query 17

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course Mth 174

11-27 4

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).

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Your solution:

For y = cos(omega * t)  y` = -omega*sin(omega *t) and y`` = -omega^2 * cos(omega * t)

Now plugging y’’ and y = cos(omega*t) into our equation

[-omega^2 * cos(omega * t)] + 9 cos(omega*t) = 0,

Now we solve for omega

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

omega^2 = 9

omega = `sqrt(9) = +- 3

confidence rating #$&*:3

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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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Self-critique (if necessary):

When you say both solutions check out you are talking about plugging these values back into the equation: -omega^2*cos(omega*t) + 9cos(omega*t) = 0

You plug y = cos(omega*t), with omega = 3, into the equation y '' + 9 y = 0.

That is, you would plug in the function y = cos( 3 t), and verify that the left-hand side is 0.

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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)

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Your solution:

For dP/dt = P(1-P)  dP/P(1-P) = dt

Int(dP/P(1-P)) = Int(dt)

using partial fractions for Int(dP/P(1-P))

1/P(1-P) = A/P + B/(1+P)

= [A + AP + BP]/P(1-P) = (A+ P(A+B))/P(1-P), no P in numerator so P = 0 and A = 1 and B = -1

1* Int(dP/P) – 1* Int(dP/1-P) = t + C

ln|P| - ln|1-P| = t + C  ln|P/1-P| = t + C

P/1-P = e^t + C = e^C + e^t = Ce^t

P = (1-P)*(Ce^t) = Ce^t – CPe^ t

P+CPe^t = Ce^t

P(1 + Ce^t) = Ce^t

P = Ce^t/(1 + Ce^t)

I’m not sure I’m approaching this problem correctly, need to look at solution.

I way over complicated things, so first we look at 1st derv. of P when P = 1 / (1 + e^-t)

P’ = [1/1+e^-t]’, which is composite fn of the quotient from

= 1’*(1+e^-t) - 1*(1+e^-t)’/(1+e^-t)^2 = 0*(1+e^-t) - 1*(-e^-t)/ (1+e^-t)^2 =

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

???Whats the real reason for proving derivative of 1/1+e^-t = 1/1+e^-t???

Plugging into equation P(1-P)

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

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

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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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Self-critique (if necessary):

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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 theequations 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) )

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Your solution:

y`` = y  e^x+e^-x

y` = e^x - e^-x and y`` = e^x+e^-x

y ' = -y ????None of these solutions seem to fit this equation????

y` = 1/y  `sqrt(2x)

y’ = [`sqrt(2x)]` = using the chain

f(z) = z^(1/2) g(x) = 2x  f`(z) = ½*(z^(-1/2)), g`(x) = 2

f`(g(x))*g`(x) = ½ * 2x^(-1/2) * 2 = 2x^(-1/2) = 1/`sqrt(2x)

y '' = -y  cos(-x) or cos(x)

for cos(-x), y` = sin(-x) and y``=-cos(-x)

for cos(x), y` = -sin(x) and y``= -cos(x)

x^2 y '' - 2 y = 0  x^2

y` = 2x and y`` = 2 so…

x^2y`` - 2y = x^2*(2) – 2*(x^2) = 2x^2 – 2x^2 = 0

Again I could not come up with solution for y ' = -y

confidence rating #$&*:3

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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.

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.

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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Self-critique (if necessary):

I am not sure so does y ' = -y not have a solution????

The equation has a solution (you should verify that y = A e^(-x) is a solution), but the solution is not represented by any of the pictures.

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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.

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Your solution:

for both solutions the asymptotes are at P=0 and P = 10.

For(0, 0) graph passes thru (0,0) and is increasing at an increasing rate until P(y axis) value is approx 2 then from P=2 to approx P=8 the slope is constant at the max slope of the solution. At P=8 the fn starts decreasing at an increasing rate as in the same manner as when it was increasing in this time fn is decreasing. Fn continues to decrease as it approaches P=10 but P =10 seems to be asymptote, so as t grows fn will continue to approach P = 10

For(1, 4) solution the fn starts on left side of P axis at approx (-1, 0) , this fn will follow the exact same pattern only instead of starting at(0, 0) this graph starts at approx (-1, 0) so that we can simply take solution for (0, 0) and shift this graph to the left by a value of 1 for t and we would have the solution for (1, 4)

confidence rating #$&*:3

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Given Solution:

RESPONSE -->

Solution through (0,0): concave up from (0,0) to about (3,3) then concave down for t>3, point of inflection at (3,3), horizontal asymptote at P=10. Graph appears to be P=0 for all t<0. Since givens include P>=0, no graph for negative P.

Solution through (1,4): concave up from (-1,0) to about (2,3) , then concave down for t>2, point of inflection at (2,3), horizontal asymptote at P=10. Graph appears to be P=0 for at t<-1. Since givens include P>=0, no graph for negative P.

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14:13:09

Query problem 11.2.10 (was 10.2.6) slope field

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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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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Self-critique (if necessary):

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self-critique rating #$&*:

Good work, but see my notes and let me know if you have additional questions.