017. `query 17

Cal 2

Question: 

Query problem 11.1.8 previously 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).

Your solution:

Confidence Assessment:

Given Solution:

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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Question: 

Query problem 11.1.18 (4th edition 11.1.14 3d edition 11.1.13) (formerly 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P)

Your solution:

Confidence Assessment:

Given Solution:

RESPONSE -->

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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Question: 

Query problem (omitted from 5th edition but should be worked and self-critiqued)  (4th edition 11.1.16 3d edition 11.1.15 formerly 10.1.1) match one of the equations y '' = y, y ' = -y, y ' = 1 / y, y '' = -y, x^2 y '' - 2 y = 0 with each of the solutions (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )

Your solution:

Confidence Assessment:

Given Solution:

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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.</h3>

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

<h3>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.

<h3>

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.

</h3>

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

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RESPONSE -->

Only solution IV corresponds to y"" = y

y = e^x + e^(-x)

y' = e^x - e^(-x)

y"" = e^x + e^(-x)

Thus, y = y""

Solution I, y = cos x, y' = -sin x, y""= -cos x, not = y

Solution II, y = cos(-x), y' = sin(-x), y"" = - cos(-x), not = y

Solution III, y = x^2, y' = 2x, y"" = 2, not = y

Solution V, y = sqrt (2x), y' = 1/sqrt(2x), y"" = -1/sqrt[(2x)^3], not = y

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

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RESPONSE -->

None of the solutions correspond to y' = -y

Solution I: y = cos x, y' = -sin x, not = -y

Solution II: y = cos(-x), y' = sin x, not = -y

Solution III: y = x^2, y' = 2x, not = -y

Solution IV: y = e^x + e^(-x), y' = e^x - e^(-x), not = -y

Solution V: y = sqrt(2x), y' = 1/sqrt(2x), not = -y

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

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RESPONSE -->

Solution V corresponds to y' = 1/y

y = sqrt(2x), y' = 1/sqrt(2x), y' = 1/y

None of the other solutions correspond (see previous responses).

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

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RESPONSE -->

Solutions I and II correspond to y"" = -y

Solution I: y = cos x, y' = -sin x, y"" = -cos x, y"" = -y

Solution II: y = cos(-x), y' = sin(-x), y"" = -cos(-x), y"" = -y

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

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RESPONSE -->

Solution III corresponds to x^2*y"" - 2y = 0

Solution III: y = x^2, y' = 2x, y"" = 2

Substituting: x^2*(2) - 2(x^2) = 2x^2 - 2x^2 = 0

None of the other solutions correspond.

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Question: 

Query problem 11.2.8 (4th edition 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.

Your solution:

Confidence Assessment:

Given Solution:

The slope of the graph on the horizontal axis is zero at every point. So the solutoin through (0, 0) simply stays on that axis. The solution function would be P(t) = 0.

Starting at (1, 4) and moving to the right the solution curve rises very quickly at first, then less and less quickly, becoming asymptotic to the line P = 10. The slope becomes 1 somewhere around (3, 9) and continues decreasing beyond that point. Moving to the left from (1, 4) the solution curve descends very quickly at first, then less and less quickly, crossing the y axis around (0, 3/2) and becoming asymptotic to the negative horizontal axis.

What are your description for the other given starting points?
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14:13:09

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Question: Query problem 11.2.14 (was 11.2.10) (previously 10.2.6) slope field

Your solution:

Confidence Assessment:

Given Solution:

<h3>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).</h3>

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14:22:17

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

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RESPONSE -->

Slope field corresponding to y' = x e^-x: Slope field III: slope increasingly negative for x<0, slope increasing positive (at decreasing rate) to about x=1, then decreasing positive, slope appears to have a limit of 0 for large x.

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14:25:41

describe the slope field corresponding to y' = sin x

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RESPONSE -->

Slope field for y' = sin x: Slope field I: Slope ranges from 0 to 1 to 0 to -1 to 0 repeatedly, with slope = 0 at (0,0), period = 2pi

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14:28:17

describe the slope field corresponding to y' = cos x

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RESPONSE -->

Slope field corresponding to y' = cos x: slope field II. Same as slope field for y' = sin x, except slope = 1 at x = 0 (cyled shifted pi/2 to left).

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14:32:27

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

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RESPONSE -->

Slope field corresponding to y' = x^2 e^-x: Slope field IV: Slope increasingly positive as x decreases from 0, slope 0 at x=0, slope increasingly positive (at decreasing rate) from x=0 to x=2, then decreasingly positive, appears to approach 0 for large x.

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14:36:20

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

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RESPONSE -->

Slope field corresponding to y' = e^-(x^2): Slope field V: slope maximum at x= 0, decreasing positive as |x| icreases, slope appears to approach 0 as |x| gets large (slope is very small with |x|>2).

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14:40:55

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

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RESPONSE -->

Slope field corresponding to y' = e^-x: Slope field VI: Slope increasing positive as x decreases, slope 1 at x=0, slope approaches 0 for large positive x (slope very small for x>about 2).

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