Overview of Differential Equations
Overview of Differential Equations

424320637 solution for a first course in differential equations with modeling applications 11th edition pdf

University

Government College University Lahore

Course

Preview text

CLICK HERE TO ACCESS FULL SOLUTIONS

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

First-Order Differential Equations

####### 2 Solution Curves Without a Solution

  1. x

1

1 2

2

x

y

–3 3

3

  1. x

–10 –5 5 10

0

5

10

x

10 y

0

  1. x

0

2

4

x

y

–4 –2 0 2 4

  1. x

–4 –2 0 2 4

0

2

4

x

y

  1. 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 FIRST-ORDER 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

  1. (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 ∞.

  1. 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

  1. (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 first-order 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.

  1. 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

  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

  1. 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

  1. Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and
    1. From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is semi-stable, and −2 is unstable (repeller).

0

2

  1. 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 FIRST-ORDER DIFFERENTIAL EQUATIONS

  1. 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

  1. 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

  1. 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 y-intervals (−∞, 0) and (c, ∞). Since f (y) < 0 for 0 < y < c, the graph of the solution is decreasing on the y-interval (0, c).

0

c

x

c

y

44 CHAPTER 2 FIRST-ORDER 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.

  1. 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.

  1. 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 y-coordinate 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.)

  2. 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 y-coordinates because between critical points they are horizontal translations of each other.

x

y

–5 5

5

  1. 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

  1. 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

  1. 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 t-axis 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.

  2. 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

  1. 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

  1. (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

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

  2. From 1 (y − 1) 2 dy = dx we obtain −

1 y − 1 = x + c or y = 1 −

1 x + c.

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

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

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

  4. From yey dy =

( e−x + e− 3 x

) dx we obtain yey − ey + e−x + 1 3 e− 3 x = c.

  1. 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.

  1. From 1 (2y + 3) 2 dy =

1 (4x + 5) 2 dx we obtain

2 2 y + 3 =

1 4 x + 5 + c.

  1. 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.

  2. 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.

  1. From e

y (ey + 1) 2

dy = −e

x (ex + 1) 3

dx we obtain − (ey + 1)− 1 = 12 (ex + 1)− 2 + c.

  1. 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.

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

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

  3. 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 .

  1. 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.

  1. 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 FIRST-ORDER DIFFERENTIAL EQUATIONS

  1. 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.

  1. 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

) .

  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.

  1. 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 initial-value problem is tan− 1 x = 4t − 3 π 4 or x = tan

( 4 t − 3 π 4

) .

  1. 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 initial-value problem is y − 1 y + 1 =

x − 1 x + 1 or y = x.

  1. 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 initial-value problem is xy = e− 1 − 1 /x or y = e−(1+1/x)/x.

  1. 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 initial-value problem is 1 − 2 y = − 4 e− 2 t or y = 2e− 2 t + 12.

  1. 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 FIRST-ORDER DIFFERENTIAL EQUATIONS

  1. 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

  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

  1. 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 , ∞).

  1. 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).

  1. 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).

  1. (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).

  1. 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.

  1. 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.

  1. The singular solution y = 1 satisfies the initial-value problem.

x

y

1

–0 –0 0 0.

54 CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS

  1. 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

  1. 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

  1. 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

You are watching: 424320637 solution for a first course in differential equations with modeling applications 11th. Info created by Bút Chì Xanh selection and synthesis along with other related topics.