Ch.1;Math240;Writtenhw

5/23/2015 Math 240 Homework

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Math 240 Home, Textbook Contents, Online Homework Home

Math 240 Written Homework 1

1. What is the order of the following equations?

a.

b.

c.

d.

e.

2. Which of the following equations are separable? Write them in separated form if possible.

a.

b.

c.

d.

e.

3. Find the general solution to




7. Solve the initial value problem


= +dy

dxx3 y2

= x − yyd3

dx3

= cos(y) − sin(x)( )dy

dx

2

+ 3 − 6y = exp(2x)yd2

dx2

dy

dx

cos ( ) = 2xydy

dx

= xydy

dx

=dy

dx

x

y

= x + ydy

dx

= x − ydy

dx

= exp(x − y)dy

dx

= 2xdy

dxey

= y(1 − y)dy

dx

x + = xdy

dxey

+ y + y = 0dy

dxx2

= − 9,dy

dxy2 y(0) = 1.

= xy − 3x,dy

dxy(0) = 0.

dy

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11. Find where is the solution to

12. Find where is the solution to

13. Find given that the solution to , satisfies

14. Find given that the solution to , satisfies

15. Find so that the integral curve for

is a straightline. You must justify your answer, which will require you to apply algebraicreasoning to the problem. Just using the slope field lab applet to find the answer by repeatedguessing will not get you full credit.

16. Find so that the integral curve for

is a straightline. You must justify your answer, which will require you to apply algebraicreasoning to the problem. Just using the slope field lab applet to find the answer by repeatedguessing will not get you full credit.

17. Find an initial value problem that is solved by .

18. Find an initial value problem that is solved by .

19. Find a differential equation whose general solution is the set of curves, Remember that is the arbitrary constant here so it should not appear in the differentialequation.

20. Find a differential equation whose general solution is the collection of all circles about theorigin, . Remember that is the arbitrary constant here so it should not appearin the differential equation.

Technical note: Since a circle is not the graph of a function as it fails the verticalline test, it would be more proper to say the general solution consists of allsemicircles of the form . But the problem is easier to understandand to solve written as above.

21. Classify whether the following equations are exact, linear, both, or neither. Youdon't have to solve them, just classify them.

= (y),dy

dxcos2 y(0) = 0.

= (y),dy

dxcos2 y(0) = π/2.

y(1) y(x) = 2y,dy

dxy(0) = 1.

y(2) y(x) − y = xy,dy

dxy(0) = 1.

y0 = xdy

dxy2 y(0) = ,y0 y(1) = 1.

y0 =dy

dxex+y y(0) = ,y0 y(1) = 2.

y0

= −dy

dx

y

2x

3y(−4) = y0

y0

= x + ydy

dxy(−4) = y0

y = ex2

y = esin(x)

+ = .ex ey C2

C

+ =x2 y2 r2 r

y = ± −r2 x2− −−−−−√



a.

b.

c.

d.

22. Classify whether the following equations are exact, linear, both, or neither. Youdon't have to solve them, just classify them.a.

b.

c.

d.





27. Solve the initial value problem ,




31. Find where is the solution to the initial value problem

Note that solving this exact equation will lead to an implicit formula for whichwill have two solutions for . However, the initial value problem has a uniquesolution and so just one of the two solutions produced from the implicit formula willbe the actual value of Justify your choice of value for You can do thisalgebraically (say by solving the implicit formula to get an explicit formula for in

= −dy

dx

y

x

= x + ydy

dx

= + ydy

dxx2

=dy

dx

y + x2

exp(y) − x

(x + y)dx + dy = 0

( + y )dx + ( − y)dy = 0x2 ex ex

(1 + y)dx + dy = 0

(x + )dx + ( + y)dy = 0y2 x2

= exp(x) − .dy

dx

y

x

= .dy

dx

exp(x) − y

x − sin(y)

(x + y)dx + dy = 0.

(2x + y)dx + (x − 2y)dy = 0.

+ 2y = exp(x)dy

dxy(0) = 2.

− 3y = x − 1dy

dxy(0) = 0.

( + 2xy)dx + ( − y)dy = 0x2 x2 y(0) = 1.

(cos(x) − )dx + (cos(y) − 2xy)dy = 0y2

y(0) = 0.

y(2) y(x)

dy

dx

y(0)

=x − y

x + 2y

= 1

yx = 2

y(2). y(2).y



terms of where the term will be specified to either be or ) orgeometrically (perhaps by arguing from the slope field that the integral curvestarting at (0,1) can only reach one of the two possible values because it can't gobelow the xaxis). Just saying you checked the slope field applet and can see whichvalue must be right won't get full credit, you must justify how you can tell theintegral curve can't go below the xaxis.

32. Find where is the solution to the initial value problem

Note that solving this exact equation will lead to an implicit formula for whichwill have two solutions for . However, the initial value problem has a uniquesolution and so just one of the two solutions produced from the implicit formula willbe the actual value of Justify your choice of value for You can do thisalgebraically (say by solving the implicit formula to get an explicit formula for interms of where the term will be specified to either be or ) orgeometrically (perhaps by arguing from the slope field that the integral curvestarting at (0,1) can only reach one of the two possible values because it can't gobelow the xaxis). Just saying you checked the slope field applet and can see whichvalue must be right won't get full credit, you must justify how you can tell theintegral curve can't go below the xaxis.


and then use your solution to evaluate Your answer should be accurate to thenearest 0.0001. Note that you will get an integral here that you can't computeanalytically. You will need to use a numerical method to complete the problem (theTI84 and similar calculators have an excellent numerical integration routine builtin, or you can use WolframAlpha.)


and then use your solution to evaluate Your answer should be accurate to thenearest 0.0001. Note that you will get an integral here that you can't computeanalytically. You will need to use a numerical method to complete the problem (theTI84 and similar calculators have an excellent numerical integration routine builtin, or you can use WolframAlpha.)


x ± + −

y(1) y(x)

dy

dx

y(0)

=2x − 2y

2x + y

= 3√

yx = 1

y(1). y(1).y

x ± + −

dy

dxy(0)

= 2xy + sin(x)

= 1

y(1).

dy

dxy(0)

= 1 − cos(x)y

= 2

y(2).

dy

dx

y(0)

= −y + exp(x)

x + cos(y)= 0

(2).

http://www.wolframalpha.com/

http://www.wolframalpha.com/



and then use your solution to evaluate Your answer should be accurate to thenearest 0.0001. In this case you will get an implicit formula that you won't be ableto solve analytically for in terms of . You can substitute for in the solutionto get an equation for which you can then solve numerically quickly usingNewton's method (which you can find described in your calculus textbook oronline). Since Newton's method has better convergence properties with a goodstarting value, you may want to approximate using Euler's method or theimproved Euler's method with a stepsize of 1 or 2 to get your initial guess. Notethat the work to finish the solution using Newton's method is much less than thework to get a numerical approximation to such a fine tolerance using the numericalmethods like these to solve the differential equation directly.


and then use your solution to evaluate Your answer should be accurate to thenearest 0.0001. In this case you will get an implicit formula that you won't be ableto solve analytically for in terms of . You can substitute for in the solutionto get an equation for which you can then solve numerically quickly usingNewton's method (which you can find described in your calculus textbook oronline). Since Newton's method has better convergence properties with a goodstarting value, you may want to approximate using Euler's method or theimproved Euler's method (a stepsize of 1 will be fine) to get your initial guess.Note that the work to finish the solution using Newton's method is much less thanthe work to get a numerical approximation to such a fine tolerance using thenumerical methods like these to solve the differential equation directly.

37. Find the general solution for . Hint: if you treat as a function of , the equation is linear, i.e. it has an integrating factor that is afunction of alone.

38. Show that the equation has an integrating factor ofthe form for some value of This will be true for any equation of theform where the value of willdepend on the values of , , and , but you don't have to justify the general case.

39. Show that any separable equation, which can by definition be written in the form will also be exact. You may observe (but you don't have to

justify) that the technique for solving separable equations is just a special case ofthe paradigm for solving exact equations.

40. Suppose solves the initial value problem

while solves the initial value problem

y(2).

y x 2 xy

y(2)

dy

dx

y(0)

=sin(x) − exp(x + y) + 2xy

3 − + exp(x + y)y2 x2

= 0

y(1).

y x 1 xy

y(1)

exp(y)dx + (2x exp(y) − 1)dy = 0x y

y

(x + 4y)dx + (3x + 6y)dy = 0(x + y)r r.

(ax + by)dx + ((a + c)x + (b + c)y)dy = 0, ra b c

N(y)dy = M(x)dx,

w(x)

+ p(x)wdw

dxw(0)

= q(x)

= 0

z(x)

+ p(x)zdz

dxz(0)

= 0

= y0

( ) = ( ) + ( )

http://en.wikipedia.org/wiki/Newton%27s_method

http://en.wikipedia.org/wiki/Newton%27s_method



Show that solves the initial value problem

The ability to cut a problem into pieces, solve each piece separately, and add upthe answers to each piece to find an answer to the overall problem is afundamental feature of "linear" equations in mathematics. We will discuss thismore in chapter 2.

In problems 41 and 42, List the technique you would use to solve the followingproblem, separable, exact, linear, Bernoulli, homogeneous, or none of the above.Note that you will lose a point for choosing homogeneous for an equation that canbe solved using the exact paradigm, even if homogeneous would work (exact is somuch easier and you are much less likely to make a mistake). You don't need tosolve the equations, just list the technique you would use.

41.

a.

b.

c.

d.

42. a.

b.

c.

d.



45. Solve the initial values problem ,

y(x) = w(x) + z(x)

+ p(x)ydy

dxy(0)

= q(x)

= y0

=dy

dx

x − y

2y − x

=dy

dx

x + y

2 − xy2

=dy

dx

x + y

2 + xy2

= y − exp(x)dy

dx

=dy

dx

+ 2xy −x2 y2

− 2xy − 3x2 y2

=dy

dx

+ 2xy −x2 y2

− + 2xy − 3x2 y2

= exp(x − y)dy

dx

= xy +dy

dxy2

+ 2y = exp(x) .dy

dxy2

= .dy

dx

3x + 13y

15x + y

=dy

dx

3y − 2x

4y − 6xy(1) = 2.

dy



46. Solve the initial values problem ,

In the Change of Variables section we went over two types of equations that can besolved by making substitutions, Bernoulli equations and homogeneous equations.There are a lot of other substitutions that apply to other types of equations. Thekeys for all the different substiution methods are to rewrite the original variable in terms of the new variable and to be sure to make the substitution everywherein the equation, including making the substitution in the derivative term, which mayrequire use of derivative rules such as the chain rule (as in Bernoulli equations) orthe product rule (as in homogeneous equations). Two other common types ofequations that can be solved by substitutions are

Any equation of the form , where , , and are

constants, can be transformed to a separable equation via the substitution .

Equations of the form are called Riccati

equations. If you know any one particular solution, , to a Riccatiequation, you can find the general solution by transforming the Riccati

equation to a linear equation with the substitution .

Using these tricks you can solve problems 4752 below. Of course, to solve aRiccati equation you must first find one particular solution, but you can find one byguess and check more often than you might imagine. In context, you may knowone obvious solution, or you might look at the slope field and be able to spot onesimple solution like or or (especially if the problemappears in an undergraduate math class).




50. Solve the initial value problem .

51. Solve the initial value problem .


53. Find and classify (as stable, unstable, or semistable) all the equilibrium solutionsto

− 3y =dy

dx

x

yy(0) = 1.

yv

= f(αx + βy + γ)dy

dxα β γ

v = αx + βy + γ

= A(x) + B(x)y + C(x)dy

dxy2

(x)y1

v =1

y − (x)y1

y = 1 y = x y = exp(x)

= .dy

dx

x − 1 + y − xy2

x

= − 2xy + .dy

dxy2 x2

= 2 exp(x) − exp(−x) .dy

dxy2

= exp(x − y) + 1,dy

dxy(0) = 1

= − 2xy + + 1,dy

dxy2 x2 y(0) = 1

= ( − x + 1) + (1 − 2x)y + ,dy

dxx2 y2

y(0) = 1.



a.

b.

54. Find and classify (as stable, unstable, or semistable) all the equilibrium solutionsto

a.

b.

55. Suppose , Explain how we can tell that for

all . Hint: sketch the slope field, specifically noting how the slope field looksalong the lines and . Explain why an integral curve can't leave thewedge in the first quadrant formed by these two lines.

56. Two students are asked if the solution to the initial value problem. ,

can ever take a negative value for Kasey argues that can't benegative because so the derivative is always nonnegative, hence isalways increasing or flat, and, since we increase from must always be

positive. On the other hand, Riley solves the problem to get and

computes hence y can be negative. Write a paragraph

explaining whether Kasey or Riley (or both or neither) is correct.

Suppose a fish population experiences logistic growth with growth rate per yearand environmental carrying capacity . The population is harvested at a rate of per year ( might be a constant or it might be variable depending on thepopulation). This situation is described by the differential equations

where is the population and is time (in years). It is desired to harvest the fishto reach the maximum sustainable yield. That is, if you harvest too few fish, thenyou aren't getting all the fish you could. But if you harvest too many fish, you willdrive the population too low (perhaps to extinction) and so your harvest won't besustainable. The next 4 problems take you through the calculations of themaximum sustainable yield for two simple harvesting models.

57. Suppose the growth rate is (20%), the carrying capacity is and the harvest is a constant fish per year.a. What are the equilibria for this model? Are they stable or unstable?b. What is the maximum value of so that the model has an equilibrium with

?Hint: as always when determining equilibria, drawing a graph of vs. may

= − 4 + 3ydy

dxy3 y2

= ( + 4y + 3)( − 1)dy

dxy2 y2

= sin(y)dy

dx

= cos(y)dy

dx

= −dy

dxx2 y2 y(1) = 0. 0 < y(x) < x

x > 1y = x y = 0

= 3dy

dxx2y2

y(0) = 1 x > 0. y3 > 0,x2y2 y

y = 1, y

y =1

1 − x3

y(2) = = − < 01

1 − 23

17

rC h

h

= rp(1 − p/C) − hdp

dt

p t

r = 0.2 C = 25, 000,H

Hp > 0

dp/dt p



be helpful.

58. Suppose the harvest is a constant fish per year. What is the maximum value of so that the model has an equilibrium with ? Your answer this time will

have as a function of and , in contrast to the previous problem where youwere given specific values for and

59. Suppose the harvest is not a constant but is a fixed proportion of the totalpopulation, so where is the proportion of the population that isharvested.a. What are the equilibria for the model now?b. Which equilibrium is stable and which is unstable?

Note that the equilibrium populations will depend on the value of , so youranswers will be functions as well as and

60. Assuming you end up at the stable equilibrium in problem 59, what proportion maximizes the harvest (recall where is now a function of , , and )?What is the maximum sustainable yield for this approach?

In the 1950's official U.S. policy was to aim for the maximum sustainable yield ascomputed above in fisheries management. However, this model is quite crude(ignoring age and size of the harvested fish among other issues) and the targetvalues developed using it often turned out to be too high in practice, leading tooverfishing.

61. As of Sept. 12, 2014, Wolfram Alpha is able to correctly solve the equation

with the initial value , but not with the initial value

. Explain why you think this is. You may find it useful to ask WolframAlpha to find the general solution without initial values.

62. Consider the initial value problem The true solution is

but here we will see how Picard iteration works for this problem.The iteration will start with Compute

a.

b. (this time you have to work out the integral to evaluate for yourself).

Answers to odd numbered exercises

If you have any problems with this page, please contact [email protected].©2010 Andrew G. Bennett

HH p > 0

H r cr C.

h = Ep E

EE r C

Eh = Ep p E r C

h

= − 5y + 6dy

dxy2 y(0) = 4

y(0) = 2

= y,dy

dxy(0) = 1.

y(x) = exp(x),(x) = 1.y0

(x) = (t)dt + 1y1 ∫ x

0y0

(x)y2

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