c b/.
32. Not exact; in the test we get 2 2y/x on the left but 1 on the right. Theorem 1 in
Sec. 1.4 gives an integrating factor depending only on x, namely, F(x) x; this follows
from
1
2y
1
R
(2 1) x
x
x 2y
by integrating R and taking exponents. The resulting exact equation is
[2xy y 2 e x(1 x)] dx (x 2 2xy) dy 0.
From it we calculate by integration with respect to y
u N dy x 2y xy 2 l(x).
We differentiate this with respect to x and equate the result to the coefficient of dx
in the exact ODE. This gives
ux 2xy y 2 l 2xy y 2 e x(x 1);
hence
l e x(x 1)
and by integration, l xe x. The general implicit solution is
u(x, y) x 2y xy 2 xe x c.
From the initial condition, u(1, 1) 1 1 e 2 e. The particular solution of
the initial value problem is
u(x, y) x 2y xy 2 xe x 2 e.
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34. In problems of this sort we need two conditions, because we must determine the
arbitrary constant c in the general solution and the constant k in the exponent. In
the present case, these are the initial temperature T(0) 10 and the temperature
T(5) 20 after 5 minutes. Newton’s law of cooling gives the model
T k(T 25).
By separation of variables and integration we obtain
T cekt 25.
The initial condition gives T(0) c 25 10; hence c 15. From the second
given condition we obtain
T(5) 15e 5k 25 20,
15e 5k 5,
k (ln _1 )/5 0.2197.
3
We can now determine the time when T reaches 24.9, namely, from
15ekt 25 24.9,
ekt 0.1/15.
Hence
t [ln (0.1/15)]/k 5.011/(0.2197) 23 [min].
36. This will give a general formula for determining the half-life H from two
measurements y1 and y2 at times t1 and t2, respectively. Accordingly, we use letters
and insert the given numeric data only at the end of the derivation. We have
y ky,
y y0 ekt
and from this
y1 y(t1) y0 ekt1,
y2 y(t2) y0 ekt2.
Taking the quotient of the two measurements y1 and y2 eliminates y0 (the initial
amount) and gives a formula for k in terms of these measurements and the
corresponding times, namely,
y2 /y1 exp [k(t 2 t1)],
k
ln (y2/y1)
.
t2 t1
Knowing k, we can now readily determine the half-life H directly from its definition
ekH 0.5.
This gives
H
ln 0.5
t2 t1
(ln 0.5)
.
k
ln (y2/y1)
For the given data we obtain from this formula
H 0.69315
10 5
12.05.
ln (0.015/0.02)
Thus the half-life of the substance is about 12 days and 1 hour.
38. Let y denote the amount of fresh air measured in cubic feet. Then the model is obtained
from the balance equation
“Inflow minus Outflow equals the rate of change”;
that is,
y 600 600
y 600 0.03y.
20000
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The general solution of this linear ODE is
y ce0.03t 20000.
The initial condition is y(0) 0 (initially no fresh air) and gives
y(0) c 20000 0;
c 20000.
hence
The particular solution of our problem is
y 20000(1 e0.03t).
This equals 90% if t is such that
e0.03t 0.1
thus if t (ln 0.1)/(0.03) 77 [min].
40. We use separation of variables. To evaluate the integral, we apply reduction by partial
fractions. This yields
dy
A
B
[
] dy k dx,
(a y)(b y)
ya
yb
where
A
1
ab
B
and
1
A.
ba
By integration,
A[ln y a ln y b] A ln j
ya
j kt c*.
yb
We multiply this on both sides by 1/A a b, obtaining
ln j
ya
j (kt c*)(a b).
yb
We now take exponents. In doing so, we can set c ec* and have
ya
ce(ab)kt.
yb
We denote the right side by E and solve algebraically for y; then
y a (y b)E,
y(1 E) a bE
and from the last expression we finally have
y
a bE
.
1E
42. Let the tangent of such a curve y(x) at (x, y) intersect the x-axis at M and the y-axis
at N, as shown in the figure. Then because of the bisection we have
OM 2x,
ON 2y,
where O is the origin. Since the slope of the tangent is the slope y(x) of the curve,
by the definition of a tangent, we obtain
y ON/OM y/x.
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By separation of variables, integration, and taking exponents, we see that
dx
dy
,
y
x
ln y ln x c*,
xy c.
This is a family of hyperbolas.
N
y
0
(x, y)
x
Section 1.7. Problem 42
M