Assignment 17

course Mth 174

I was just wondering whenever you get my second test in the mail could you e-mail me my results. i dont know if i done as good as the first one but i really did try hard. I am trying my best here with this math its just hard to learn over the net and on my own. I really just want my C so i can graduate.

•ÎðØ´zÒÅÃ`Þ‹˜«£ñÄÊêCÎçH€±ýÞassignment #017

I did receive a note from your proctor indicating that the test has been mailed. Hopefully it will be in this morning's mail.

017. `query 17

Cal 2

11-12-2007

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21:52:53

Query problem 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).{}{}Give your solution.

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k we have to first substitue the the function of y = cos(omega*t) into the differential equation of

d^2*y / dt^2 + 9y = 0

so if we do that we end up getting

2(cos(omega*t)) + 9(cos(omega*t)) = 11(cos(omega*t))

Substitute y = cos(omega * t) into the equation. The goal is to find the value of omega that satisfies the equation:

We first calculate y ‘’

y = cos(omega*t) so

y' = -omega*sin(omega*t) and

y"" = -omega^2*cos(omega*t)

Now substituting in y"" + 9y = 0 we obtain

-omega^2*cos(omega*t) + 9cos(omega*t) = 0 , which can be easily solved for omega:

-omega^2*cos(omega*t) = -9cos(omega*t)

omega^2 = 9

omega = +3, -3

Both solutions check in the original equation.

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21:58:46

Query problem 11.1.14 (3d edition 11.1.13) (was 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P)

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we must so that P = 1 / (1+e^-t) is a solution to the equation of dP/dt = P(1 - P)

so if we do this we look at the equation first which is

dP/dt = P(1 - P) and then we decide wheather the function of P = 1 / (1 + e^-t) is a solution of this so we substitue it into the differential equation of

dP/dt = P(1 -P) and we get the equation of once we substitute of

1/ (1+e^-t) * (1 - (1/ (1+e^-t)) so when the math is down we just end up with it equallying one because when you do the right half of the equation involving

(1 - (1/(1+e^-t)) we just get a positive number there which is one and on the right side we get one so one time one is just equal to 1 so this is a solution.

Good start but there's a lot more algebra to it and you have to take the derivative of P:

P = 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have

P = e^t/(e^t+1) , which is a form that makes the algebra a little easier.

dP/dt = [ (e^t)’ ( e^t + 1) – (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to

dP/dt = [ e^t ( e^t + 1) – e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2.

Substituting:

P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt:

e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) – (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator

e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] – (e^t)^2 / (e^t + 1 ) ^ 2

= (e^(2 t) + e^t) / (e^t + 1)^2 – e^(2 t) / (e^t + 1) ^ 2

= (e^(2 t) + e^t – e^(2 t) ) / (e^t + 1)^2

= e^t / (e^t + 1)^2.

This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 – P).

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21:58:56

how did you show that the given function satisfies the given equation?

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shown in the first screen

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21:59:02

What is the derivative dP/dt?

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first screen

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21:59:10

Does P(1-P) simplify to the same expression? If you have already shown the details, show them now.

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first screen

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22:03:07

Query problem 11.1.16 (3d edition 11.1.15) (was 10.1.1) match equation with solution (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )

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well we will look at each differential equation first so

A) y'' = y and its solution can be (III) y = x^2

y '' = y would give us (x^2) '' = x^2. Since (x^2)'' = 2 we would have the equation 2 = x^2. This is not true, so y = x^2 is not a solution to this equation.

B) y' = -y and its solution can be (II) y = cos(-x)

C) y' = 1/y and its solution can be (V) y = sqrt(2x)

D) y'' = -y and its solution can be (II) y = cos(-x)

E) x^2y'' - 2y = 0 and its solution can be (IV) y = e^x + e^-x

If y = e^x + e^-x then y '' = (e^x) '' + (e^-x) '' = e^x + e^-x and this equation would give us

x^2 ( e^x + e^-x) - 2 ( e^x + e^-x) = 0.

This is not so; therefore y = e^x + e^-x is not a solution to this equation.

The function y = cos(x) has derivative y ‘ = -sin(x), which in turn has derivative y ‘’ = - cos(x). If we plug these functions into the equation y ‘’ = -y we get –cos(x) = - ( cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = cos(-x) has derivative y ‘ = sin(-x), which in turn has derivative y ‘’ = - cos(-x). If we plug these functions into the equation y ‘’ = -y we get –cos(-x) = - (cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = x^2 has derivative y ‘ = 2 x , which in turn has derivative y ‘’ = 2. Plugging into the equation x^2 y ‘’ – 2 y = 0 we get x^2 * 2 – 2 ( x^2) = 0, or 2 x^2 – 2 x^2 = 0, which is true.

The function y = e^x + e^(-x) has derivative y ‘ = e^x – e^(-x), which in turn has derivative y ‘’ = e^x + e^(-x). It is clear that y ‘’ and y are the same, so this function satisfies the differential equation y ‘’ = y.

The function y = sqrt(2x) has derivative y ‘ = 2 * 1 / (2 sqrt(2x)) = 1 / sqrt(2x) and second derivative y ‘’ = -1 / (2x)^(3/2). It should be clear that y ‘ = 1 / sqrt(2x) is the reciprocal of y = sqrt(2x), so this function satisfies the differential equation y ‘ = 1 / y.

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22:03:12

which solution(s) correspond to the equation y'' = y and how can you tell?

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22:03:17

which solution(s) correspond to the equation y' = -y and how can you tell

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22:03:19

which solution(s) correspond to the equation y' = 1/y and how can you tell

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22:03:21

which solution(s) correspond to the equation y''=-y and how can you tell

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22:03:22

which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell

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22:08:39

Query problem 11.2.5 (3d edition 11.2.4) The slope field is shown. Plot solutions through (0, 0) and (1, 4).{}{}Describe your graphs. Include in your description any asymptotes or inflection points, and also intervals where the graph is concave up or concave down.

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there is a line that the lines do not cross in this graph and those are the lines of x = 0 and x = 4 so the graph is just like in the s formation between 0 and 4 all the way from the negatives to the positives at the bottom of the graph it is concave down and at the top of it the graph is concave up.

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22:13:54

Query problem 11.2.10 (was 10.2.6) slope field

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A) y' = xe^-x matches the slope field of (VI)

B) y' = sinx matches the slope field of (II)

C) y' = cosx matches the slope field of (I)

D) y' = x^2e^-x matches the slope field of (III)

E) y' = e^-x^2 matches the slope field of (V)

F) y' = e^-x matches the slope field of (IV)

You don't give your reasoning, but despite some errors the pattern of your answers appears to demonstrate a degree of understanding. Consider the following:

Field I alternates positive and negative slopes periodically and is symmetric with respect to the y axis, so slope at –x is negative of slope at x, implying an odd function. The periodic odd function is sin(x). So the field is for y ‘ = sin(x).

Field II similar but slope at –x is equal to slope at x, implying an even function. The periodic even function is cos(x). So the field is for y ‘ = cos(x).

Field III is negative for negative x, approaching vertical as we move to the left, then is positive for positive x, slope increasing as x becomes positive then decreasing for larger positive x. The equation is therefore y ‘ = x e^(-x).

Field IV has all non-negative slopes, with near-vertical slopes for large negative x, slope 0 on the y axis, slope increasing as x becomes positive then decreasing for larger positive x. These slopes represent the equation y ‘ = x^2 e^(-x).

Field V has all non-negative slopes, about slope 1 on the x axis, slopes approaching 0 as we move away from the x axis. This is consistent with the equation y ‘ = e^(-x^2).

Field VI has all positive slopes, with slope approaching zero as x approaches infinity, slope approaching infinity (vertical segments) for large negative x. The field therefore represents y ‘ = e^(-x).

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22:14:01

describe the slope field corresponding to y' = x e^-x

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first page

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22:14:03

describe the slope field corresponding to y' = sin x

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22:14:04

describe the slope field corresponding to y' = cos x

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22:14:06

describe the slope field corresponding to y' = x^2 e^-x

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22:14:07

describe the slope field corresponding to y' = e^-(x^2)

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22:14:09

describe the slope field corresponding to y' = e^-x

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22:14:13

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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See my notes and let me know if you have questions.