424320637 solution for a first course in differential equations with modeling applications 11th edition pdf
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SOLUTIONS
Solutions Manual for A First Course in Differential Equations
with Modeling Applications 11th Edition by Zill
CLICK HERE TO ACCESS FULL SOLUTIONS
Chapter 2
FirstOrder Differential Equations
####### 2 Solution Curves Without a Solution
 x
1
1 2
2
x
y
–3 3
3
 x
–10 –5 5 10
0
5
10
x
10 y
0
 x
0
2
4
x
y
–4 –2 0 2 4
 x
–4 –2 0 2 4
0
2
4
x
y
 x
–4 –2 0 2 4
0
2
4
x
y 6. x
–4 –2 0 2 4
0
2
4
x
y
36
38 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
(b) The isoclines have the form x 2 + y 2 = c, which are circles centered at the origin.
y
1
1
2
2
x
 (a) When x = 0 or y = 4, dy/dx = −2 so the lineal elements have slope −2. When y = 3 or y = 5, dy/dx = x − 2, so the lineal elements at (x, 3) and (x, 5) have slopes x − 2.
(b) At (0, y 0 ) the solution curve is headed down. If y → ∞ as x increases, the graph must eventually turn around and head up, but while heading up it can never cross y = 4 where a tangent line to a solution curve must have slope −2. Thus, y cannot approach ∞ as x approaches ∞.
 When y < 12 x 2 , y′ = x 2 − 2 y is positive and the portions of solution curves “outside” the nullcline parabola are increasing. When y > 12 x 2 , y′ = x 2 − 2 y is negative and the portions of the solution curves “inside” the nullcline parabola are decreasing.
–3 –2 –1 0 1 2 3
0
1
2
3
x
y
 (a) Any horizontal lineal element should be at a point on a nullcline. In Problem 1 the nullclines are x 2 − y 2 = 0 or y = ±x. In Problem 3 the nullclines are 1 − xy = 0 or y = 1/x. In Problem 4 the nullclines are (sin x) cos y = 0 or x = nπ and y = π/2 + nπ, where n is an integer. The graphs on the next page show the nullclines for the equations in Problems 1, 3, and 4 superimposed on the corresponding direction field.
 4 –2 0 Problem 3
2 4

4
0
2
4
x
y
–4 –

4
0 2 4 Problem 4
0
2
4
x
y
–3 –2 –1 0 Problem 1
1 2 3
0
1
2
3
x
y
2 Solution Curves Without a Solution 39
(b) An autonomous firstorder differential equation has the form y′ = f (y). Nullclines have the form y = c where f (c) = 0. These are the graphs of the equilibrium solutions of the differential equation.
 Writing the differential equation in the form dy/dx = y(1 − y)(1 + y) we see that critical points are y = −1, y = 0, and y = 1. The phase portrait is shown at the right.
(a) x
1 2 x
5 4 3 2 1
y (b) x
x
1
y
–2 –1 1 2
(c) x –2 –1 1 2 x
y (d) x 1 2 x
y
0
1
 Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that critical points are y = −1, y = 0, and y = 1. The phase portrait is shown at the right.
(a) x
1 2 x
5 4 3 2 1
y (b) x
x
1
y
–2 –1 1 2
(c) x
–2 –1 1 2 x
y (d) x
–2 – –
y x
1
0
1
2 Solution Curves Without a Solution 41
 Solving 10 + 3y − y 2 = (5 − y)(2 + y) = 0 we obtain the critical points −2 and 5. From the phase portrait we see that 5 is asymptotically stable (attractor) and −2 is unstable (repeller).
5
 Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and
 From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is semistable, and −2 is unstable (repeller).
0
2
 Solving y(2 − y)(4 − y) = 0 we obtain the critical points 0, 2, and 4. From the phase portrait we see that 2 is asymptotically stable (attractor) and 0 and 4 are unstable (repellers).
0
2
4
42 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
 Solving y ln(y+2) = 0 we obtain the critical points −1 and 0. From the phase portrait we see that −1 is asymptotically stable (attractor) and 0 is unstable (repeller).
0
 Solving yey − 9 y = y(ey − 9) = 0 (since ey is always positive) we obtain the critical points 0 and ln 9. From the phase portrait we see that 0 is asymptotically stable (attractor) and ln 9 is unstable (repeller).
0
1n 9
 The critical points are 0 and c because the graph of f (y) is 0 at these points. Since f (y) > 0 for y < 0 and y > c, the graph of the solution is increasing on the yintervals (−∞, 0) and (c, ∞). Since f (y) < 0 for 0 < y < c, the graph of the solution is decreasing on the yinterval (0, c).
0
c
x
c
y
44 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
graph of any nonconstant solution must lie entirely on one side of any equilibrium solution. Since f is continuous it can only change signs at a point where it is 0. But this is a critical point. Thus, f (y) is completely positive or completely negative in each region Ri. If y(x) is oscillatory or has a relative extremum, then it must have a horizontal tangent line at some point (x 0 , y 0 ). In this case y 0 would be a critical point of the differential equation, but we saw above that the graph of a nonconstant solution cannot intersect the graph of the equilibrium solution y = y 0.
 By Problem 33, a solution y(x) of dy/dx = f (y) cannot have relative extrema and hence must be monotone. Since y′(x) = f (y) > 0, y(x) is monotone increasing, and since y(x) is bounded above by c 2 , limx→∞ y(x) = L, where L ≤ c 2. We want to show that L = c 2. Since L is a horizontal asymptote of y(x), limx→∞ y′(x) = 0. Using the fact that f (y) is continuous we have
f (L) = f
( x lim→∞ y(x)
) = lim x→∞ f (y(x)) = lim x→∞ y′(x) = 0.
But then L is a critical point of f. Since c 1 < L ≤ c 2 , and f has no critical points between c 1 and c 2 , L = c 2.

Assuming the existence of the second derivative, points of inflection of y(x) occur where y′′(x) = 0. From dy/dx = f (y) we have d 2 y/dx 2 = f ′(y) dy/dx. Thus, the ycoordinate of a point of inflection can be located by solving f ′(y) = 0. (Points where dy/dx = 0 correspond to constant solutions of the differential equation.)

Solving y 2 − y − 6 = (y − 3)(y + 2) = 0 we see that 3 and − 2 are critical points. Now d 2 y/dx 2 = (2y − 1) dy/dx = (2y − 1)(y − 3)(y + 2), so the only possible point of inflection is at y = 12 , although the concavity of solutions can be different on either side of y = −2 and y = 3. Since y′′(x) < 0 for y < −2 and 12 < y < 3, and y′′(x) > 0 for − 2 < y < 12 and y > 3, we see that solution curves are concave down for y < −2 and 12 < y < 3 and concave up for − 2 < y < 12 and y > 3. Points of inflection of solutions of autonomous differential equations will have the same ycoordinates because between critical points they are horizontal translations of each other.
x
y
–5 5
5
 If (1) in the text has no critical points it has no constant solutions. The solutions have neither an upper nor lower bound. Since solutions are monotonic, every solution assumes all real values.
2 Solution Curves Without a Solution 45
 The critical points are 0 and b/a. From the phase portrait we see that 0 is an attractor and b/a is a repeller. Thus, if an initial population satisfies P 0 > b/a, the population becomes unbounded as t increases, most probably in finite time, i. P (t) → ∞ as t → T. If 0 < P 0 < b/a, then the population eventually dies out, that is, P (t) → 0 as t → ∞. Since population P > 0 we do not consider the case P 0 < 0. 0
b a

From the equation dP/dt = k (P − h/k) we see that the only critical point of the autonomous differential equationis the positive number h/k. A phase portrait shows that this point is unstable, that is, h/k is a repeller. For any initial condition P (0) = P 0 for which 0 < P 0 < h/k, dP/dt < 0 which means P (t) is monotonic decreasing and so the graph of P (t) must cross the taxis or the line P − 0 at some time t 1 > 0. But P (t 1 ) = 0 means the population is extinct at time t 1.

Writing the differential equation in the form
dv dt =
k m
( mg k − v
)
we see that a critical point is mg/k. From the phase portrait we see that mg/k is an asymptotically stable critical point. Thus, lim t→∞ v = mg/k.
mg k
 Writing the differential equation in the form
dv dt =
k m
( mg k − v
2 ) = k m
(√ mg k − v
) (√ mg k + v
)
we see that the only physically meaningful critical point is
√ mg/k. From the phase portrait we see that
√ mg/k is an asymptotically stable critical point. Thus, lim t→∞ v =
√ mg/k.
mg √k
 (a) From the phase portrait we see that critical points are α and β. Let X(0) = X 0. If X 0 < α, we see that X → α as t → ∞. If α < X 0 < β, we see that X → α as t → ∞. If X 0 > β, we see that X(t) increases in an unbounded manner, but more specific behavior of X(t) as t → ∞ is not known.
β
α
2 Separable Variables 47

From dy = −e− 3 x dx we obtain y = 13 e− 3 x + c.

From 1 (y − 1) 2 dy = dx we obtain −
1 y − 1 = x + c or y = 1 −
1 x + c.

From 1 y dy = 4 x dx we obtain ln y = 4 ln x + c or y = c 1 x 4.

From 1 y 2 dy = − 2 x dx we obtain − 1 y = −x 2 + c or y = 1 x 2 + c 1 .

From e− 2 y dy = e 3 xdx we obtain 3e− 2 y + 2e 3 x = c.

From yey dy =
( e−x + e− 3 x
) dx we obtain yey − ey + e−x + 1 3 e− 3 x = c.
 From
( y + 2 + 1 y
) dy = x 2 ln x dx we obtain y
2 2 + 2y + ln y =
x 3 3 ln x −
1 9 x
3 + c.
 From 1 (2y + 3) 2 dy =
1 (4x + 5) 2 dx we obtain
2 2 y + 3 =
1 4 x + 5 + c.

From 1 csc y dy = − 1 sec 2 x dx or sin y dy = − cos 2 x dx = − 12 (1 + cos 2x) dx we obtain − cos y = − 12 x − 14 sin 2x + c or 4 cos y = 2x + sin 2x + c 1.

From 2y dy = − sin 3x cos 3 3 x dx or 2y dy = − tan 3x sec
2 3 x dx we obtain y 2 = − 1 6 sec 2 3 x + c.
 From e
y (ey + 1) 2
dy = −e
x (ex + 1) 3
dx we obtain − (ey + 1)− 1 = 12 (ex + 1)− 2 + c.
 From y (1 + y 2 ) 1 / 2
dy = x (1 + x 2 ) 1 / 2
dx we obtain
( 1 + y 2
) 1 / 2 =
( 1 + x 2
) 1 / 2 + c.

From 1 S dS = k dr we obtain S = cekr.

From 1 Q − 70 dQ = k dt we obtain ln Q − 70  = kt + c or Q − 70 = c 1 ekt.

From 1 P − P 2 dP =
( 1 P +
1 1 − P
) dP = dt we obtain ln P  − ln  1 − P  = t + c so that
ln
∣∣ ∣∣ P 1 − P
∣∣ ∣∣ = t + c or P 1 − P = c 1 et. Solving for P we have P = c 1 e
t 1 + c 1 et .
 From N dN 1 =
( tet+2 − 1
) dt we obtain ln N  = tet+2 − et+2 − t + c or N = c 1 etet+2−et+2−t.
 From y − 2 y + 3 dy = x − 1 x + 4 dx or
( 1 − 5 y + 3
) dy =
( 1 − 5 x + 4
) dx we obtain
y − 5 ln y + 3 = x − 5 ln x + 4 + c or
( x + 4 y + 3
) 5 = c 1 ex−y.
48 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
 From y + 1 y − 1 dy = x + 2 x − 3 dx or
( 1 + 2 y − 1
) dy =
( 1 + 5 x − 3
) dx we obtain
y + 2 ln y − 1  = x + 5 ln x − 3  + c or (y − 1)
2 (x − 3) 5 = c 1 e
x−y.
 From x dx = 1 √ 1 − y 2
dy we obtain 12 x 2 = sin− 1 y + c or y = sin
( x 2 2 + c 1
) .
 From 1 y 2 dy = 1 ex + e−x dx = e
x (ex) 2 + 1 dx we obtain − 1 y = tan− 1 ex + c or
y = − 1 tan− 1 ex + c.
 From 1 x 2 + 1 dx = 4 dt we obtain tan
− 1 x = 4t + c. Using x(π/4) = 1 we find c = − 3 π/4. The
solution of the initialvalue problem is tan− 1 x = 4t − 3 π 4 or x = tan
( 4 t − 3 π 4
) .
 From 1 y 2 − 1 dy = 1 x 2 − 1 dx or 1 2
( 1 y − 1 − 1 y + 1
) dy = 1 2
( 1 x − 1 − 1 x + 1
) dx we obtain
ln y − 1  − ln y + 1 = ln x − 1  − ln x + 1 + ln c or y − 1 y + 1 =
c(x − 1) x + 1. Using y(2) = 2 we find c = 1. A solution of the initialvalue problem is y − 1 y + 1 =
x − 1 x + 1 or y = x.
 From 1 y dy = 1 − x x 2 dx =
( 1 x 2 − 1 x
) dx we obtain ln y = − 1 x − ln x = c or xy = c 1 e− 1 /x. Using y(−1) = −1 we find c 1 = e− 1. The solution of the initialvalue problem is xy = e− 1 − 1 /x or y = e−(1+1/x)/x.
 From 1 1 − 2 y dy = dt we obtain −
1 2 ln  1 − 2 y = t + c or 1 − 2 y = c 1 e− 2 t. Using y(0) = 5/2 we find c 1 = −4. The solution of the initialvalue problem is 1 − 2 y = − 4 e− 2 t or y = 2e− 2 t + 12.
 Separating variables and integrating we obtain
√ dx 1 − x 2
− √dy 1 − y 2
= 0 and sin− 1 x − sin− 1 y = c.
Setting x = 0 and y =
√ 3 /2 we obtain c = −π/3. Thus, an implicit solution of the initial value problem is sin− 1 x − sin− 1 y = π/3. Solving for y and using an addition formula from trigonometry, we get
y = sin
( sin− 1 x + π 3
) = x cos π 3 +
√ 1 − x 2 sin π 3 = x 2 +
√ 3
√ 1 − x 2 2 .
50 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
 Separating variables and then proceeding as in Example 5 we get
dy dx = y
2 sin (x 2 )
1 y 2
dy dx = sin (x 2 ) ˆ x
− 2
1 y 2 (t)
dy dt dt =
ˆ x
− 2
sin (t 2 ) dt
− 1 y(t)
∣∣ ∣
x − 2 =
ˆ x − 2
sin (t 2 ) dt
− 1 y(x) + 1 y(−2) =
ˆ x
− 2
sin (t 2 ) dt
− 1 y(x) + 3 =
ˆ x
− 2
sin (t 2 ) dt
y(x) =
[ 3 −
ˆ x
− 2
sin (t 2 ) dt
]− 1
 Separating variables we get dy dx = 2x + 1 2 y 2 y dy = (2x + 1) dx ˆ 2 y dy =
ˆ (2x + 1) dx
y 2 = x 2 + x + c
The condition y(−2) = −1 implies c = −1. Thus y 2 = x 2 + x − 1 and y = −√x 2 + x − 1 in order for y to be negative. Moreover for an interval containing −2 for values of x such that x 2 + x − 1 > 0 we get
( −∞, − 1 2 −
√ 5 2
) .
2 Separable Variables 51
 Separating variables we get
(2y − 2) dy dx = 3x 2 + 4x + 2
(2y − 2) dy =
( 3 x 2 + 4x + 2
) dx ˆ (2y − 2) dy =
ˆ ( 3 x 2 + 4x + 2 ) dx ˆ 2 (y − 1) dy =
ˆ ( 3 x 2 + 4x + 2
) dx
(y − 1) 2 = x 3 + 2x 2 + 2x + c
The condition y(1) = −2 implies c = 4. Thus y = 1 − √x 3 + 2x 2 + 2x + 4 where the minus sign is indicated by the initial condition. Now x 3 + 2x 2 + 2x+ 4 = (x + 2)
( x 2 + 1
) > 0 implies x > −2, so the interval of definition is (− 2 , ∞).
 Separating variables we get
ey dx − e−x dy = 0 ey dx = e−x dy ex dx = e−y dy ˆ ex dx =
ˆ e−y dy
ex = −e−y + c
The condition y(0) = 0 implies c = 2. Thus e−y = 2 − ex. Therefore y = − ln (2 − ex). Now we must have 2 − ex > 0 or ex < 2. Since ex is an increasing function this imples x < ln 2 and so the interval of definition is (−∞, ln 2).
 Separating variables we get
sin x dx + y dy = 0 ˆ sin x dx +
ˆ y dy =
ˆ 0 dx
− cos x + 1 2 y 2 = c
The condition y(0) = 1 implies c = − 12. Thus − cos x + 12 y 2 = − 12 or y 2 = 2 cos x − 1. Therefore y =
√ 2 cos x − 1 where the positive root is indicated by the initial condition. Now we must have 2 cos x − 1 > 0 or cos x > 12. This means −π/ 3 < x < π/3, so the the interval of definition is (−π/ 3 , π/3).
 (a) The equilibrium solutions y(x) = 2 and y(x) = −2 satisfy the initial conditions y(0) = 2
2 Separable Variables 53
(a) Setting x = 0 and y = 1 we have 1 = 1/(1 − 0), which is true for all values of c 1. Thus, solutions passing through (0, 1) are y = 1/(1 − c 1 x).
(b) Setting x = 0 and y = 0 in y = 1/(1 − c 1 x) we get 0 = 1. Thus, the only solution passing through (0, 0) is y = 0.
(c) Setting x = 12 and y = 12 we have 12 = 1/(1 − 12 c 1 ), so c 1 = −2 and y = 1/(1 + 2x).
(d) Setting x = 2 and y = 14 we have 14 = 1/(1 − 2 c 1 ), so c 1 = − 32 and y = 1/(1 + 32 x) = 2/(2 + 3x).
 Singular solutions of dy/dx = x
√ 1 − y 2 are y = −1 and y = 1. A singular solution of (ex + e−x)dy/dx = y 2 is y = 0.
 Differentiating ln (x 2 + 10) + csc y = c we get
2 x x 2 + 10 − csc y cot y dy dx = 0, 2 x x 2 + 10 −
1 sin y ·
cos y sin y
dy dx = 0, or
2 x sin 2 y dx − (x 2 + 10) cos y dy = 0.
Writing the differential equation in the form
dy dx =
2 x sin 2 y (x 2 + 10) cos y
we see that singular solutions occur when sin 2 y = 0, or y = kπ, where k is an integer.
 The singular solution y = 1 satisfies the initialvalue problem.
x
y
1
–0 –0 0 0.
54 CHAPTER 2 FIRSTORDER DIFFERENTIAL EQUATIONS
 Separating variables we obtain dy (y − 1) 2 = dx. Then
− 1 y − 1 = x + c and y = x + c − 1 x + c .
Setting x = 0 and y = 1 we obtain c = −100. The solution is y = x − 101 x − 100 .
–0 –0 0 0. x
1 y
 Separating variables we obtain dy (y − 1) 2 + 0 = dx. Then
10 tan− 1 10(y − 1) = x + c and y = 1 + 1 10 tan x + c 10 .
Setting x = 0 and y = 1 we obtain c = 0. The solution is
y = 1 + 1 10 tan x 10.
–0 –0 0 0. x
y
 Separating variables we obtain dy (y − 1) 2 − 0 .01 = dx. Then, with u = y − 1 and a = 101 , we get
5 ln
∣∣ ∣∣ 10 y − 11 10 y − 9
∣∣ ∣∣ = x + c.
Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The solution is 5 ln
∣∣ ∣∣ 10 y − 11 10 y − 9
∣∣ ∣∣ = x.
–0 –0 0 0 x
y
Solving for y we obtain y = 11 + 9e
x/ 5 10 + 10ex/ 5 .
Alternatively, we can use the fact that ˆ dy (y − 1) 2 − 0 .01 = − 1 0 .1 tanh
− 1 y − 1 0. 1 = −10 tanh− 1 10(y − 1).
(We use the inverse hyperbolic tangent because y − 1  < 0. 1 or 0. 9 < y < 1 .1. This follows from the initial condition y(0) = 1.) Solving the above equation for y we get y = 1 + 0 tanh (x/10).
424320637 solution for a first course in differential equations with modeling applications 11th edition pdf