query 17

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course Mth 174

6/3 2:00 pm

017. `query 17 Cal 2

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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:

We are asked to find a value of omega such that y’’ + 9y=0 is satisfied by y=cos(omega *t)

We want to make omega satisfy the equation y=cos(omega *t) so we need to find y’’ and substitute in.

So we have y= cos(omega * t) using the chain rule we can find

y’=-sin(omega *t)*omega pulling constants out front and using the chain rule again we find

y’’= -omega^2 cos(omega*t)

now that we have y’’ we want to substitute into the equation y’’+9y=0 values for y’’ and y

-omega^2 cos(omega 8t) + 9 cos(omega *t) =0 so now we solve to find the value of omega

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

-omega^2 =-9

Omega = +-3

We plug the values of 3 and -3 in to ensure they work getting

-(-3^2) cos (-3t) + 9 cos(-3t) =0

-9cos(-3t)=-9cost(- 3t) and

-(3^2) cos (3t) + 9 cos(3t) =0

9cos3t=9cost 3t

Both values check out so the value of omega meeting the conditions are +-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):

ok

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Self-critique Rating:

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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:

We begin with P=1/(1+e^-t) and want to satisfy dPdt=P(1-P)

We start by trying to find the derivative of P. to make things easier we multiply the top and bottom of the P expression by e^t to get rid of the e^-t in the denominator. After doing this our new expression for P is

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

So now we see we can use the quotient rule to take the derivative with

g(t)= e^t g’(t)= e^t h(t)=(e^t+1) h’(t)=e^t so we have

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

Simplifiying we get

e^t^2 + e^t -e^t^2/(e^t+1)^2 which is just e^t/(e^+1)

so now we know that P=e^t/(e^t+1) and we know dP/dt=e^t(e^t+1)^2 so we try to plug into dP/dt=P(P-1) so we have

e^t/(e^t+1)^2 is this equal to e^t/(e^t+1)(1+(e^t/(e^t+1))

simplifying the right side e^t/(et+1)_ - (e^t/(e^t+1))^2

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

so we can see that we need to multiply the left term by (e^t+1) to get the common denominator of (e^t+1)^2 doing so we get

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

Simplifying we get

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

Which is just e^t/(e^t+1)^2 whish is equal to e^t/(e^t+1)^2 so the equations match.

confidence rating #$&*: 3

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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):

Ok

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Self-critique Rating:3

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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) )

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Your solution:

we are given the following information:

Equations Solutions

1. y’’=y 1. Cos x

2. y’=-y 2. Cos(-x)

3. y’=1/y 3. x^2

4. y’’=-y 4. e^x+e^-x

5. x^2y’’-2y=0 5. Sqaureroot (2x)

we are trying to find which of these are solutions match with each equation.

I will begin by find the 1st and 2nd derivatives of each solution and trying to find correlations and fill in to see which equation they are the solution for.

1. y=cos (x)

y’=-sin(x)

y’’=-cos(x) the relationship here is that y’’=-y so we know that the equation y’’=-y is the equation for which -cos(x) is a solution.

2. y=cos(-x)

y’=sin(-x)

y’’=-cos(-x) so here we can see that y’’=-y as well so equation for this solution is y’’=-y

3. y=x^2

y’=2x

y’’=2 here there isn’t a clear correlation between the first 4 equations so I will try substituting into the 5th one

x^2 y’’-2y=0

x^2(2)-2(x^2)=0

2x^2 = 2x^2 so x^2y’’-2y=0 is the equation for the solution y=x^2

4. y=e^x + e^-x

y’= e^x + e^-x

y’’= e^x + e^-x it is clear that these are the same so the equation that this is the solution for is y’’=y

5. y=squareroo of (2x)

so y=2x^(1/2) by chain rule

y’=(2x)^-1/2 by chain rule

y’’=-1/(2x)^3/2 here we can see that y’ is the reciprocal of y so the equation this is the solution for is y’=1/y

Every equation had a solution except for y’=-y

and y’’=-y had 2 solutions

confidence rating #$&*: 3

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Given Solution:

.

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):

Excited to say I understood all of this easily!!!

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Self-critique Rating:3

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This is very good. This material actually occurs about halfway through most differential equations courses (those courses contain a lot of stuff that isn't included here, but this chapter gives you an excellent introduction to the subject).

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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:

We are given that dP/dt= 0.1P(10-P) for p greater than or equal to 0.

We know that we can get the slope of graph at a given point because dP/dt is the value of the slope at point P for example

Slope t(0) is 0.1(0)(10-0)=0

t(1)=.9

t(2)=1.6

t(3)=2.1

t(4)=2.4

t(5)=2.5

t(6)=2.4

t(7)=2.1

t(8)=1.6

t(9)=.9

t(10)=0

@&

It took me a minute to figure out that the above lines indicated the slopes. The t( ) notation was a little misleading, but it's not easy to come up with a good notation for this.

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We are given these slopes as tick marks in the text but it is more clear to me to have the actual slopes per unit.

So if we begin at (0,0) we know the graph increases by:

.9 from t =0 to t=1, 1.6 from t= to t=2, 2.1 from t=2 to t=3, 2.4 from y=3 to t=4, and 2.5 from t=4 to t=5. So from t=0 to t=5 the graph is concave up and increasing. At t=5 we have a point of inflection and the graph begins having an increasing slope at a decreasing rate and is concave down.

@&

The slope is not increasing, but the graph is increasing. The graph increases and the slope decreases, which makes the graph concave down, as you say.

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The graph increases by 2.4 from t=5 to t=6, 2.1 from t=6 to t=7, 1.6 from t=7 to t=8, .9 from t=8 to t=9 and 0 at t=10 so the horizontal asymptote approaches t=10.

######It seems my solution is different than the given solution below, did I work the wrong problem or am I wrong in my understanding of how to construct the graph???????

@&

The only differences I see are that you are estimating the slopes numerically, and probably coming out with low estimate on most. So your graph extends a bit further to the right.

If you were to trace a graph on the given slope field, you would probably find that the graph becomes very nearly vertical closer to t = 3 than to t = 5.

However your estimates are OK, and you clearly understand what you are doing.

*@

confidence rating #$&*: 2

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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

#####In my book, 5th edition, 11-2.10 asks to sketch the slope field for the given equation y’=x-y at points y=1 and x=1

I construct a table

X y x-y (slope)

1 -3 4

1 -2 3

1 -1 2

1 0 1

1 1 0

1 2 -1

1 3 -2

-3 1 -4

-2 1 -3

-1 1 -2

0 1 -1

1 1 0

2 1 1

3 1 2

####Now that I have the slopes I can put slope marks at each point and then graph. At this point I need to find the equation for a solution that passes through the point (0, 1) I know that the slope at point (0,1) is -1 and that y’=x-y. I know that I need to put in slope marks and attempt to graph the function in order to find the equation but I am stuck and do not know how to proceed???????

@&

I pretty clearly mislabeled this one. The only task on Problem 10, after sketching the slope field, is to sketch the graph through the given point.

The correct problem is 11.2.14. It's definitely worth a look.

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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):

####The last little bit of information above does not match what I had in the text so I wasn’t sure which problem I should be working. I did try to work out section 11-2 problem number 10 and had a question which I indicated above and need help on how to proceed to find the equation of the function at a certain point when sketching a slope field.?????????

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Self-critique Rating:3

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Self-critique rating:

#*&!

@&

Very good. I apologize for the confusion on that one problem.

Check my notes and let me know if you have questions.

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