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#### 1)Consider the graph   2 − 2 − 4 2 4 6 − 2 x y   Using the above graph of y(x), choose the equation whose solution satisfies the initial condition y(0) = 3

###### Math

1)Consider the graph

 2
 −
 2
 −
 4
 2
 4
 6
 −
 2
 x
 y

Using the above graph of y(x), choose the equation whose solution satisfies the initial condition y(0) = 3.5.

1. y= y3 x3
2. y= y − 1
3. y= y x
4. y= y2 x2
5. y= y x2

2. From this, we can see that the given graph is indeed a solution to the differential equation y= y x.

Consider the graph

 2
 −
 2
 −
 4
 2
 4
 6
 −
 2
 x
 y

Using the above graph of y(x), choose the equation whose solution is graphed and satisfies the initial condition y(0) = 1.5.

1. y= y x − 1
2. y= y − 1
3. y= −y + 2
4. y= y3 x2
5. y= y2 x
6. y= y x
7. y= y2 x2

 4
 −
 2
 2
 −
 2
 2
 x
 y
3. From this, we can see that the given graph is indeed a solution to the differential equation y= y − 1.

Use the direction field of the differential equation y= y3 to sketch a solution curve that passes through the point (0, 1).

 −
 2
 2
 −
 2
 2
 x
 y
 1
 −
 2
 2
 −
 2
 2
 x
 y
 2
 −
 2
 2
 −
 2
 2
 x
 y
5.

 6
 −
 2
 2
 −
 2
 2
 x
 y

:

1. Choose the equation whose solution is graphed and satisfies the initial condition y(0) = 1.4.
1. y= xsin(2y)
2. y= y sin(4x)
3. y= y cos(2x)
4. y= −y sin(2x)
5. y= y sin(2x)
6. y= y sin(x)

1. As we can see from the above plot, our graph of y is indeed a solution to the differential equation y= y sin(2x).

A solution y(t) of the differential equation

y= y3 − 4y

satisfies the initial condition y(0) = 1. Find lim y(t). t→−∞

1. Doesn’t exist.
2. −2
3. 2
4. −∞
5. 0

.

1. Use Euler’s method with step size 0.5 to compute the approximate y-values y1, y2, and y3 of the solution of the initial-value problem

y= 1 + 5x − 3y, y(1) = 2.

1. y1 = 2.000, y2 = 3.350, y3 = 3.945
2. y1 = 1.900, y2 = 2.860, y3 = 3.602
3. y1 = 2.000, y2 = 3.220, y3 = 3.663
4. y1 = 2.000, y2 = 3.020, y3 = 3.966
5. y1 = 2.000, y2 = 3.250, y3 = 3.875 cor-

rect

In Euler’s Method, we predict the value of yi using, and dx with the following equation yi = yi−1 + yi1 dx.

Recall that y(x,y) = 1 + 5x − 3y and y(1) = 2. Hence, in this problem 2) = 0 and dx = 0.5.

In order to visualize this better, let us use the following table to display our values

 i x y y′ 0 1.0 2.000 0.000 1 1.5 2.000 2.500 2 2.0 3.250 1.250 3 2.5 3.875 1.875

Notice in this table, y is computed in each row using the values for y and yin the previous row while yin each row is determined using y and x in the current row. Hence,

 y1 = 2.000, y2 = 3.250, y3 = 3.875

.

1. Use Euler’s method with step size 0.3 to estimate y(0.9), where y(x) is the solution of the initial-value problem y= x2 y2, y(1) = 1.
1. y(0.9) ≈ 11.219
2. y(0.9) ≈ 2.485
3. y(0.9) ≈ −6.492
4. y(0.9) ≈ 1.538
5. y(0.9) ≈ −8.384

.

1. Use Euler’s method with step size 0.3 to estimate y(0.9), where y(x) is the solution of the initial-value problem , y(1) = 3.
1. y(0.9) ≈ 12.083
2. y(0.9) ≈ 8.305
3. y(0.9) ≈ 5.76
4. y(0.9) ≈ 12.236
5. y(0.9) ≈ 7.138

.

1. For the differential equation

,

find its general solution.

1.

2.

3.

4.

5.

1. Find the particular solution y0 such that

3.

4.

5.

1. For the particular solution y0 in (ii), find the value of y0(3).
1. y0(3) = 901/3
2. y0(3) = 124−1/4
3. y0(3) = 1251/4
4. y0(3) = 89−1/3
5. y0(3) = 116−1/3

1. If y0 satisfies the equations

,

find the value of y0(2).

1. y0(2) = 521/5
2. y0(2) = 32−1/4
3. y0(2) = 361/4
4. y0(2) = 01/4
5. y0(2) = 121/5

1. For the differential equation

,

find the general solution.

1. y = (C − 3lnx)1/5
2. y = (C + 3lnx)1/6
3. y = (C − 3lnx)1/6
4. y = (C + 3lnx)1/5
5. y = (C + 4lnx)1/6

1. Then find the particular solution y1 such that y1(1) = 2.
1. y = (25 − 3lnx)1/5
2. y = (26 + 3lnx)1/6
3. y = (25 + 3lnx)1/5
4. y = (24 + 3lnx)1/6
5. y = (26 − 3lnx)1/6

1. For the particular solution y1 in part 2, determine the value of y1(e).
1. y1(e) = 641/5
2. y1(e) = 631/6
3. y1(e) = 621/5
4. y1(e) = 611/6
5. y1(e) = 651/6

1. If y = y0(x) satisfies the equations

dy           x+3,        y(−3) = 0,

2y= e dx

and y > 0, find the value of y0(0).

1. y0(0) = 2(e3 − 1)1/2
2. y0(0) = (e3 + 1)1/2
3. y0(0) = 2(e3 + 1)1/2
4. y0(0) = (e3 − 1)1/2
5. y0(0) = (e4 − 1)1/2

1. Consider the differential equation

,

Find its general solution.

1.

2. 3.

4.

18. Find the particular solution y1 such that y1(8) = 5.

1                         2

1. y1(x) = 5 − (x + 4)(x − 8)

9

1                         2

1. y1(x) = 5 + (x + 4)(x − 8)

3

4.

5.

1. For the particular solution y1 in part 2, determine the value of y1(4).

83

1. y1(4) =

9

83

1. y1(4) = −

9

113

1. y1(4) = −

3

143

1. y1(4) =

3

.

1. If y = y0(x) is the solutionof the differential equation

dy                       2),

y= x(36 + y dx

which satisfies y(0) = 0, find the value of y0(1).

1.

2.

3.

4.

5.

1. If y0 is the particular solution of the differential equation

dy

+ 2xy + 4x = 0,

dx such that y(0) = 3, find the value of

1.

2.

3.

4.

5.

1. For the differential equation

,

(i) First find the general solution.

1      x+1 + A)

1. y =(e

7

1. y = 7(ex+1 + A)−1/2
2. y = 7(ex+1 + A) 4. y = 7(ex+1 + A)1/2

5.

1. 23. Then find the particular solution y1 such that y(−1) = 0.

1               1    x+1 − 1)1/2

1. y (x) =(e

7

1. y1(x) = 7(ex+1 − 1)

1

1. y1(x) = (ex+1 − 1)

7

1. y1(x) = 7(ex+1 − 1)1/2
2. y1(x) = 7(ex+1 − 1)−1/2

1. 24. For the particular solution y1 in (ii), determine the value of y1(0).

1.

1. y1(0) = 7(e − 1)1/2
2. y1(0) = 7(e − 1)

1

1. y1(0) = (e − 1)

7

1. y1(0) = 7(e − 1)−1/2

25. For the differential equation

(i) first find its general solution.

1.

2.

3.

1.
2. y = 4lnx + 4(lnx)2 + C

1. 26. Then find the particular solution y1 such that y1(e) = 6.

2.

3.

4.

5.

1. 27. For the particular solution y1 in (ii), determine the value of y1(1).
1. y1(1) = 441/2
2. y1(1) = 301/2
3. y1(1) = 421/2
4. y1(1) = 381/2
5. y1(1) = 291/2

28. If y1 satisfies the equations

dy 2 lnx, y(1) = e, 2xy= 3 + 16x dx

determine the value of y1(e).

1. y1(e) = 7 + 5e2
2. y1(e) = (7 + 5e2)1/2
3. y1(e) = (11 + 5e2)1/2
4. y1(e) = 7 − 5e2
5. y1(e) = (11 − 5e2)1/2

29. If y1 satisfies the equations

1

for x > , determine the value of y1(e).

2

1

1. y1(e) =

13

1

1. y1(e) =

12

1

1. y1(e) =

9

1

1. y1(e) =

11

30. A differentiable function f has the property that

holds for all x, y in the domain of f.               It is

known also that

.

(i) Use the definition of the derivative of f to determine f(x) in terms of f(x).

1. f(x) = 1 − f(x)2
2. f(x) = 1 + f(x)2
3. f(x) = 6f(x)2
4. f(x) = 6(1 + f(x)2)
5. f(x) = 6(1 − f(x)2)

31.  (ii) By solving the differential equation in Part

(i) for f, find the value of

1.

2.

3.

4.

5.

32. An initial deposit of \$P is made into an account that earns 5% interest compounded continuously. Money is then withdrawn at a constant rate of \$3000 per year.

Set up the differential equation for the amount A = A(t) (in thousands of dollars) in the account after t years.

dA

1.  = 5A + 3 dt dA
2.  = 5A − 3 dt dA
3.  = 0.05A + 3 dt dA
4.  = A − 3 dt
5.

33. Solve the differential equation in part 1.

1.

2.

3.

4.

5.

 A
 (
 t
 )=60+
 P
 1000
 −
 60
 e
 0
 .
 05
 t
 .

34. If the account balance becomes zero after 10 years, what was the amount of the initial deposit \$P? (This value of P is often called the Present Value of the regular withdrawals \$3000 over a period of n years.)

1. \$25383
2. \$21742
3. \$23608
4. \$27071
5. \$28677

35. In a West Texas school district the school year began on August 1 and lasted until May

31. On August 1 a Soft Drink company installed soda machines in the school cafeteria. It found that after t months the machines generated income at a rate of

dollars per month. Find the total income produced during the first semester ending on December 31.

1. income = \$287.16
2. income = \$327.16
3. income = \$267.16
4. income = \$347.16
5. income = \$307.16

36. A flea collar for dogs contains an active ingredient that evaporates at a rate proportional to the amount still present. Half of the ingredient evaporates in the first 18 days after the collar is removed from the protective packaging.

If the collar becomes ineffective after 60% of the active ingredient has evaporated, how many days does the collar remain effective?

1. Days effective = 24.79
2. Days effective = 22.79
3. Days effective = 23.79
4. Days effective = 20.79
5. Days effective = 21.79

37.  A population is modeled by the differential equation.

For what values of P is the population increasing?

1. P > 2000
2. 0 < P < 1.1
3. 0 < P < 4000
4. P > 4000
5. P > 1.1

38. A common inhabitant of human intestines is the bacterium Escherichiacoli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes, and its initial population is 65 cells.

Find the number of cells after 4 hours.

1. P = 2.66684 × 105
2. P = 2.66057 × 105
3. P = 2.6624 × 105
4. P = 2.66798 × 105
5. P = 2.6657 × 105

39. Experiments show that if the chemical reaction

N

takes place at 45?C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows

.

How long will the reaction take to reduce the concentration of N2O5 to 70% of its original value?

1. t = 2014 s
2. t = 2066 s
3. t = 2055 s
4. t = 2077 s
5. t = 2044 s

40. Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.         Suppose that a roast turkey is taken from an oven when its temperature has reached 160?F and is placed on a table in a room where the temperature is 70?F.

If u(t) is the temperature of the turkey after t minutes, then Newton’s Law of Cooling implies that

.

This could be solved as a separable differential equation. Another method is to make the change of variable y = u − 70?F. ?

If the temperature of the turkey is 130 F after half an hour, what is the temperature after 20 min?

1. t = 139?F
2. t = 134?F
3. t = 135?F
4. t = 132?F
5. t = 138?F

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