First Order Differential Equations 2.1 for Autonomous Equations, Brannan, Boyce

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Chapter II Assignment2
(Mixing Experiment)

First Order Differential Equations 2.1 for Autonomous Equations, Brannan, Boyce

Where Q is a measure of the total salt in the tank at any time and r is the flow rate in and out of the well mixed tank. We illustrate an autonomous equation with a mixing example from Chapter 2.2. As you adjust the parameter value you are changing the initial conditions in the equation. You can follow the resulting solution path and watch the color of the mixture change in the dynamic graphic in real time.

Open the animation for this example of an autonomous differential equations:

This example provides a flow rate of 3 gallons per minute with a mixture of 1/4 pounds (lbs) of salt for each gallon. Note that the x, y location of the initial condition is given by the point on the vector field. To move this location drag the point with your mouse. Based on your knowledge of autonomous equations and solution techniques answer the following:

1. In the mixture flowing into the tank is 1/5 lbs of salt per gallon as t  the amount of salt in the 100 gallon tank is?

a)Q(t)→50 lbs.

b)Q(t) →25 lbs.

c)Q(t) →20 lbs.

d)Q(t) →0 lbs.

2. The resulting equation when the mixture flowing into the tank is 1/5 lbs of salt per gallon with a flow rate of 3 gallons per minute?

a) dQ/dt +(3Q(t) )/25= r/4.

b)dQ/dt +(3Q(t) )/50= r/4.

c)dQ/dt +(3Q(t) )/100= r/5.

d)dQ/dt +(3Q(t) )/500= r/5.

3. If the tank initially had 1/2 lbs of salt per gallon solution already in it, the resulting initial conditions would be?

a)Q(0) = 1/2 lbs/gal.

b)Q(0) =2lbs/gal.

c)Q(0) = 50lbs.

d)Q(0) = 25lbs.

e)None of the above.

4. If you were to solve the equation given in the text (example 1. sect 2.2) using an integration factor; that integration factor would be?

a)I(x) = e3t/100.

b)I(x) = e-3t/100.

c)I(x) = et/30.

d)Cannot be done with an integration factor.

5. If the flow rate in was given as 4 gallons per minute and the salt per gallon was 1 lbs/gallon The equation would become?

a)dQ/dt +Q(t) /25=1.

b)dQ/dt +Q(t) /50=1/2.

c)dQ/dt +Q(t) /100= ¼.

d)dQ/dt +Q(t) /500=4.

6. In the mixture flowing into the tank is 1 lbs of salt per gallon as t  the amount of salt in the tank is?

a)Q(t)→50 lbs.

b)Q(t)→25 lbs.

c)Q(t)→20 lbs.

d)Q(t)→100 lbs.

7. If the tank initially had 1/2 lbs per gallon salt solution and the flow rate was 4 gallons per minute of fresh water the resulting equation would be?

a)dQ/dt +Q(t) /25=0.

b)dQ/dt +Q(t) /100=4.

c)dQ/dt = 4.

d)dQ/dt +Q(t) /400=4.

8. If the tank initially had 50 lbs of salt in the solution and the flow rate was 4 gallons per minute of fresh water the resulting equilibrium state would be?

a) Tank with 1/2 lbs per gallon salt.

b) Fresh water only.

c) Tank with 2 lbs per gallon salt.

d) None of the above.

9. If the tank initially had 25 lbs salt in it and the flow rate was 4 gallons per minute of water with 1/5 lbs Salt per gallon the resulting equilibrium state would be?

a) Tank with 1/4 lbs per gallon salt.

b) Fresh water only.

c) Tank with 20 lbs of salt in it.

d) None of the above.

10. For problem 9 the solution form will?

a) Exponentially increase to equilibrium.

b) Exponentially decrease to equilibrium.

c) Remain in equilibrium.

d) Will not approach an equilibrium.