**Unformatted text preview: **SIGUENOS EN: LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS GRATIS EN
DESCARGA DIRECTA
VISITANOS PARA DESARGALOS GRATIS. Table of Contents
Part I Ordinary Differential Equations
1 Introduction to Diﬀerential Equations 1 2 First-Order Diﬀerential Equations 22 3 Higher-Order Diﬀerential Equations 99 4 The Laplace Transform 198 5 Series Solutions of Linear Diﬀerential Equations 252 6 Numerical Solutions of Ordinary Diﬀerential Equations 317 Part II Vectors, Matrices, and Vector Calculus
7 Vectors 339 8 Matrices 373 9 Vector Calculus 438 Part III Systems of Differential Equations
10 Systems of Linear Diﬀerential Equations 551 11 Systems of Nonlinear Diﬀerential Equations 604 Part IV Fourier Series and Partial Differential Equations
12 Orthogonal Functions and Fourier Series 634 13 Boundary-Value Problems in Rectangular Coordinates 680 14 Boundary-Value Problems in Other Coordinate Systems 755 15 Integral Transform Method 793 16 Numerical Solutions of Partial Diﬀerential Equations 832 Part V Complex Analysis
17 Functions of a Complex Variable 854 18 Integration in the Complex Plane 877 19 Series and Residues 896 20 Conformal Mappings 919 Appendices
Appendix II Gamma function 942 Projects
3.7 Road Mirages 944 3.10 The Ballistic Pendulum 946 8.1 Two-Ports in Electrical Circuits 947 8.2 Traﬃc Flow 948 8.15 Temperature Dependence of Resistivity 949 9.16 Minimal Surfaces 950 14.3 The Hydrogen Atom 952 15.4 The Uncertainity Inequality in Signal Processing 955 15.4 Fraunhofer Diﬀraction by a Circular Aperture 958 16.2 Instabilities of Numerical Methods 960 Part I Ordinary Differential Equations Introduction to
Differential Equations 1 EXERCISES 1.1
Definitions and Terminology 1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2 6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of x˙ 2
9. Writing the diﬀerential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 .
However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the diﬀerential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u.
11. From y = e−x/2 we obtain y = − 12 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0.
12. From y = 6
5 − 65 e−20t we obtain dy/dt = 24e−20t , so that
dy
+ 20y = 24e−20t + 20
dt 6 6 −20t
− e
5 5
= 24. 13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y = 5e3x cos 2x − 12e3x sin 2x, so that
y − 6y + 13y = 0.
14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y = 1 + 2(x + 2)−1/2 we have
(y − x)y = (y − x)[1 + (2(x + 2)−1/2 ]
= y − x + 2(y − x)(x + 2)−1/2
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8. 1 1.1 Deﬁnitions and Terminology
An interval of deﬁnition for the solution of the diﬀerential equation is (−2, ∞) because y is not deﬁned at
x = −2. 16. Since tan x is not deﬁned for x = π/2 + nπ, n an integer, the domain of y
{x 5x = π/2 + nπ} or {x x =
π/10 + nπ/5}. From y = 25 sec2 5x we have = 5 tan 5x is y = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .
An interval of deﬁnition for the solution of the diﬀerential equation is (−π/10, π/10). Another interval is
(π/10, 3π/10), and so on.
17. The domain of the function is {x 4 − x2 = 0} or {x x = −2 or x = 2}. From y = 2x/(4 − x2 )2 we have
y = 2x 1
4 − x2 2
= 2xy. An interval of deﬁnition for the solution of the diﬀerential equation is (−2, 2). Other intervals are (−∞, −2)
and (2, ∞).
√
18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain
is {x x = π/2 + 2nπ}. From y = − 12 (1 − sin x)−3/2 (− cos x) we have
2y = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.
An interval of deﬁnition for the solution of the diﬀerential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2),
and so on.
19. Writing ln(2X −1)−ln(X −1) = t and diﬀerentiating implicitly we obtain X 2
dX
1 dX
−
=1
2X − 1 dt
X − 1 dt
2
dX
1
−
=1
2X − 1 X − 1 dt 4
2 2X − 2 − 2X + 1 dX
=1
(2X − 1)(X − 1) dt -4 dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt
Exponentiating both sides of the implicit solution we obtain -2 2 4 t -2
-4 2X − 1
= et
X −1
2X − 1 = Xet − et
(et − 1) = (et − 2)X
X= et − 1
.
et − 2 Solving et − 2 = 0 we get t = ln 2. Thus, the solution is deﬁned on (−∞, ln 2) or on (ln 2, ∞). The graph of the
solution deﬁned on (−∞, ln 2) is dashed, and the graph of the solution deﬁned on (ln 2, ∞) is solid. 2 1.1 Deﬁnitions and Terminology 20. Implicitly diﬀerentiating the solution, we obtain y dy
dy
−2x2
− 4xy + 2y
=0
dx
dx
−x2 dy − 2xy dx + y dy = 0 4
2 2xy dx + (x2 − y)dy = 0.
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get
√
√
y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions
√
√
are y1 = x2 + x4 + 1 and y2 = x2 − x4 + 1 . Both solutions are deﬁned
on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -4 -2 2 4 x -2
-4 21. Diﬀerentiating P = c1 et / (1 + c1 et ) we obtain
dP
(1 + c1 et ) c1 et − c1 et · c1 et
c1 et [(1 + c1 et ) − c1 et ]
=
=
2
dt
1 + c1 et
1 + c1 et
(1 + c1 et )
c1 et
c1 et
1
−
= P (1 − P ).
=
1 + c1 et
1 + c1 et x
2
2
2
22. Diﬀerentiating y = e−x
et dt + c1 e−x we obtain
0
−x2 x2 y =e e −x2 x − 2xe −x2 t2 e dt − 2c1 xe −x2 x = 1 − 2xe 0 et dt − 2c1 xe−x .
2 2 0 Substituting into the diﬀerential equation, we have x
−x2
t2
−x2
−x2
e dt − 2c1 xe
+ 2xe
y + 2xy = 1 − 2xe
0 23. From y = c1 e2x + c2 xe2x we obtain x et dt + 2c1 xe−x = 1.
2 2 0 dy
d2 y
= (2c1 + c2 )e2x + 2c2 xe2x and
= (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx
dx2 d2 y
dy
+ 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
−4
2
dx
dx
24. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2 and d3 y
= −6c1 x−4 − c3 x−2 ,
dx3 so that
x3 d3 y
d2 y
dy
+ y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
+ 2x2 2 − x
3
dx
dx
dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
= 12x2 . 25. From y = −x2 , x < 0
we obtain y =
x2 ,
x≥0 −2x, x < 0
so that xy − 2y = 0.
2x,
x≥0 3 1.1 Deﬁnitions and Terminology 26. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y (x) does not
x→0− x→0+ exist at x = 0.
27. (a) From y = emx we obtain y = memx . Then y + 2y = 0 implies
memx + 2emx = (m + 2)emx = 0.
Since emx > 0 for all x, m = −2. Thus y = e−2x is a solution.
(b) From y = emx we obtain y = memx and y = m2 emx . Then y − 5y + 6y = 0 implies
m2 emx − 5memx + 6emx = (m − 2)(m − 3)emx = 0.
Since emx > 0 for all x, m = 2 and m = 3. Thus y = e2x and y = e3x are solutions.
28. (a) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then xy + 2y = 0 implies
xm(m − 1)xm−2 + 2mxm−1 = [m(m − 1) + 2m]xm−1 = (m2 + m)xm−1
= m(m + 1)xm−1 = 0.
Since xm−1 > 0 for x > 0, m = 0 and m = −1. Thus y = 1 and y = x−1 are solutions.
(b) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then x2 y − 7xy + 15y = 0 implies
x2 m(m − 1)xm−2 − 7xmxm−1 + 15xm = [m(m − 1) − 7m + 15]xm
= (m2 − 8m + 15)xm = (m − 3)(m − 5)xm = 0.
Since xm > 0 for x > 0, m = 3 and m = 5. Thus y = x3 and y = x5 are solutions.
In Problems 29–32, we substitute y = c into the diﬀerential equations and use y = 0 and y = 0
29. Solving 5c = 10 we see that y = 2 is a constant solution.
30. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
31. Since 1/(c − 1) = 0 has no solutions, the diﬀerential equation has no constant solutions.
32. Solving 6c = 10 we see that y = 5/3 is a constant solution.
33. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain
dx
= −2e−2t + 18e6t
dt and dy
= 2e−2t + 30e6t .
dt Then
x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t = dx
dt 5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t = dy
.
dt and 34. From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain and dx
1
= −2 sin 2t + 2 cos 2t + et
dt
5 and dy
1
= 2 sin 2t − 2 cos 2t − et
dt
5 d2 x
1
= −4 cos 2t − 4 sin 2t + et
2
dt
5 and d2 y
1
= 4 cos 2t + 4 sin 2t − et .
2
dt
5 Then and 1
1
d2 x
4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2
5
5
dt 4 1.1 Deﬁnitions and Terminology 1
1
d2 y
4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 .
5
5
dt
35. (y )2 + 1 = 0 has no real solutions because (y )2 + 1 is positive for all functions y = φ(x).
36. The only solution of (y )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y )2 + y 2 ≥ y 2 > 0.
37. The ﬁrst derivative of f (x) = ex is ex . The ﬁrst derivative of f (x) = ekx is kekx . The diﬀerential equations are
y = y and y = ky, respectively.
38. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding diﬀerential
equation is y − y = 0. Functions of the form y = c sin x or y = c cos x have second derivatives that are the
negatives of themselves. The diﬀerential equation is y + y = 0.
√
39. We ﬁrst note that 1 − y 2 = 1 − sin2 x = cos2 x = | cos x|. This prompts us to consider values of x for
which cos x < 0, such as x = π. In this case
dy
d
=
= cos xx=π = cos π = −1,
(sin x)
dx
dx
x=π x=π but
√
1 − y 2 |x=π = 1 − sin2 π = 1 = 1.
Thus, y = sin x will only be a solution of y = 1 − y 2 when cos x > 0. An interval of deﬁnition is then
(−π/2, π/2). Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on.
40. Since the ﬁrst and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that a linear
combination of these functions, A sin t + B cos t, could be a solution of the diﬀerential equation. Using y =
A cos t − B sin t and y = −A sin t − B cos t and substituting into the diﬀerential equation we get
y + 2y + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t
= (3A − 2B) sin t + (2A + 3B) cos t = 5 sin t.
Thus 3A − 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we ﬁnd A =
10
particular solution is y = 15
13 sin t − 13 cos t. 15
13 and B = − 10
13 . A 41. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solution
is given by the lower portion of the graph, also with domain approximately (0, 2.6).
42. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together
with the lower part of the graph in the ﬁrst quadrant. A second solution, with domain approximately (0, 1.6)
is the upper part of the graph in the ﬁrst quadrant. The third solution, with domain (0, ∞), is the part of the
graph in the fourth quadrant.
43. Diﬀerentiating (x3 + y 3 )/xy = 3c we obtain
xy(3x2 + 3y 2 y ) − (x3 + y 3 )(xy + y)
=0
x2 y 2
3x3 y + 3xy 3 y − x4 y − x3 y − xy 3 y − y 4 = 0
(3xy 3 − x4 − xy 3 )y = −3x3 y + x3 y + y 4
y = y 4 − 2x3 y
y(y 3 − 2x3 )
.
=
2xy 3 − x4
x(2y 3 − x3 ) 44. A tangent line will be vertical where y is undeﬁned, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0
and 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get 5 1.1 Deﬁnitions and Terminology
1
1
x3 + x3 = 3x
x
2
21/3
3 3
3
x = 1/3 x2
2
2
x3 = 22/3 x2
x2 (x − 22/3 ) = 0.
Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ). Since 22/3 ≈ 1.59, the
estimates of the domains in Problem 42 were close.
√
√
45. The derivatives of the functions are φ1 (x) = −x/ 25 − x2 and φ2 (x) = x/ 25 − x2 , neither of which is deﬁned
at x = ±5.
46. To determine if a solution curve passes through (0, 3) we let t = 0 and P = 3 in the equation P = c1 et /(1+c1 et ).
This gives 3 = c1 /(1 + c1 ) or c1 = − 32 . Thus, the solution curve
P = −3et
(−3/2)et
=
1 − (3/2)et
2 − 3et passes through the point (0, 3). Similarly, letting t = 0 and P = 1 in the equation for the one-parameter family
of solutions gives 1 = c1 /(1 + c1 ) or c1 = 1 + c1 . Since this equation has no solution, no solution curve passes
through (0, 1).
47. For the ﬁrst-order diﬀerential equation integrate f (x). For the second-order diﬀerential equation integrate twice. In the latter case we get y = ( f (x)dx)dx + c1 x + c2 .
48. Solving for y using the quadratic formula we obtain the two diﬀerential equations
1 1 2 + 2 1 + 3x6
2 − 2 1 + 3x6 ,
and y =
y =
x
x
so the diﬀerential equation cannot be put in the form dy/dx = f (x, y).
49. The diﬀerential equation yy − xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a
solution of the ﬁrst diﬀerential equation but not a solution of the second.
50. Diﬀerentiating we get y = c1 + 3c2 x2 and y = 6c2 x. Then c2 = y /6x
y
xy
y= y −
x+
x3 = xy −
2
6x and c1 = y − xy /2, so
1 2
x y
3 and the diﬀerential equation is x2 y − 3xy + 3y = 0.
51. (a) Since e−x is positive for all values of x, dy/dx > 0 for all x, and a solution, y(x), of the diﬀerential equation
must be increasing on any interval.
2
2
dy
dy
(b) lim
= lim e−x = 0 and lim
= lim e−x = 0. Since dy/dx approaches 0 as x approaches −∞
x→−∞ dx
x→−∞
x→∞ dx
x→∞
and ∞, the solution curve has horizontal asymptotes to the left and to the right.
2 (c) To test concavity we consider the second derivative
2
d2 y
d −x2 d dy
= −2xe−x .
=
e
=
dx2
dx dx
dx
Since the second derivative is positive for x < 0 and negative for x > 0, the solution curve is concave up on
(−∞, 0) and concave down on (0, ∞). 6 1.1
(d) Deﬁnitions and Terminology y x 52. (a) The derivative of a constant solution y = c is 0, so solving 5 − c = 0 we see that c = 5 and so y = 5 is a
constant solution.
(b) A solution is increasing where dy/dx = 5−y > 0 or y < 5. A solution is decreasing where dy/dx = 5−y < 0
or y > 5.
53. (a) The derivative of a constant solution is 0, so solving y(a − by) = 0 we see that y = 0 and y = a/b are
constant solutions.
(b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > 0 or 0 < y < a/b. A solution is decreasing
where dy/dx = by(a/b − y) < 0 or y < 0 or y > a/b.
(c) Using implicit diﬀerentiation we compute
d2 y
= y(−by ) + y (a − by) = y (a − 2by).
dx2
Solving d2 y/dx2 = 0 we obtain y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0 for
a/2b < y < a/b, the graph of y = φ(x) has a point of inﬂection at y = a/2b.
(d) y y=aêb y=0
x 54. (a) If y = c is a constant solution then y = 0, but c2 + 4 is never 0 for any real value of c.
(b) Since y = y 2 + 4 > 0 for all x where a solution y = φ(x) is deﬁned, any solution must be increasing on any
interval on which it is deﬁned. Thus it cannot have any relative extrema.
(c) Using implicit diﬀerentiation we compute d2 y/dx2 = 2yy = 2y(y 2 + 4). Setting d2 y/dx2 = 0 we see that
y = 0 corresponds to the only possible point of inﬂection. Since d2 y/dx2 < 0 for y < 0 and d2 y/dx2 > 0
for y > 0, there is a point of inﬂection where y = 0. 7 1.1 Deﬁnitions and Terminology (d) y x 55. In Mathematica use
Clear[y]
y[x ]:= x Exp[5x] Cos[2x]
y[x]
y''''[x] − 20y'''[x] + 158y''[x] − 580y'[x] +841y[x]//Simplify
The output will show y(x) = e5x x cos 2x, which veriﬁes that the correct function was entered, and 0, which
veriﬁes that this function is a solution of the diﬀerential equation.
56. In Mathematica use
Clear[y]
y[x ]:= 20Cos[5Log[x]]/x − 3Sin[5Log[x]]/x
y[x]
xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x] − 78y[x]//Simplify
The output will show y(x) = 20 cos(5 ln x)/x − 3 sin(5 ln x)/x, which veriﬁes that the correct function was
entered, and 0, which veriﬁes that this function is a solution of the diﬀerential equation. EXERCISES 1.2
Initial-Value Problems
1. Solving −1/3 = 1/(1 + c1 ) we get c1 = −4. The solution is y = 1/(1 − 4e−x ).
2. Solving 2 = 1/(1 + c1 e) we get c1 = −(1/2)e−1 . The solution is y = 2/(2 − e−(x+1) ) .
3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = −1. The solution is y = 1/(x2 − 1). This solution is
deﬁned on the interval (1, ∞).
4. Letting x = −2 and solving 1/2 = 1/(4 + c) we get c = −2. The solution is y = 1/(x2 − 2). This solution is
√
deﬁned on the interval (−∞, − 2 ).
5. Letting x = 0 and solving 1 = 1/c we get c = 1. The solution is y = 1/(x2 + 1). This solution is deﬁned on the
interval (−∞, ∞). 8 1.2 Initial-Value Problems
6. Letting x = 1/2 and solving −4 = 1/(1/4 + c) we get c = −1/2. The solution is y = 1/(x2 − 1/2) = 2/(2x2 − 1).
√
√
This solution is deﬁned on the interval (−1/ 2 , 1/ 2 ).
In Problems 7–10, we use x = c1 cos t + c2 sin t and x = −c1 sin t + c2 cos t to obtain a system of two equations in
the two unknowns c1 and c2 .
7. From the initial conditions we obtain the system
c1 = −1
c2 = 8.
The solution of the initial-value problem is x = − cos t + 8 sin t.
8. From the initial conditions we obtain the system
c2 = 0
−c1 = 1.
The solution of the initial-value problem is x = − cos t.
9. From the initial conditions we obtain √ 3
1
1
c1 + c2 =
2
2
2
√
3
1
c2 = 0.
− c1 +
2
2 Solving, we ﬁnd c1 = √
3/4 and c2 = 1/4. The solution of the initial-value problem is
√
x = ( 3/4) cos t + (1/4) sin t. 10. From the initial conditions we obtain √
√
2
2
c1 +
c2 = 2
2
2
√
√
√
2
2
c1 +
c2 = 2 2 .
−
2
2
Solving, we ﬁnd c1 = −1 and c2 = 3. The solution of the initial-value problem is x = − cos t + 3 sin t.
√ In Problems 11–14, we use y = c1 ex + c2 e−x and y = c1 ex − c2 e−x to obtain a system of two equations in the two
unknowns c1 and c2 .
11. From the initial conditions we obtain
c1 + c2 = 1
c1 − c2 = 2.
Solving, we ﬁnd c1 = 3
2 and c2 = − 12 . The solution of the initial-value problem is y = 32 ex − 12 e−x . 12. From the initial conditions we obtain ec1 + e−1 c2 = 0
ec1 − e−1 c2 = e. Solving, we ﬁnd c1 = 1
2 and c2 = − 12 e2 . The solution of the initial-value problem is
y= 13. From the initial conditions we obtain 1 x 1 2 −x
1
1
e − e e = ex − e2−x .
2
2
2
2
e−1 c1 + ec2 = 5
e−1 c1 − ec2 = −5. 9 1.2 Initial-Value Problems
Solving, we ﬁnd c1 = 0 and c2 = 5e−1 . The solution of the initial-value problem is y = 5e−1 e−x = 5e−1−x . 14. From the initial conditions we obtain
c1 + c2 = 0
c1 − c2 = 0.
Solving, we ﬁnd c1 = c2 = 0. The solution of the initial-value problem is y = 0.
15. Two solutions are y = 0 and y = x3 .
16. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.)
∂f
2
= y −1/3 . Thus, the diﬀerential equation will have a unique solution in any
∂y
3
rectangular region of the plane where y = 0.
√
18. For f (x, y) = xy we have ∂f /∂y = 12 x/y . Thus, the diﬀerential equation will have a unique solution in any
17. For f (x, y) = y 2/3 we have region where x > 0 and y > 0 or where x < 0 and y < 0.
19. For f (x, y) =
where x = 0. ∂f
1
y
we have
= . Thus, the diﬀerential equation will have a unique solution in any region
x
∂y
x 20. For f (x, y) = x + y we have ∂f
= 1. Thus, the diﬀerential equation will have a unique solution in the entire
∂y plane.
21. For f (x, y) = x2 /(4 − y 2 ) we have ∂f /∂y = 2x2 y/(4 − y 2 )2 . Thus the diﬀerential equation will have a unique
solution in any region where y < −2, −2 < y < 2, or y > 2.
x2
−3x2 y 2
∂f
=
we
have
2 . Thus, the diﬀerential equation will have a unique solution in
1 + y3
∂y
(1 + y 3 )
any region where y = −1. 22. For f (x, y) = y2
2x2 y
∂f
=
we
have
2 . Thus, the diﬀerential equation will have a unique solution in
x2 + y 2
∂y
(x2 + y 2 )
any region not containing (0, 0). 23. For f (x, y) = 24. For f (x, y) = (y + x)/(y − x) we have ∂f /∂y = −2x/(y − x)2 . Thus the diﬀerential equation will have a unique
solution in any region where y < x or where y > x.
y 2 ...

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