Assignment 17

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course MTH 174

3:24 p.m. 7/21/2010

¦‚ω‰ηΰ―Jξ„δ™Π”ζ|‘ΔYΫ°«mϊassignment #017

017. `query 17

Cal 2

07-21-2010

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14:13:10

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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RESPONSE -->

First we find y'' = -omega^2 cos(omega*t). Substitute into the equation: [-omega^2 cos(omega*t)] + 9 cos (omega*t)=0 So add the first term to both sides

and then divide by cos (omega*t) to leave: omega^2 = 9 which can easily be solved to give: omega = +- 3.

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14:21:13

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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RESPONSE -->

Hera again we find dP/dt and it is: (e^-t)/(1+e^-t)^2. Substitution then produces: e^-t / (1+e^-t)^2 = 1/(1+e^-t) * (1 - 1/(1+e^-t)). Now I used several

steps of algebra and rearranging: first I multiplied both sides by: 1+e^-t, then I added 1/(1+e^-t) to both sides and factored out the 1/(1+e^-t). This is

what it looks like at this point: 1/(1+e^-t) (e^-t+1)= 1. I multiply both sides by 1/(1+e^-t) and this makes: e^-t + 1 = 1+e^-t. These expressions are

equivalent and so we shown that the given P function does indeed satisfy the logistic equation.

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14:21:45

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

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RESPONSE -->

I used algebra to rearrange the equation into equivalent expressions as shown above.

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14:22:12

What is the derivative dP/dt?

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RESPONSE -->

dP/dt = e^-t/ (1+e^-t)^2.

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14:22:45

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

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RESPONSE -->

I simplified to equivalent expressions so yes. I showed the details in my work above.

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14:23:34

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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RESPONSE -->

I will answer these individually in the next few questions.

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14:26:01

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

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RESPONSE -->

y''=y could only be satisfied by IV: y= e^x+e^-x. The only part that changes in the derivatives of this equation is the e^-x. It alternates between + and

-. y' is negative and then y'' will be +, which is the original function.

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14:26:51

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

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RESPONSE -->

There are no functions in the list that satisfy this equation. I can tell by taking the first derivative of each function and none are -y.

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14:30:31

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

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RESPONSE -->

This is satisfied by V: y=sqrt(2x). The derivative of function is easily found to be: y'=1/sqrt(2x) = 1/y

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14:32:40

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

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RESPONSE -->

This is satisfied by two functions, both I and II. The second derivative of cos x is -cos x. Cos x and cos -x are already equivalent expressions and so

follows that its second derivative is also equivalent to cos x.

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14:34:59

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

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RESPONSE -->

We've already found a few second derivatives and none wil fit here. The last function we haven't looked at is y= x^2. Here y''= 2. Substitution then

produces: x^2 (2)- 2(x^2)=0 This can be rearranged to: 2 x^2= 2x^2.

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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14:41:31

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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RESPONSE -->

The solution through the point (0,0) is merely a line along the x-axis. The (1,4) solution is concave up for (-inf, 1) interval and concave down for the

(1,inf) interval. The point (1,4) is a point of inflection and the graph is asymptotic at y=7.

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14:41:57

Query problem 11.2.10 (was 10.2.6) slope field

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RESPONSE -->

I will answer these in the individual questions below.

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14:51:23

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

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RESPONSE -->

The slope field is figure (III). For values of x <0 the slopes increase in negativity from left to right. Around x= 1 the slopes become positive and then

flatten out as you move to the right.

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15:00:24

describe the slope field corresponding to y' = sin x

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RESPONSE -->

The corresponding slope field is figure (I). The slopes follow the sinusoidal pattern alternating between 0 and 1. The slopefield differs from the cos x

slopefield in that the x=0 is -1 for this function and 0 for cos x. This also has a period 2pi.

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15:03:35

describe the slope field corresponding to y' = cos x

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RESPONSE -->

Here we have the figure to be (II). This one is the one that is also periodic and has the point (0,0). The period here pi/2.

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15:08:00

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

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RESPONSE -->

The figure here is IV. This graph is very similar to VI. The slopes for x<-0 are increasingly positive. Around x=0 The slopes are decreasingly positive

until they are nearly 0 at x=0. Then they increase positively again for x>0.

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15:13:59

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

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RESPONSE -->

The figure here is (V). The slopes here are nearly 0 for x<-3. Between x=-3 and x=0 the slopes are increasingly positive and then 0

decreasingly positive and then nearly 0 again.

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15:18:01

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

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RESPONSE -->

The figure for this is (VI). This graph is very similar to IV. The slopes for decreasing x values are increasingly positive and nearly asymptotic as x->-

inf. They are decreasingly positive starting about x=-3 and as x grows larger the level out to nearly 0.

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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15:18:54

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

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RESPONSE -->

Differential equations have always interested me for some reason. I may not feel the same way by the end of this chapter.

THis chapter actually gives you a very solid introduction to differential equations. More, I'm afraid, than the differential equations courses taught at some 4-year schools.

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Good work. I've inserted the 'given solutions' to a couple of the problems, for comparison.

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