Mohammad Al-saad

Renaut MAT 275 ONLINE A Fall 2022

Assignment Section 1.1 Direction Fields due 08/20/2022 at 11:59pm MST

Problem 1. (1 point)

Match the following equations with their direction field. Clicking

on each picture will give you an enlarged view. While you can

probably solve this problem by guessing, it is useful to try to predict characteristics of the direction field and then match them to

the picture. Here are some handy characteristics to start with â€“

you will develop more as you practice.

Problem 2. (1 point)

Match the following equations with their direction field. Clicking

on each picture will give you an enlarged view.

1. y0 = xeâˆ’2x âˆ’ 2y

2. y0 = 2x âˆ’ 1 âˆ’ y2

x3

y3

3. y0 = âˆ’ y âˆ’

6

6

4. y0 = x + 2y

(i) Set y equal to zero and look at how the derivative behaves along

the x-axis.

(ii) Do the same for the y-axis by setting x equal to 0

(iii) Consider the curve in the plane defined by setting y0 = 0 â€“ this

should correspond to the points in the picture where the slope is

zero.

(iv) Setting y0 equal to a constant other than zero gives the curve of

points where the slope is that constant. These are called isoclines,

and can be used to construct the direction field picture by hand.

1. y0 = eâˆ’x âˆ’ y

(2x + y)

2. y0 = âˆ’

(2y)

3. y0 = âˆ’2 + x âˆ’ y

A

B

A

B

C

D

Answer(s) submitted:

â€¢

â€¢

â€¢

â€¢

C

Answer(s) submitted:

â€¢

â€¢

â€¢

(incorrect)

(incorrect)

1

â€¢

Problem 3. (1 point)

Match the following equations with their direction field. Clicking

on each picture will give you an enlarged view. While you can

probably solve this problem by guessing, it is useful to try to predict characteristics of the direction field and then match them to

the picture. Here are some handy characteristics to start with â€“

you will develop more as you practice.

(incorrect)

A. Set y equal to zero and look at how the derivative behaves

along the x-axis.

B. Do the same for the y-axis by setting x equal to 0

C. Consider the curve in the plane defined by setting y0 = 0 â€“

this should correspond to the points in the picture where

the slope is zero.

D. Setting y0 equal to a constant other than zero gives the

curve of points where the slope is that constant. These

are called isoclines, and can be used to construct the direction field picture by hand.

Problem 4. (1 point)

Consider the direction field below for a differential equation y0 =

y(y âˆ’ 2)(y âˆ’ 6). Use the graph and equation to answer the following questions.

1. y0 = y + 2

2. y0 = 2 sin(x) + 1 + y

3. y0 = 2y + x2 e2x

4. y0 = âˆ’2 + x âˆ’ y

1. Find the equilibrium solutions for the differential equation.

Answer (separate by commas): y =

2. If the initial condition is y(0) = c, for what values of c is

lim y(t) finite?

tâ†’âˆž

Answer (as an interval):

A

B

Note: You can click on the graph to enlarge the image.

Answer(s) submitted:

â€¢

â€¢

(incorrect)

C

D

Answer(s) submitted:

â€¢

â€¢

â€¢

2

Problem 5. (1 point)

Problem 6. (1 point)

Consider the two slope fields shown, in figures 1 and 2 below.

Consider the slope field shown.

(a) For the solution that satisfies y(0) = 0, sketch the

solution curve and estimate the following:

and y(âˆ’1) â‰ˆ

y(1) â‰ˆ

(b) For the solution that satisfies y(0) = 1, sketch the

solution curve and estimate the following:

y(0.5) â‰ˆ

and y(âˆ’1) â‰ˆ

figure 1

(c) For the solution that satisfies y(0) = âˆ’1, sketch the

solution curve and estimate the following:

and y(âˆ’1) â‰ˆ

y(1) â‰ˆ

figure 2

On a print-out of these slope fields, sketch for each three solution curves to the differential equations that generated them. Then

complete the following statements:

For the slope field in figure 1, a solution passing through the point

(3,-3) has a

â€¢?

â€¢ positive

â€¢ negative

â€¢ zero

â€¢ undefined

slope.

For the slope field in figure 1, a solution passing through the point

(-3,-2) has a

â€¢?

â€¢ positive

â€¢ negative

â€¢ zero

â€¢ undefined

slope.

Answer(s) submitted:

â€¢

â€¢

â€¢

â€¢

â€¢

â€¢

For the slope field in figure 2, a solution passing through the point

(1,2) has a

â€¢?

â€¢ positive

â€¢ negative

â€¢ zero

â€¢ undefined

slope.

(incorrect)

For the slope field in figure 2, a solution passing through the point

(0,3) has a

â€¢?

â€¢ positive

â€¢ negative

â€¢ zero

â€¢ undefined

slope.

Answer(s) submitted:

3

â€¢

â€¢

â€¢

â€¢

Problem 9. (1 point)

Match each differential equation to a function which is a solution.

FUNCTIONS

A. y = 3x + x2 ,

B. y = eâˆ’7x ,

C. y = sin(x),

1

D. y = x 2 ,

E. y = 3 exp(4x),

DIFFERENTIAL EQUATIONS

(incorrect)

Problem 7. (1 point)

The slope field for the equation y0 = âˆ’x + y is shown below

1. y00 + 14y0 + 49y = 0

2. y00 + y = 0

3. xy0 âˆ’ y = x2

4. y0 = 4y

Answer(s) submitted:

â€¢

â€¢

â€¢

â€¢

On a print out of this slope field, sketch the solutions that pass

through the points

(i) (0,0);

(ii) (-3,1); and

(iii) (-1,0).

(incorrect)

Problem 10. (1 point)

Find the value of k for which the constant function x(t) = k is a

dx

solution of the differential equation 5t 4 + 8x âˆ’ 3 = 0.

dt

From your sketch, what is the equation of the solution to the differential equation that passes through (-1,0)? (Verify that your

solution is correct by substituting it into the differential equation.)

y=

Answer(s) submitted:

â€¢

(incorrect)

Problem 11. (1 point)

Answer(s) submitted:

â€¢

For what values of r does the function y = 8erx satisfy the differential equation y00 + y0 âˆ’ 42y = 0?

The smaller one is

.

The larger one (possibly the same) is

.

(incorrect)

Problem 8. (1 point)

For what positive values of k does the function y = sin(kt) satisfy

the differential equation y00 + 144y = 0?

Answer(s) submitted:

â€¢

â€¢

For what negative values of k does the function y = cos(kt) satisfy

the differential equation y00 + 144y = 0?

(incorrect)

Problem 12. (1 point)

Find k such that x(t) = 10t is a solution of the differential equation

dx

= kx.

dt

k=

.

Answer(s) submitted:

â€¢

â€¢

Answer(s) submitted:

â€¢

(incorrect)

(incorrect)

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4

Mohammad Al-saad

Renaut MAT 275 ONLINE A Fall 2022

Assignment Section 1.2 Solutions of Some DE due 08/20/2022 at 11:59pm MST

Problem 1. (1 point)

Find the solution to

Problem 4. (1 point)

A pond contains 2640 L of pure water and an uknown amount of

an undesirable chemical. Water contaninig 0.08 kg of this chemical per liter flows into the pond at a rate of 9 L/h. The mixture

flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed

throughout the pond.

Let Q(t) be the amount of chemical (in kg) in the pond at time t

hours.

dy

= 4y

dt

satisfying

y(7) = 9

y=

Answer(s) submitted:

â€¢

(a) Write a differential equation for the amount of chemical in the

pond?

at any time time (enter Q for Q(t):

dQ

=

dt

(incorrect)

Problem 2. (1 point)

Find the general solution to the differential equation modeling

how a person learns:

dy

= 100 âˆ’ y.

dt

Then find the particular solutions with the following initial conditions:

y(0) = 10 : y =

y(0) = 120 : y =

(b) How much chemical will be in the pond after a long time?

Qâˆž =

(kg)

(c) Does the limiting value in part (b) depend on the amount

that was present initially? [?/yes/no]

Answer(s) submitted:

â€¢

â€¢

â€¢

Plot the slope field of this differential equation and sketch the solutions with y(0) = 10 and y(0) = 120.

(incorrect)

Which of these two particular solutions could represent how a person learns?

â€¢ A. y(0) = 10

â€¢ B. y(0) = 120

â€¢ C. either of these

â€¢ D. none of the above

Answer(s) submitted:

â€¢

â€¢

â€¢

(incorrect)

Problem 3. (1 point)

Solve the differential equation

dP

= 4P + a.

dt

Assume a is a non-zero constant, and use C for any constant of

integration that you may have in your answer.

P=

Answer(s) submitted:

â€¢

(incorrect)

1

Problem 5. (1 point)

Supposed a body of mass 4 kg is falling in the atmosphere near

sea level.

Problem 6. (1 point)

Supposed a body of mass 31 kg is falling in the atmosphere near

sea level.

Let v(t)m/s be the velocity of the body at time t in seconds.

Assume that v is positive in the downward direction – that is,

when the object is falling. We assume that the forces acting on

the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with

magnitude cv(t) where c = 0.25 kgs . The gravitational constant is

g = 9.8m/s2 .

Let v(t) m/s be the velocity of the body at time t in seconds.

Assume that v is positive in the downward direction – that is,

when the object is falling. We assume that the forces acting on

the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with

magnitude proportional to the square of the velocity. Let c be the

constant of proportionality.

The gravitational constant is g = 9.8m/s2 .

a) Find a differential equation for the velocity v:

dv

=

dt

a) Find a differential equation for the velocity v:

dv

=

dt

(b) Find the equilibrium solution of the differential equation,

that is, find the limiting velocity.

limiting velocity =

(b) Determine the limiting velocity after a long time. Your answer should be an expression in c.

m/s

limiting velocity =

Answer(s) submitted:

â€¢

â€¢

m/s

(c) Find the drag coefficient, c, so that the limiting velocity is

42 m/s.

(incorrect)

c=

Answer(s) submitted:

â€¢

â€¢

â€¢

(incorrect)

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