Assignment 17

course Mth 174

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017. `query 17

Cal 2

08-08-2009

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assignment #017

017. `query 17

Cal 2

08-08-2009

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14:22:50

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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a possible solution for the given function is omega=3.

You have asserted this solution but you haven't shown how it was obtained.

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(omegat) so
y' = -omegasin(omegat) and
y"" = -omega^2cos(omegat)

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

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

-omega^2cos(omegat) = -9cos(omega*t)
omega^2 = 9
omega = +3, -3

Both solutions check in the original equation.

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14:31:58

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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14:32:39

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

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by finding the derivative of the given function and then setting the derivative equal to the equation with the function substituted in for P.

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14:33:19

What is the derivative dP/dt?

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1/(4(cosh(t/2))^2)

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14:40:49

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

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yes it does. Since 1/(1+e^(-t)) = e^(t)/e^t+1 = P, P-P^2 = 1/(4(cosh(t/2))^2) which is a hyperbolic function.

There are several missing steps between the exponential form and the hyperbolic cosine, which is however unnecessary for this solution.

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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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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14:55:42

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

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solution 4 which was y=e^x+e^(-x) corresponds to the equation because the number e does not change when the derivative is taken, even twice.

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15:01:12

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

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no solutions correspond to the given equation because none of the derivatives of these solutions are equal to the negative of the solution.

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15:08:34

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

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The solution y=sqrt(2x) corresponds to that equation. One can tell by working out that 1/sqrt(2x) = the derivative of the function.

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16:08:19

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

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The solutions that correspond to that equation are both y = cosx and y = cos(-x) because the second derivative of each is equal to the opposite of the solution.

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16:09:20

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

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The solution that corresponds to the equation is y = x^2. One can tell be simply substituting the function in the 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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16:54:11

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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The first graph is a horizontal line on the x-axis. The second graph has two horizontal asymptotes, one at P=10 and one at P=0 and there is an inflection point at P=5.

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17:03:05

Query problem 11.2.10 (was 10.2.6) slope field

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17:16:56

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

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The field corresponding to the given equation is field III with slopes that start off as being vertical, but near x=1.8, begin to slant back until they are horizontal at x=0. After the y axis, they continue their rotation until x=1.5 where they become horizontal once more.

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17:23:05

describe the slope field corresponding to y' = sin x

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The slope field for this function is simply the slopes of the sine wave, with the horizontal slope at x=0.

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17:24:08

describe the slope field corresponding to y' = cos x

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The slope field of this function is the slopes of the cosine wave function.

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17:29:52

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

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The slope field of this function is number IV which starts off, at the left side of the graph, with vertical slope lines which bend towards horizontal at about x=-1.2. Slopes become horizontal at x=-.2 and then tilt up slightly at x=.8 until x=3.8 where the slopes again become horizontal.

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17:32:06

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

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The slope field for this function is number V where the slopes are horizontal from -t until -.6 where the slopes become slightly positive until returning to horizontal at x=.9.

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17:52:48

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

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The slope field for this function is number VI with vertical slopes from -infinity to about x=-2 where the slope becomes positive and bends towards horizontal, which is reaches at about x=1.4 and continues horizontally.

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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17:52:51

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

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Assignment 17

&#Your work looks good. See my notes. Let me know if you have any questions. &#