Assignment 17

course Mth 174

ωꑊω“oΰ«ΜΡW‘Š‰…α½QψBzξΐ’θ…ΰΝι„assignment #017

017. `query 17

Cal 2

06-25-2009

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11:13:35

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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y = cos(wt)

y ' = -wcos(wt)

y '' = -w^2cos(wt)

y '' +9y = 0

-w^2 cos(wt) + 9 cos(wt) = 0

w = 3

Good, but note that the symbol is the Greek letter omega, not w.

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11:20:48

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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P = 1 / ( 1 + e^-t)

P ' = ln (1 + e^-t)

P' = P(1 - P)

ln (1 + e^-t) = 1 / ( 1 + e^-t) (1 -1 / ( 1 + e^-t))

I wasn't sure what to do from here

You did not use the chain rule in taking your derivative; this will lead to an error.

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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11:20:54

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

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RESPONSE -->

P = 1 / ( 1 + e^-t)

P ' = ln (1 + e^-t)

P' = P(1 - P)

ln (1 + e^-t) = 1 / ( 1 + e^-t) (1 -1 / ( 1 + e^-t))

I wasn't sure what to do from here

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11:21:22

What is the derivative dP/dt?

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dp/dt = ln(1 + e^-t)

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11:22:24

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

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11:29:27

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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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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11:31:29

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

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for y '' = y I found that y = e^x + e^-x is a solution. I found this by taking the second derivitive of y and substituting into the equation y '' = y.

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11:33:38

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

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for y ' = -y I found the solution to be y = sqrt (2x) becaue the derivitive of y = (2x)^1/2 which equals y.

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11:34:18

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

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I was unable to figure out which one was this solution equation.

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11:35:07

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

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For y '' = -y I found the solution equations y = cosx and y =cos(-x). The second derivitve of both equations is equal to the negative original equation.

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11:36:29

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

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For x^2y'' - 2y = 0 I found the solution equation y = x^2 .

y''= 2 so

2x^2 - 2x^2 = 0

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11:42:04

Query problem 11.2.10 (was 10.2.6) slope field

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For (0,0) the graph is first decreasing at an increasing rate and concave up. Then just before point (0,0) the graph is concave down. At point (0,0) the graph begins increasing at an increasing rate and is concae up. As y increases the graph begins to be concave down. The graph looks the same at point (1,4) except it begins at x = 1.

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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11:45:27

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

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y' = xe^-x graph has a slope of all vertical lines on the left side of the graph that begin to equal to the y value as x increases.

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11:46:01

describe the slope field corresponding to y' = sin x

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This slope field increases and decreases at a constant rate from - infinity to + infinity.

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11:46:56

describe the slope field corresponding to y' = cos x

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This slope field looks similar to y' = sinx. I'm not sure how to tell the difference.

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11:49:42

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

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I had a hard time understanding these. I wasn't sure how to figure out the slope fields after integrating them.

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11:49:45

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

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11:49:53

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

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11:50:56

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

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I had a hard time with 11.2 # 10. I understood the notes for that class but in these examples there are very slight changes in some of the graphs and I'm not sure how to differentiate between them.

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&#Good work. See my notes and let me know if you have questions. &#