Ed Azoff Math2700 Azoff F11 - University of Georgiamath.uga.edu/~azoff/courses/ww1108.pdf · Ed Azoff Math2700 Azoff F11 WeBWorK assignment number wwassign1 is due : 08/22/2011 at

Ed Azoff Math2700 Azoff F11WeBWorK assignment number wwassign1 is due : 08/22/2011 at 10:00am EDT.The course homepage http://www.math.uga.edu/ azoff/MATH2700 contains the syllabus, grading policy and other information.

1. (1 pt) An unknown radioactive element decays into non-radioactive substances. In 220 days the radioactivity of a sampledecreases by 57 percent.

(a) What is the half-life of the element?half-life: (days)

(b) How long will it take for a sample of 100 mg to decayto 77 mg?time needed: (days)

Correct Answers:• 180.684582416496• 68.1306720859227

2. (1 pt) A cell of some bacteria divides into two cells every20 minutes.The initial population is 3 bacteria.

(a) Find the size of the population after t hoursy(t) =(function of t)

(b) Find the size of the population after 3 hours.y(3) =

(c) When will the population reach 9?T =

Correct Answers:• 3*(2ˆ(3*t))• 1536• 0.528320833573719

3. (1 pt) The doubling period of a baterial population is 20minutes. At time t = 60 minutes, the baterial population was4568. What was the initial population at time t = 0?Find the size of the baterial population after 5 hours.

Correct Answers:• 571• 18710528

4. (1 pt) Each curve in the figure below represents the bal-ance in a bank account into which a single deposit was made attime zero.

Assuming continuously compounded interest, find (entereach answer as value or a list, e.g, I or II,III) : :(a) The curves representing the same initial deposit :(b) The curve(s) representing the largest initial deposit :(c) The curves representing the same interest rate :(d) The curve(s) representing the largest interest rate :

Correct Answers:

• I, IV• II• I, III• IV

5. (1 pt) Warfarin is a drug used as an anticoagulant. Afteradministration of the drug is stopped, the quantity remaining ina patient’s body decreases at a rate proportional to the quantityremaining. Suppose that the half-life of warfarin in the body is35 hours.

Sketch the quantity, Q, of warfarin in a patient’s body as afunction of the time, t (in hours), since stopping administrationof the drug. Mark 35 hours on your graph.

Write a differential equation satisfied by Q:dQdt =

(Your equation should not involve any undetermined constants.)How many days does it take for the drug level in the body to

be reduced to 15 percent of its original level?time =(include units)

Correct Answers:

• -ln(2)*Q/35• 95.7938 hr

6. (1 pt) Suppose that a population develops according to thelogistic equation

dPdt

= 0.05P−0.0002P2

where t is measured in weeks.

(a) What is the carriying capacity?

(b) Is the solution increasing or decreasing when P is between 0and the carriying capacity? ?

(c) Is the solution increasing or decreasing when P is greaterthan the carriying capacity? ?

Correct Answers:

• 250• INCREASING• DECREASING

1

7. (1 pt) Any population, P, for which we can ignore immi-gration, satisfies

dPdt

= Birth rate−Death rate.

For organisms which need a partner for reproduction but relyon a chance encounter for meeting a mate, the birth rate is pro-portional to the square of the population. Thus, the populationof such a type of organism satisfies a differential equation of theform

dPdt

= aP2−bP with a, b > 0.

This problem investigates the solutions to such an equation.(a) Sketch a graph of dP/dt against P. Note when dP/dt is

positive and negative.dP/dt < 0 when P is indP/dt > 0 when P is in(Your answers may involve a and b. Give your answers as aninterval or list of intervals: thus, if dP/dt is less than zero for Pbetween 1 and 3 and P greater than 4, enter (1,3),(4,infinity).)

(b) Use this graph to sketch the shape of solution curves withvarious initial values: use your answers in part (a), and wheredP/dt is increasing and decreasing to decide what the shape ofthe curves has to be. Based on your solution curves, why isP = b/a called the threshold population?

If P(0) > b/a, what happens to P in the long run?P→

If P(0) = b/a, what happens to P in the long run?P→

If P(0) < b/a, what happens to P in the long run?P→

Correct Answers:

• (0,1.74796)• (1.74796,infinity)• infinity• b/a• 0

8. (1 pt) It is of considerable interest to policy makers tomodel the spread of information through a population. For ex-ample, various agricultural ministries use models to help themunderstand the spread of technical innovations or new seed typesthrough their countries. Two models, based on how the informa-tion is spread, are given below. Assume the population is of a

constant size M, and let P (a function of time, t) be the num-ber of people in that population who have the information ofinterest.

(a) If the information is spread by mass media (TV, radio,newspapers), the rate at which information is spread is believedto be proportional (with constant of proportionality k > 0) to thenumber of people not having the information at that time. Writea differential equation for the number of people having the in-formation by time t.dPdt =

Sketch a solution assuming that no one (except the mass me-dia) has the information initially. What is the limiting value ofthe population that knows the information?P→

At what value of P is the number of people who know theinformation increasing the fastest?When P =

(b) If the information is spread by word of mouth, the rate ofspread of information is believed to be proportional (again, withconstant of proportionality k > 0) to the product of the numberof people who know and the number who don’t. Write a differ-ential equation for the number of people having the informationby time t.dPdt =

Sketch the solution for the cases in which (i) no one; (ii) 5percent of the population; and (iii) 75 percent of the populationinitially knows the information.

What is the limiting value for the population that knows theinformation in each case?If no one initially knows, P→If 5 percent initially know, P→If 75 percent initially know, P→

In the latter two cases, when is the information spreadingfastest?If 5 percent initially know, when P =If 75 percent initially know, when P =

Correct Answers:

• k*(M-P)• M• 0• k*P*(M-P)• 0• M• M• M/2• 0.75*M

Generated by the WeBWorK system c©WeBWorK Team, Department of Mathematics, University of Rochester

2


1. (1 pt) Find a positive value of k for which y = sin(kt)satisfies

d2ydt2 +9y = 0.

k =Correct Answers:

• 3

2. (1 pt) Match the solutions to the differential equations. Ifthere is more than one solution to an equation, select the answerthat includes all solutions.

1. dydx =−2y

2. d2ydx2 = 4y

3. dydx = 2y

4. d2ydx2 =−4y

A. y = e−2x or y = e2x

B. y = sin(2x) or y = 2sin(x)C. y = e−2x

D. y = 2sin(x)E. y = sin(2x)F. y = e2x

Correct Answers:

• C• A• F• E

3. (1 pt) Let A and k be positive constants.Which of the given functions is a solution to dy

dt = k(A− y)?

• A. y =−A+Cekt

• B. y =−A+Ce−kt

• C. y = A+Cekt

• D. y = A−1 +Ce−Akt

• E. y = A+Ce−kt

• F. y = A−1 +CeAkt

Correct Answers:

• E

4. (1 pt)Match the following equations with their direction field. Click-ing on each picture will give you an enlarged view. While youcan probably solve this problem by guessing, it is useful to tryto predict characteristics of the direction field and then matchthem to the picture. Here are some handy characteristics to startwith – you will develop more as you practice.

A. Set y equal to zero and look at how the derivative be-haves along the x-axis.

B. Do the same for the y-axis by setting x equal to 0C. Consider the curve in the plane defined by setting y′ = 0

– this should correspond to the points in the picturewhere the slope is zero.

D. Setting y′ equal to a constant other than zero gives thecurve of points where the slope is that constant. Theseare called isoclines, and can be used to construct thedirection field picture by hand.

Go to this page to launch the phase plane plotter to check youranswers. (Choose the ”integral curves utility” from the appletmenu, enter dx/dt = 1 to identify the variables x and t and thenenter the function you want for dy/dx = dy/dt = . . . ).

1. y′ = y+22. y′ = e−x− y3. y′ = 3sin(x)+1+ y

A B C

Correct Answers:

• C• A• B

5. (1 pt) The functions

y = x2 +cx2

are all solutions of equation:

xy′+2y = 4x2, (x > 0).

Find the constant c which produces a solution which also satis-fies the initial condition y(3) = 8.c =

Correct Answers:

• -9

1

6. (1 pt) Find f (x) if y = f (x) satisfies

dydx

= 54yx5

and the y-intercept of the curve y = f (x) is 5.f (x) = .

Correct Answers:

• 5 * exp(9 * (xˆ6 ) )

7. (1 pt) Find the solution to the differential equation

dydt

= y2(4+ t),

y = 3 when t = 1.y =Correct Answers:

• 2*3/(2 + (1-t)*(1 + 2*4 + t)*3)

8. (1 pt) Dead leaves accumulate on the ground in a forest ata rate of 4 grams per square centimeter per year. At the sametime, these leaves decompose at a continuous rate of 70 percentper year.

A. Write a differential equation for the total quantity Q ofdead leaves (per square centimeter) at time t:dQdt =

B. Sketch a solution to your differential equation showingthat the quantity of dead leaves tends toward an equilibriumlevel. Assume that initially (t = 0) there are no leaves on theground.What is the initial quantity of leaves? Q(0) =What is the equilibrium level? Qeq =Does the equilibrium value attained depend on the initial condi-tion?

• A. yes• B. no

Correct Answers:

• 4 - 0.7*Q• 0• 4/0.7• B

9. (1 pt) Water leaks from a vertical cylindrical tank througha small hole in its base at a rate proportional to the square rootof the volume of water remaining. The tank initially contains150 liters and 24 liters leak out during the first day.

A. When will the tank be half empty? t =(include units)

B. How much water will remain in the tank after 5 days? vol-ume =(include units)

Correct Answers:• 3.50833929977397 day• 50.909166052992 L

10. (1 pt) A thermometer is taken from a room where thetemperature is 25oC to the outdoors, where the temperature is−5oC. After one minute the thermometer reads 15oC.(a) What will the reading on the thermometer be after 3 moreminutes?

,(b) When will the thermometer read −4oC?

minutes after it was taken to the outdoors.Correct Answers:• 0.925925925925925• 8.38838487861903

11. (1 pt) A tank contains 1040 L of pure water. Solution thatcontains 0.09 kg of sugar per liter enters the tank at the rate 5L/min, and is thoroughly mixed into it. The new solution drainsout of the tank at the same rate.(a) How much sugar is in the tank at the begining?y(0) = (kg)

(b) Find the amount of sugar after t minutes.y(t) = (kg)

(c) As t becomes large, what value is y(t) approaching ? Inother words, calculate the following limit. lim

t→∞y(t) = (kg)

Correct Answers:• 0• 93.6*(1-2.71828182845905ˆ(-0.00480769230769231*t))• 93.6

12. (1 pt) A young person with no initial capital invests kdollars per year in a retirement account at an annual rate of re-turn 0.09. Assume that investments are made continuously andthat the return is compounded continuously.

Determine a formula for the sum S(t) – (this will involve theparameter k):S(t) =

What value of k will provide 4135000 dollars in 40 years?k =

Correct Answers:• (k/0.09)*(exp(0.09*t) - 1 )• 10454.1701524215


2


1. (1 pt) This problem is harder, and doesn’t give you cluesas to which matches you have right. Study the previous prob-lem, if you are having trouble.Go to this page to launch the phase plane plotter to check youranswers.Match the following equations with their direction field. Click-ing on each picture will give you an enlarged view.

1. y′ =−y(5− y)2. y′ = y(4− y)3. y′ = x+2y4. y′ = xe−2x−2y

A B

C D

Correct Answers:

• B• D• A• C

2. (1 pt) Consider the solution of the differential equationy′ = 2y passing through y(0) = 1.

A. Sketch the slope field for this differential equation, andsketch the solution passing throughthe point (0,1).

B. Use Euler’s method with step size ∆x = 0.2 to estimate thesolution at x = 0.2,0.4, . . . ,1, using these to fill in the followingtable. (Be sure not to round your answers at each step!)

x = 0 0.2 0.4 0.6 0.8 1.0y≈ 1

C. Plot your estimated solution on your slope field. Comparethe solution and the slope field. Is the estimated solution an overor under estimate for the actual solution?

• A. over• B. under

D. Check that y = e2x is a solution to y′ = 2y with y(0) = 1.Correct Answers:

• 1.4• 1.96• 2.744• 3.8416• 5.37824• B

3. (1 pt) Use Euler’s method with step size 0.25 to computethe approximate y-values y1, y2, y3, and y4 of the solution of theinitial-value problem

y′ = 2+2x+4y, y(1) = 1.

y1 = ,y2 = ,y3 = ,y4 = .

Correct Answers:

• 3• 7.125• 15.5• 32.375

4. (1 pt) Use Euler’s method to solvedBdt

= 0.08B

with initial value B = 800 when t = 0 .A. ∆t = 1 and 1 step: B(1)≈B. ∆t = 0.5 and 2 steps: B(1)≈C. ∆t = 0.25 and 4 steps: B(1)≈

1

D. Suppose B is the balance in a bank account earning inter-est. Be sure that you can explain why the result of your calcu-lation in part (a) is equivalent to compounding the interest oncea year instead of continuously. Then interpret the result of yourcalculations in parts (b) and (c) in terms of compound interest.

Correct Answers:

• (1 + 0.08)*800• 800*(1 + 0.08/2)ˆ2• 800*(1 + 0.08/4)ˆ4

5. (1 pt) How many years would it take your money to dou-ble:(a) At 10% interest compounded yearly.

years.(b) At 10% interest compounded weekly.

years and weaks.(c) At 10% interest compounded continuously.

years.Correct Answers:

• 8• 6• 49• 6.93147180559945

6. (1 pt) The slope field for the equation dP/dt =0.0333333P(30−P), for P≥ 0, is shown below.

On a print out of this slope field, sketch the solutions thatpass through (0,0); (3,12); (12,3); (−14.5,3); (−6,36); and(−6,30).

For which positive values of P are the solutions increasing?Increasing for:(Give your answer as an interval or list of intervals, e.g., if Pis increasing between 1 and 5 and between 7 and infinity, enter(1,5),(7,Inf).)

For what positive values of P are the solutions decreasing?Decreasing for:(Again, give your answer as an interval or list of intervals, e.g.,if P is decreasing between 1 and 5 and between 7 and infinity,enter (1,5),(7,Inf).)

What is the equation of the solution to this differential equa-tion that passes through (0,0)?P =

If the solution passes through a value of P > 0, what is thelimiting value of P as t gets large?P→

Correct Answers:

• (0,30)

• (30,infinity)• 0• 30

7. (1 pt) A population P obeys the logistic model. It satisfiesthe equationdPdt

=5

1100P(11−P) for P > 0.

(a) The population is increasing when < P <

(b) The population is decreasing when P >

(c) Assume that P(0) = 2. Find P(62).P(62) =

Correct Answers:• 0• 11• 11• 9.14592515065347

8. (1 pt) Suppose that a population develops according to thelogistic equation

dPdt

= 0.1P−0.0005P2

where t is measured in weeks.

(a) What is the carriying capacity?

(b) Is the solution increasing or decreasing when P is between 0and the carriying capacity? ?

(c) Is the solution increasing or decreasing when P is greaterthan the carriying capacity? ?

Correct Answers:• 200• INCREASING• DECREASING

9. (1 pt) A tank contains 1340 L of pure water. Solution thatcontains 0.01 kg of sugar per liter enters the tank at the rate 5L/min, and is thoroughly mixed into it. The new solution drainsout of the tank at the same rate.(a) How much sugar is in the tank at the begining?y(0) = (kg)

(b) Find the amount of sugar after t minutes.y(t) = (kg)

(c) As t becomes large, what value is y(t) approaching ? Inother words, calculate the following limit. lim

t→∞y(t) = (kg)

Correct Answers:• 0• 13.4*(1-2.71828182845905ˆ(-0.00373134328358209*t))• 13.4

2

10. (1 pt)

A country has 10 billion dollars in paper currency in circu-lation, and each day 57 million dollars comes into the country’sbanks. The government decides to introduce new currency byhaving the banks replace old bills with new ones whenever oldcurrency comes into the bank. Let x = x(t) denote the numberof new dollars in circulation after t days with units in billionsand x(0) = 0.

A. Determine a differential equation which describes the rateat which x is growing:dxdt =

B. Solve the differential equation subject to the initial con-ditions given above.x(t) =

C. How many days will it take for the new bills to accountfor 90 percent of the currency in circulation?

Correct Answers:

• (57/1000)*(10-x)/10• 10 - 10*exp(-(57/10000)*t)• 403.962297016499


3


1. (1 pt)The function y(t) satisfies the differential equation

dydt

= y4 +5y3−14y2

List the (distinct) constant solutions (y = c) to the differen-tial equation in ascending order. (If there are fewer than foursolutions, leave the latter blanks empty)

For what values of y (in interval notation) is y increasing?Use the strings ”plus infinity” or ”minus infinity” if appropri-ate, and if there is only one interval, leave the second one blank.Finally, list your intervals so that the first interval is to the leftof the second (on the real line).

Interval 1: ( , )Interval 2:( , )

Correct Answers:• -7• 0• 2•• MINUS_INFINITY• -7• 2• PLUS_INFINITY

2. (1 pt) The graph of the function f (x) is

(the hori-zontal axis is x.)Consider the differential equation x′(t) = f (x(t)).

List the constant (or equilibrium) solutions to this differential

equation in increasing order and indicate whether or not theseequations are stable, semi-stable, or unstable.

????

Correct Answers:• -3• UNSTABLE• -2• STABLE• -0.5• UNSTABLE• 1.5• STABLE

3. (1 pt) The graph of the function f (x) is

(the hori-zontal axis is x.)Given the differential equation x′(t) = f (x(t)).List the constant (or equilibrium) solutions to this differentialequation in increasing order and indicate whether or not theseequations are stable, semi-stable, or unstable.

????

Correct Answers:• -3.5• UNSTABLE• -2.5• SEMI-STABLE

1

• -0.5• STABLE• 1.5• UNSTABLE

4. (1 pt) Given the differential equation x′ = (x + 3) ∗ (x +1.5)3(x−0.5)2(x−2.5).List the constant (or equilibrium) solutions to this differen-tial equation in increasing order and indicate whether or notthese equations are stable, semi-stable, or unstable. (It helpsto sketch the graph. xFunctions will plot functions as well asphase planes. )

????

Correct Answers:

• -3• UNSTABLE• -1.5• STABLE• 0.5• SEMI-STABLE• 2.5• UNSTABLE

5. (1 pt) The slope field for a population P modeled bydP/dt = 4P−4P2 is shown in the figure below.

(a) On a print-out of the slope field, sketch three solutioncurves showing different types of behavior for the population P.Give an initial condition that will produce each:

P(0) = ,P(0) = , andP(0) = .

(b) Is there a stable value of the population? If so, give thevalue; if not, enter none:

Stable value =(c) Considering the shape of solutions for the population,

give any intervals for which the following are true. If no suchinterval exists, enter none, and if there are multiple intervals,give them as a list. (Thus, if solutions are increasing when Pis between 1 and 3, enter (1,3) for that answer; if they are de-creasing when P is between 1 and 2 or between 3 and 4, enter(1,2),(3,4). Note that your answers may reflect the fact that P isa population.)

P is increasing when P is inP is decreasing when P is in

Think about what these conditions mean for the population,and be sure that you are able to explain that.

In the long-run, what is the most likely outcome for the pop-ulation?P→(Enter infinity if the population grows without bound.)

Are there any inflection points in the solutions for the popu-lation? If so, give them as a comma-separated list (e.g., 1,3); ifnot, enter none.

Inflection points are at P =Be sure you can explain what the meaning of the inflection

points is for the population.(d) Sketch a graph of dP/dt against P. Use your graph to

answer the following questions.When is dP/dt positive?

When P is inWhen is dP/dt negative?When P is in(Give your answers as intervals or a list of intervals.)

When is dP/dt zero?When P =(If there is more than one answer, give a list of answers, e.g.,1,2.)

When is dP/dt at a maximum?When P =

Be sure that you can see how the shape of your graph ofdP/dt explains the shape of solution curves to the differentialequation.

Correct Answers:• 0.5• 1• 2• 1• (0,1)• (1,infinity)• 1• 1/2• (0,1)• (1,infinity)• 0, 1• 1/2

6. (1 pt) Find the particular solution of the differential equa-tion

x2

y2−2dydx

=12y

satisfying the initial condition y(1) =√

3.Answer: y = .Your answer should be a function of x.

Correct Answers:• sqrt(eˆ(1-1/x)+2)

7. (1 pt) The count in a bateria culture was 300 after 15 min-utes and 1900 after 30 minutes. What was the initial size of theculture? Find the doubling period. Find the popu-lation after 65 minutes. When will the population reach10000.

2

Correct Answers:• 47.3684210526316• 5.6328190300424• 141003.234618038• 43.4958326426623

8. (1 pt) A tank initially contains 200 gallons of brine, with50 pounds of salt in solution. Brine containing 2 pounds of saltper gallon is entering the tank at the rate of 4 gallons per minuteand is is flowing out at the same rate. If the mixture in the tankis kept uniform by constant stirring, find the amount of salt inthe tank at the end of 40 minutes.The amount of salt in the tank at the end of 40 minutes is

pounds.Correct Answers:• 242.734862537726

9. (1 pt) A drug is administered intravenously at a constantrate of r mg/hour and is excreted at a rate proportional to thequantity present, with constant of proportionality k > 0.

(Set up and) Solve a differential equation for the quantity, Q,in milligrams, of the drug in the body at time t hours. Assumethere is no drug in the body initially. Your answer will contain rand k.Q =

Graph Q against t. What is Q∞, the limiting long-run valueof Q?Q∞ =

If r is doubled (to 2r), by what multiplicative factor is Q∞

increased?Q∞ (for 2r) = Q∞ (for r)

Similarly, if r is doubled (to 2r), by what multiplicative fac-tor is the time it takes to reach half the limiting value, 1

2 Q∞,changed?t (to 1

2 Q∞), for 2r) = t (to 12 Q∞), for r)

If k is doubled (that is, we use 2k instead of k), by what mul-tiplicative factor is Q∞ increased?Q∞ (for 2k) = Q∞ (for k)

On the time to reach 12 Q∞?

t (to 12 Q∞), for 2k) = t (to 1

2 Q∞), for k)Correct Answers:• (r/k)*(1 - eˆ(-k*t))• (r/k)• 2• 1

• 1/2• 1/2

10. (1 pt) The total number of people infected with a virusoften grows like a logistic curve. Suppose that 25 people origi-nally have the virus, and that in the early stages of the virus (withtime, t, measured in weeks), the number of people infected is in-creasing exponentially with k = 1.8. It is estimated that, in thelong run, approximately 4250 people become infected.

(a) Use this information to find a logistic function to modelthis situation.P =

(b) Sketch a graph of your answer to part (a). Use your graphto estimate the length of time until the rate at which people arebecoming infected starts to decrease. What is the vertical coor-dinate at this point?vertical coordinate =

Correct Answers:

• 4250/(1 + (4250 - 25)*eˆ(-1.8*t)/25)• 4250/2

11. (1 pt) A. Solve the following initial value problem:

(t2−16t +28)dydt

= y

with y(8) = 1. (Find y as a function of t.)y = .

B. On what interval is the solution valid?Answer: It is valid for < t < .

C. Find the limit of the solution as t approaches the left endof the interval.(Your answer should be a number or the word ”infinite”.)Answer: .

D. Similar to C, but for the right end.Answer: .

Correct Answers:

• ((14-t)/(t-2))**(1/(2*6))• 2• 14• infinite• 0


3


1. (1 pt) Solve the following initial value problem:

tdydt

+7y = 5t

with y(1) = 1.

Find the integrating factor, u(t) = ,and then find y(t) = .

Correct Answers:

• tˆ(7)• ( 5 * t / (1 +7)) + 0.375 * (t**(-7 ))

2. (1 pt) Find the general solution to the differential equation

x2−1xy+ xdydx

= 0

Find the integrating factor, u(x) = .

Find y(x) = .Use C as the unknown constant.

Correct Answers:

• eˆ(-1 x)• C eˆ(1 x) + x/1 + 1/1ˆ2

3. (1 pt) Find the particular solution of the differential equa-tion

dydx

+ ycos(x) = 5cos(x)

satisfying the initial condition y(0) = 7.Answer: y(x)= .

Correct Answers:

• 5 + 2*eˆ(-sin(x))

4. (1 pt) Solve the following initial value problem:

dydt

+0.7ty = 3t

with y(0) = 2.y = .

Correct Answers:

• (3 / 0.7 ) + -2.28571428571429 * exp( -0.7 * t * t /2)

5. (1 pt) Solve the initial value problem

8(t +1)dydt−6y = 12t,

for t >−1 with y(0) = 14.y = .

Correct Answers:

• 6*t +8 + (6 * ((t + 1)**0.75))

6. (1 pt)Find the solution to the initial value problem

(1+ x2)y′+2x1y = 10x8

subject to the condition y(0) = 2.

Correct Answers:

• 10*x**(8+1)/((8+1)*(1+x**(2))) + 2/(1+x**(2))


dydt

+2y = 25sin(t)+20cos(t)

with y(0) = 9.y = .

(Show the student hint after 4 attempts: )Reminder: To find the anti-derivative of eu sin(u), the trick is todo integration by parts twice.

Correct Answers:

• (2*5 + 4 )* sin(t) + (2*4 - 5 )* cos(t) + 6 * exp(-2* t)


dxdt

+5x = cos(4t)

with x(0) = 1.x(t) = .

Correct Answers:

• 0.878048780487805*eˆ(- 5*t) + 5/41*cos(4*t) + 4/41*sin(4*t)

1


dydt− y = 3et +49e8t

with y(0) = 7.y = .

Correct Answers:• (7 - 7 )* exp(t) + 3 * t * exp(t) + 7 * exp(8 *t)

10. (1 pt) Here is a multipart example on finance. Be patientand careful as you work on this problem. You will probablybe surprised to find how long it takes to get all of the details ofthe solution of a realistic problem exactly right, even when youknow how to do each of the steps. Use the computer to checkthe steps for you as you go along. There is partial credit on thisproblem.

A recent college graduate borrows 100000 dollars at an (an-nual) interest rate of 9.75 per cent. Anticipating steady salaryincreases, the buyer expects to make payments at a monthly rateof 750(1 + t/100) dollars per month, where t is the number ofmonths since the loan was made.

Let y(t) be the amount of money that the graduate owes t monthsafter the loan is made.y(0) = (dollars)

With y representing the amount of money in dollars at timet (in months) write a differential equation which models thissituation.y′ = f (t,y) =

.Note: Use y rather than y(t) since the latter confuses the com-puter. Remember to check your units, but don’t enter units forthis equation – the computer won’t understand them.

Find an equation for the amount of money owed after t months.y(t) =

Next we are going to think about how many months it willtake until the loan is paid off. Remember that y(t) is the amountthat is owed after t months. The loan is paid off when y(t) =

Once you have calculated how many months it will take topay off the loan, give your answer as a decimal, ignoringthe fact that in real life there would be a whole number ofmonths. To do this, you should use a graphing calculator oruse a plotter on this page to estimate the root. If you use thethe xFunctions plotter, then once you have launched xFunc-tions, pull down the Multigaph Utility from the menu in theupper right hand corner, enter the function you got for y (usingx as the independent variable, sorry!), choose appropriate rangesfor the axes, and then eyeball a solution.

The loan will be paid off in months.

If the borrower wanted the loan to be paid off in exactly 20years, with the same payment plan as above, how much couldbe borrowed?Borrowed amount =

Correct Answers:• 100000• 0.008125*y - 750*(1+t/100)• 205917.159763314 + 923.076923076923*t - 105917.159763314*e**(0.008125*t)• 0• 142.678886227308• 145101.30685658


2

Ed Azoff Math2700 Azoff F11WeBWorK assignment number wwassign6 is due : 11/07/2011 at 10:00am EST.The course homepage http://www.math.uga.edu/ azoff/MATH2700 contains the syllabus, grading policy and other information.

1. (1 pt) Find the equilibrium solution for

x′1(t) =−7.2+1.2x1−0.8x2

x′2(t) =−13.8+2.1x1−1.2x2

x1(0) = 12; x2(0) = 8

Equilibrium: xe1 = ,

xe2 = .

[Note– you’ll probably want to view the phase plotter atphase plotter

(right click to open in a new window).Select the ”integral curves utility” from the main menu. ]Plot with xmin = -5, xmax = 20, ymin = -5, ymax = 20, start x= 12, start y = 8 (then select new curve).Clear the above curve and try starting at the equilibrium solutionfor x and y.

? 1. Describe the trajectory.

? 1. What kind of interaction do we observe?Correct Answers:• 10• 6• Ellipse counterclockwise• Predator-prey


x′1(t) =−6.2+1.1x1−0.8x2

x′2(t) =−13.8+2.1x1−1.2x2

x1(0) = 11; x2(0) = 4


xe2 = .


(right click to open in a new window).Select the ”integral curves utility” from the main menu. ]

Plot with xmin = -5, xmax = 20, ymin = -5, ymax = 20, start x= 11, start y = 4 (then select new curve).Click on the graph to try other starting points.


? 1. What kind of interaction do we observe?

Correct Answers:

• 10• 6• Spiral inward counterclockwise• Predator-prey

3. (1 pt) Let w be the number of worms (in millions) and rthe number of robins (in thousands) living on an island. Sup-pose w and r satisfy the following differential equations, whichcorrespond to the slope field shown below.

dwdt

= w−wr,drdt

=−r +wr.

Assume w = 3 and r = 4 when t = 0.Does the number of worms increase, decrease, or stay the sameat first? ?Does the number of robins increase, decrease, or stay the sameat first? ?

What happens in the long run??

Correct Answers:

• decreases• increases• w and r oscillate

1

4. (1 pt) Solve the systemdxdt

=[

1 -44 -7

]x

with x(0) =[

32

].

Give your solution in real form.x1 = ,x2 = .

[Note– you’ll probably want to view the phase plotter atphase plotter (right click to open in a new window).

Select the ”integral curves utility” from the main menu. ]If y′ = Ay is a differential equation, how would the solution

curves behave?

• A. All of the solutions curves would converge towards0. (Stable node)• B. All of the solution curves would run away from 0.

(Unstable node)• C. The solution curves converge to different points.• D. The solution curves would race towards zero and

then veer away towards infinity. (Saddle)

Correct Answers:

• eˆ(-3*t) * (3+4*t)• eˆ(-3*t) * (2+4*t)• A

5. (1 pt) Consider the interaction of two species of animalsin a habitat. We are told that the change of the populations x(t)and y(t) can be modeled by the equations

dxdt

= 1.6x+0.75y,

dydt

= 1.66666666666667x−0.4y.

For this system, the smaller eigenvalue is andthe larger eigenvalue is .




curves behave?

• A. All of the solution curves would run away from 0.(Unstable node)• B. The solution curves would race towards zero and

then veer away towards infinity. (Saddle)• C. The solution curves converge to different points.

• D. All of the solutions curves would converge towards0. (Stable node)

The solution to the above differential equation with initialvalues x(0) = 3, y(0) = 2 isx(t) = ,y(t) = .

Correct Answers:

• -0.9• 2.1• Symbiosis• B• -0*0.75*exp(-0.9*t)+4*0.75*exp(2.1*t)• -0*(-0.9-1.6)*exp(-0.9*t)+4*(2.1-1.6)*exp(2.1*t)


dxdt

= 0.4x+0.5y,

dydt

= 1.5x−0.6y.





curves behave?

• A. All of the solution curves would run away from 0.(Unstable node)• B. The solution curves would race towards zero and

then veer away towards infinity. (Saddle)• C. All of the solutions curves would converge towards

0. (Stable node)• D. The solution curves converge to different points.


Correct Answers:

• -1.1• 0.9• Symbiosis• B• -2.5*0.5*exp(-1.1*t)+6.5*0.5*exp(0.9*t)• -2.5*(-1.1-0.4)*exp(-1.1*t)+6.5*(0.9-0.4)*exp(0.9*t)

2


dxdt

= 0.1x−0.8y,

dydt

= −0.2x+0.7y.





curves behave?

• A. The solution curves converge to different points.• B. All of the solution curves would run away from 0.

(Unstable node)• C. All of the solutions curves would converge towards

0. (Stable node)• D. The solution curves would race towards zero and



Correct Answers:

• -0.1• 0.9• Competition• D• -17*-0.8*exp(-0.1*t)+5.75*-0.8*exp(0.9*t)• -17*(-0.1-0.1)*exp(-0.1*t)+5.75*(0.9-0.1)*exp(0.9*t)

8. (1 pt) Calculate the eigenvalues of this matrix:

A =[

90 36-90 -108

]smaller eigenvalue =associated eigenvector = ( , )larger eigenvalue =associated, eigenvector = ( , )



curves behave?

• A. The solution curves converge to different points.

• B. All of the solutions curves would converge towards0. (Stable node)• C. All of the solution curves would run away from 0.

(Unstable node)• D. The solution curves would race towards zero and


Correct Answers:

• The eigenvalues and eigenvectors are {72,-90} and the eigenvectors are ( -4 ,2 ) and ( 1 ,-5 )• The eigenvalues and eigenvectors are {72,-90} and the eigenvectors are ( -4 ,2 ) and ( 1 ,-5 )• D


dxdt

= −0.8x+2y,

dydt

= 3x−0.8y.





curves behave?

• A. All of the solution curves would run away from 0.(Unstable node)• B. The solution curves converge to different points.• C. All of the solutions curves would converge towards

0. (Stable node)• D. The solution curves would race towards zero and


Correct Answers:

• -2.84948974278318• 2.04948974278318• Symbiosis• D


x′1(t) =−9.7+1.2x1−0.5x2

x′2(t) =−9.8+1.4x1−0.8x2

x1(0) = 20; x2(0) = 343


xe2 = .


Select the ”integral curves utility” from the main menu. ]Plot with xmin = -5, xmax = 50, ymin = -5, ymax = 50, start x= 20, start y = 34 (then select new curve).Click on the graph to try other starting points, including theequilibrium solution.

If y′ = Ay is a differential equation, how would the solutioncurves behave?

• A. All of the solution curves would run away from theequilibrium point. (Unstable node)• B. The solution curves converge to different points.• C. All of the solutions curves would converge towards

the equilibrium point. (Stable node)• D. The solution curves would race towards the equilib-

rium point and then veer away towards infinity. (Sad-dle)

Correct Answers:

• 11• 7• D

11. (1 pt) Consider the system of differential equations

dxdt

= −1.4x+1y,

dydt

= 1.25x−3.4y.




curves behave?

• A. The solution curves would race towards zero andthen veer away towards infinity. (Saddle)• B. All of the solutions curves would converge towards

0. (Stable node)• C. The solution curves converge to different points.• D. All of the solution curves would run away from 0.

(Unstable node)

The solution to the above differential equation with initialvalues x(0) = 8, y(0) = 6 is

x(t) = ,y(t) = .

Correct Answers:

• -3.9• -0.9• B• -0.666666666666667*1*exp(-3.9*t)+8.66666666666667*1*exp(-0.9*t)• -0.666666666666667*(-3.9 + 1.4)*exp(-3.9*t)+8.66666666666667*(-0.9 + 1.4)*exp(-0.9*t)


dxdt

= 1.6x−0.75y,

dydt

= −1.66666666666667x+3.6y.





curves behave?

• A. The solution curves would race towards zero andthen veer away towards infinity. (Saddle)• B. All of the solution curves would run away from 0.

(Unstable node)• C. The solution curves converge to different points.• D. All of the solutions curves would converge towards

0. (Stable node)


Correct Answers:

• 1.1• 4.1• Competition• B• -9.33333333333333*-0.75*exp(1.1*t)+1.33333333333333*-0.75*exp(4.1*t)• -9.33333333333333*(1.1-1.6)*exp(1.1*t)+1.33333333333333*(4.1-1.6)*exp(4.1*t)

13. (1 pt) Liam opens a bank account with an initial balanceof 2000 dollars. Let b(t) be the balance in the account at time t.Thus b(0) = 2000. The bank is paying interest at a continuousrate of 5% per year. Liam makes deposits into the account at acontinuous rate of s(t) dollars per year. Suppose that s(0) = 800and that s(t) is increasing at a continuous rate of 6% per year(Liam can save more as his income goes up over time).

4

(a) Set up a linear system of the formdbdt

= m11b+m12s,

dsdt

= m21b+m22s.

m11 = ,m12 = ,m21 = ,m22 = .

(b) Find b(t) and s(t).b(t) = ,s(t) = .

Correct Answers:• 0.05• 1• 0• 0.06• (2000 - 800/(0.06 - 0.05)) * eˆ(0.05*t) + 800/(0.06 - 0.05) * eˆ(0.06*t)• 800 * eˆ(0.06*t)

14. (1 pt) Write the given second order equation as its equiv-alent system of first order equations.

u′′−8u′−4.5u =−3.5sin(3t), u(1)=−5, u′(1)=−3.5

Use v to represent the ”velocity function”, i.e. v = u′(t).Use v and u for the two functions, rather than u(t) and v(t). (Thelatter confuses webwork. Functions like sin(t) are ok.)u′ =v′ =Now write the system using matrices:ddt

[uv

]=[ ] [

uv

]+[ ]

and the initial value for the vector valued function is:[u(1)v(1)

]=[ ]

.

Correct Answers:• v• + 8v + 4.5u - 3.5sin(3t)• 0• 1• 4.5• 8• 0• -3.5sin(3t)• -5• -3.5

15. (1 pt) Write the given second order equation as its equiv-alent system of first order equations.

u′′+4u′+4u = 0

Use v to represent the ”velocity function”, i.e. v = u′(t).Use v and u for the two functions, rather than u(t) and v(t). (Thelatter confuses webwork. Functions like sin(t) are ok.)

u′ =v′ =Now write the system using matrices:ddt

[uv

]=[ ] [

uv

].

Correct Answers:

• v• -4v - 4u• 0• 1• -4• -4

16. (1 pt) Calculate the eigenvalues of this matrix:[Note– you’ll probably want to use a graphing calculator to

estimate the roots of the polynomial which defines the eigenval-ues. You can use the web version at xFunctions.

If you select the ”integral curves utility” from the main menu,will also be able to plot the integral curves of the associated dif-fential equations. ]

A =[

-6 -24-3 0

]smaller eigenvalue =associated eigenvector = ( , )larger eigenvalue =associated, eigenvector = ( , )

If y′ = Ay is a differential equation, how would the solutioncurves behave?

• A. All of the solutions curves would converge towards0. (Stable node)• B. All of the solution curves would run away from 0.

(Unstable node)• C. The solution curves would race towards zero and

then veer away towards infinity. (Saddle)• D. The solution curves converge to different points.

Correct Answers:

• -12• ( -4, -1 )• 6• ( 2, -1 )• C


=[

-19 7-42 16

]x

with the initial value x(0) =[

512

].

x(t) =[ ]

.

Correct Answers:

• 3*eˆ(-5*t)*1 + 2*eˆ(2*t)*1• 3*eˆ(-5*t)*2 + 2*eˆ(2*t)*3

5


=[

12 336 9

]x


4-23

].

x(t) =[ ]

.

Correct Answers:• -1*eˆ(21*t)*1 + -5*-1• -1*eˆ(21*t)*3 + -5*4


6



=[

5 -61 0

]x


1-1

].

x(t) =[ ]

.

Correct Answers:

• 3*eˆ(3*t)*3 + -4*eˆ(2*t)*2• 3*eˆ(3*t)*1 + -4*eˆ(2*t)*1


=[

-16 8-8 4

]x


5-2

].

x(t) =[ ]

.

Correct Answers:

• 4*eˆ(-12*t)*2 + -3*1• 4*eˆ(-12*t)*1 + -3*2


dxdt

= 0.4x+0.5y,

dydt

= 1.5x−0.6y.





curves behave?

• A. The solution curves would race towards zero andthen veer away towards infinity. (Saddle)• B. All of the solution curves would run away from 0.

(Unstable node)• C. The solution curves converge to different points.• D. All of the solutions curves would converge towards

0. (Stable node)


Correct Answers:• -1.1• 0.9• SYMBIOSIS• A• 0.5*0.5*exp(-1.1*t)+9.5*0.5*exp(0.9*t)• 0.5*(-1.1-0.4)*exp(-1.1*t)+9.5*(0.9-0.4)*exp(0.9*t)


=[

3 1-10 -3

]x

with x(0) =[

36

].



(right click to open in a new window).Select the ”integral curves utility” from the main menu. ]


? 1. What kind of interaction do we observe?Correct Answers:• 3*(- (-3/1)*sin(1*t) + cos(1*t)) + (6/1)*(sin(1*t))• 3*(-(-3*-3/1)*sin(1*t)-1*sin(1*t)) + 6*(cos(1*t) + (-3/1)*sin(1*t))• ELLIPSE CLOCKWISE• PREDATOR-PREY


=[

2 -55 2

]x

with x(0) =[

3-3

].


? 1. Describe the trajectory.Correct Answers:• eˆ(2*t) * (3*cos(5*t) - -3*sin(5*t))• eˆ(2*t) * (3*sin(5*t) + -3*cos(5*t))• SPIRAL OUTWARD COUNTERCLOCKWISE

1

6. (1 pt) Find the general solution to the homogeneous dif-ferential equation

d2ydt2 −4

dydt

= 0

The solution can be written in the form

y = C1er1t +C2er2t

withr1 < r2

Using this form, r1 = and r2 =Correct Answers:• 0• 4

7. (1 pt)Find the general solution to the homogeneous differential equa-tion

d2ydt2 +15

dydt

+50y = 0

The solution can be written in the form

y = C1er1t +C2er2t

withr1 < r2

Using this form, r1 = and r2 =Correct Answers:• -10• -5

8. (1 pt) Find y as a function of t if

y′′+6y′+58y = 0, y(0) = 3, y′(0) = 5.

y =Note: This problem cannot interpret complex numbers. Youmay need to simplify your answer before submitting it.

Correct Answers:• (3) *(exp((-3)*t))*(cos((7)*t)) + (2) *(exp((-3)*t))*(sin((7)*t))

9. (1 pt) Which of the following functions are solutions ofthe differential equation y′′+2y′+1y = 0?

• A. y(x) = ex

• B. y(x) = xe−x

• C. y(x) = 0• D. y(x) = e−x

• E. y(x) =−ex

• F. y(x) =−x• G. y(x) = xex

Correct Answers:• BCD

10. (1 pt) Find the solution to initial value problem

d2ydt2 +8

dydt

+16y = 0, y(0) = 4,y′(0) = 7

The solution is .Correct Answers:• 4*exp(-4*t) + 23*t*exp(-4*t)


2


1. (1 pt) Use the method of undetermined coefficients to findone solution of

y′′+ y′−5y = 4e2t .

y =(It doesn’t matter which specific solution you find for this prob-lem.)

Correct Answers:• ( 4 ) * exp((2)*t) + c*eˆ(1.79128784747792*t) +d*eˆ(-2.79128784747792*t)

2. (1 pt) Find a particular solution to

y′′+4y = 8sin(2t).

yp =Correct Answers:• -2 * t * cos(2*t) + a*sin(2*t) + b*cos(2*t)

3. (1 pt) Find the solution of

y′′+3y′+2y = 6e1t

with y(0) = 9 and y′(0) = 2.y =

Correct Answers:• ( 1 ) * exp(t) + (4 - 13) *exp((-3/2 - 1/2)*t) + (4 + 13) *exp((-3/2 + 1/2)*t)


y′′+8y′ = 512 sin(8t)+256 cos(8t)

with y(0) = 6 and y′(0) = 5.y =

Correct Answers:• (-6 )*1*cos(8 *t) + (-2 )*1*sin(8 *t) + (6 - 69/8) *exp((-4 - 4)*t) + (6 + 69/8) *exp((-4 + 4)*t)

5. (1 pt) Use the method of undetermined coefficients to findone solution of

y′′−10y′+57y = 16e5t cos(6t)+16e5t sin(6t)+2e3t .

(It doesn’t matter which specific solution you find for this prob-lem.)y =

Correct Answers:• (-4 )*exp((5)*t)*cos(6 *t) + (-4 )*exp((5)*t)*sin(6 *t) + ( 1/18 ) * exp((3)*t) + c*eˆ(5*t)cos(5.65685424949238*t) +d*eˆ(5*t)*sin(5.65685424949238*t)

6. (1 pt) Find a particular solution to

y′′+5y′+4y = 15te3t .

yp =Correct Answers:

• (0.535714285714286 * t + -0.210459183673469) * ((2.71828182845905)**(3*t)) + a*eˆ(-1*t) + b*e**(-4*t)


y′′−4y′+4y = 343e9t

with y(0) = 1 and y′(0) = 9.y =

Correct Answers:

• ( 7 ) * exp((9)*t) + (-6) *exp((2)*t) + (-42)*t *exp((2)*t)

8. (1 pt) This problem is an example of critically dampedharmonic motion.

A hollow steel ball weighing 4 pounds is suspended from aspring. This stretches the spring 1

8 feet.The ball is started in motion from the equilibrium position

with a downward velocity of 9 feet per second. The air resis-tance (in pounds) of the moving ball numerically equals 4 timesits velocity (in feet per second) . Suppose that after t secondsthe ball is y feet below its rest position. Find y in terms of t.

Take as the gravitational acceleration 32 feet per second persecond. (Note that the positive y direction is down in this prob-lem.)

y =

(Show the student hint after 1 attempts: )When using English units (lb, ft, etc.) you need to be a bitcareful with equations involving mass. This is causing someconfusion — I hope this hint helps:

Pounds (lb) is a unit of force, not mass. Using mg=F andg=32 ft/sec2weseethatanob jectatthesur f aceo f theearthwhichweighs32lbs(i.e.the f orceonitis32lbs)willhaveamasso f 1(theunito f massintheEnglishunitsiscalledtheslug−−really!)Sooneslugweighs32lbsatthesur f aceo f theearth(orlb =(1/32)∗ slug∗ f t/sec2).

When using metric units, kilogram is a unit of mass not forceor weight. A 1 kilogram mass will weigh 1 * 9.8 newtons on thesurface of the earth. (g= 9.8 m/sec2andnewton = kg∗m/sec2).

Saying that a mass ”weighs” 1 kilogram is technically incor-rect useage, but it is often used. What one really means is that ithas 1 kilogram of mass and therefore weighs 9.8 newtons.

1

Correct Answers:

• (0) *exp((-16)*t) + (9)*t *exp((-16)*t)


2


1. (1 pt) Find the inverse Laplace transform of

4s+8s2−49

s > 7

y(t) = .Correct Answers:• 4*cosh(7*t)+1.14285714285714*sinh(7*t)


4s+8s2 +13

s > 0

y(t) = .Correct Answers:• 4*cos(3.60555127546399*t)+2.21880078490092*sin(3.60555127546399*t)

3. (1 pt) Use the Laplace transform to solve the followinginitial value problem:

y′′+3y′ = 0 y(0) =−3, y′(0) = 7First, using Y for the Laplace transform of y(t), i.e., Y =

L{y(t)},find the equation you get by taking the Laplace transform of thedifferential equation

= 0Now solve for Y (s) =

and write the above answer in its partial fraction decomposition,Y (s) = A

s+a + Bs+b where a < b

Y (s) = +Now by inverting the transform, find y(t) =

.Correct Answers:• sˆ2*Y--3*s-7+3*(s*Y--3)• (-3*s+-2)/(sˆ2+3*s)• -0.666666666666667/s• -2.33333333333333/(s+3)• -0.666666666666667+-2.33333333333333*exp(-3*t)


y′′−3y′−28y = 0 y(0) =−1, y′(0) = 2

First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},find the equation you get by taking the Laplace transform of thedifferential equation

= 0

Now solve for Y (s) =and write the above answer in its partial fraction decomposition,Y (s) = A

s+a + Bs+b where a < b

Y (s) = +

Now by inverting the transform, find y(t)= .Correct Answers:

• sˆ2*Y + 1*s-2 - 3*(s*Y + 1) - 28*Y• (-1*s+5)/(sˆ2 - 3*s - 28)• -0.181818181818182/(s-7)• -0.818181818181818/(s+4)• -0.181818181818182*exp(7*t) - 0.818181818181818*exp(-4*t)


y′′−10y′+41y = 0 y(0) = 0, y′(0) = 4

First, using Y for the Laplace transform of y(t), i.e., Y =L{y(t)},find the equation you get by taking the Laplace transform of thedifferential equation

= 0Now solve for Y (s) =By completing the square in the denominator and inverting

the transform, findy(t) = .

Correct Answers:

• sˆ2*Y-4-10*s*Y+41*Y• 4/(sˆ2-10*s+41)• exp(5*t)*sin(4*t)

6. (1 pt) Consider the following initial value problem:

y′′−7y′−8y = sin(3t) y(0) = 1, y′(0) = 4

Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},find the equation you get by taking the Laplace transform of thedifferential equation and solve forY (s) =

Correct Answers:

• (1*s - 3)/(sˆ2 - 7*s - 8)+3/((sˆ2 - 7*s - 8)*(sˆ2+9))

1


y′′+−6y′+9y = 0 y(0) = 3, y′(0) =−3

First, using Y for the Laplace transform of y(t), i.e., Y =L{y(t)},find the equation you get by taking the Laplace transform of thedifferential equation

= 0

Now solve for Y (s) =and write the above answer in its partial fraction decomposition,Y (s) = A

s+a + B(s+a)2

Y (s) = +

Now by inverting the transform, find y(t)= .Correct Answers:

• sˆ2*Y-3*s--3+-6*(s*Y-3)+9*Y• (3*s+-21)/(sˆ2+-6*s+9)• 3/(s+-3)• -12/(s+-3)ˆ2• 3*exp(3*t)+-12*t*exp(3*t)

8. (1 pt) Find the Laplace transform of

f (t) =−2u2(t)−1u5(t)+2u8(t)

F(s) = .Correct Answers:• -2*exp(-2*s)/s - 1*exp(-5*s)/s+2*exp(-8*s)/s

9. (1 pt) Find the Laplace transform of

f (t) =

{0, t < 8(t−8)3, t ≥ 8

F(s) = .Correct Answers:• exp(-8*s)*6/sˆ4


F(s) =e−6s

s2−2s−8

f (t) = . (Use step(t-c) foruc(t) .)

Correct Answers:• step(t-6)*(-0.166666666666667*exp(-2*(t-6))+0.166666666666667*exp( + 4*(t-6)))


2

Ed Azoff Math2700 Azoff F11 - University of Georgiamath.uga.edu/~azoff/courses/ww1108.pdf · Ed Azoff Math2700 Azoff F11 WeBWorK assignment number wwassign1 is due : 08/22/2011 at

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