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**???How to know if equilibrium points are stable or not. Is my solution correct**

## Homework Statement

## Homework Equations

## The Attempt at a Solution

solution above

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solution above

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HallsofIvy

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For large t, and y close to either -1 or 1, t^2- y- 2 is positive.

For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasing

For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasing

For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasing

Since for starting values close to -1, y then tends to -1, -1 is a

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ok i followed what your saying. can you tell me if my solution below is correct? for part c and d

For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasingtoward-1.

For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasingtoward-1 andaway from1.

For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasingaway from1.

Since for starting values close to -1, y then tends to -1, -1 is astableequilibrium. Since for starting values close to 1, y then tends away from 1, 1 is anunstableequilibrium.

i said C is stable, and so is D

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Mark44

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It would be helpful if you pasted the text into the reply text box instead of uploading a scanned image. A scanned image makes it harder for responders such as myself to identify a particular line. I have to identify the line where there's a problem, and then point out what the problem is.

For c) you said "A fixed point is stable, if when we sub values from either side of the fixed point, we see that the gradient y', are opposite symbols."

This is not true.

From a previous thread, you gave a definition of stability that involved the second derivative. If x

In terms of the first derivative, if x

It is not enough to notice that the derivative changes sign. It has to change sign in a certain way for the fixed point to be stable fixed point, and it has to change sign in the opposite way for it to be an unstable fixed point.

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actually, the other time i was dealing with potential energy, and the notes said V(x)>0 is stable (minimum)

For c) you said "A fixed point is stable, if when we sub values from either side of the fixed point, we see that the gradient y', are opposite symbols."

This is not true.

From a previous thread, you gave a definition of stability that involved the second derivative. If x_{c}is a fixed point (i.e., f'(x_{c}) = 0), x_{c}is stable if f''(x_{c}) > 0. x_{c}is an unstable fixed point if f''(x_{c}) < 0.

In terms of the first derivative, if x_{c}is a fixed point, x_{c}is a stable fixed point if the derivative f' changes sign from negative to positive, moving left to right. x_{c}is an unstable fixed point if the derivative f' changes sign from positive to negative, moving left to right.

It is not enough to notice that the derivative changes sign. It has to change sign in a certain way for the fixed point to be stable fixed point, and it has to change sign in the opposite way for it to be an unstable fixed point.

where V=potential energy equation

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Mark44

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If I recall, it said x was a stable fixed point if V''(x) > 0, not V(x) > 0.

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you said (1-y)^2=(y-1)(y+1)

For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasingtoward-1.

For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasingtoward-1 andaway from1.

For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasingaway from1.

Since for starting values close to -1, y then tends to -1, -1 is astableequilibrium. Since for starting values close to 1, y then tends away from 1, 1 is anunstableequilibrium.

but that is false

(1-y)^2=-(y-1)(y+1)=(1-y)(y+1)

see here: http://www.wolframalpha.com/input/?i=%281-y^2%29

so doesnt that mean your solution is wrong?

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yes sorry V''(x)>0 is stable , as it is a minimum, but still i dont think this has got anything to do with this question. because the other 1 was related to potential energy.If I recall, it said x was a stable fixed point if V''(x) > 0, not V(x) > 0.

or am i wrong?

calculating y'' gives -2y+3y^2+4y. this equation contains all y's? not x's? so how can we sub x, it would be V(y)???

if i sub y=1 i get -2+3+4=5

since V''(1)=5>0, means x=1 is stable

V''(-1) gives 2+3+4=7

V''(-1)=7>0, means x=-1 is stable

so agrees with my solution

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furthermore, when y'=y^2-1,this does not work

as y''=2y

we have fixed points -1 and 1

V''(-1)=-2<0, so unstable

but -1 is actually stable. according to website below. see last question.

http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf [Broken]

but my method of showing symbols change, shows -1 is stable (just like above website). the above website has error. it says x=1 is stable. but the graph they sketched clearly shows its unstable !

again the fixed point 1 gives V''(1)=2>0, so is stable. but is actually unstable !

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- #10

Mark44

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HallsOfIvy, you are mistakenly working with y

For y< -1, both y- 1 and y+ 1 are negative so (t^2- y- 2)(y- 1)(y+ 1) is the product of one positive and two negative numbers- that is, y' is positive and y is increasingtoward-1.

For -1< y< 1, y- 1 is negative but y+ 1 is positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of two positive and one negative number- that is, y' is negative and y is decreasingtoward-1 andaway from1.

For y> 1, both y- 1 and y+ 1 are positive so (t^2- y- 2)(y- 1)(y+ 1) is the product of three positive numbers- that is y' is positive and y is increasingaway from1.

Since for starting values close to -1, y then tends to -1, -1 is astableequilibrium. Since for starting values close to 1, y then tends away from 1, 1 is anunstableequilibrium.

- #11

vela

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You mean 1-yyou said (1-y)^2=(y-1)(y+1)

but that is false

(1-y)^2=-(y-1)(y+1)=(1-y)(y+1)

see here: http://www.wolframalpha.com/input/?i=%281-y^2%29

so doesnt that mean your solution is wrong?

The only thing that does is flip some of the signs in HallsofIvy's analysis. If you understand what he said, you should be able to tell us how that changes the conclusions, if it does at all.

You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.yes sorry V''(x)>0 is stable , as it is a minimum, but still i dont think this has got anything to do with this question. because the other 1 was related to potential energy.

or am i wrong?

I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).calculating y'' gives -2y+3y^2+4y. this equation contains all y's? not x's? so how can we sub x, it would be V(y)???

Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.furthermore, when y'=y^2-1,this does not work

as y''=2y

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HallsofIvy

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Thanks. I started out intending to say "-(y-1)(y+1)" and lost the negative sign.HallsOfIvy, you are mistakenly working with y^{2}- 1 instead of 1 - y^{2}, so some of your results have the wrong sign.

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ok, for the question in the opening. the one about stability.You mean 1-y^{2}= (1-y)(1+y), not (1-y)^{2}, which equals 1-2y+y^{2}.

The only thing that does is flip some of the signs in HallsofIvy's analysis. If you understand what he said, you should be able to tell us how that changes the conclusions, if it does at all.

You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.

I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).

Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.

if i differentiate with respect to t. i get

-2ty^2+2t. and t=1 gives

-2y^2+2=0

and y is imaginery. does this mean at t=1, point is unstable?

also at t=-1, i get 2y^2-2==>y^2=1

i got y=+/- 1?? how do i know if it is stable or not?

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also, for the last question on this pageYou mean 1-y^{2}= (1-y)(1+y), not (1-y)^{2}, which equals 1-2y+y^{2}.

The only thing that does is flip some of the signs in HallsofIvy's analysis. If you understand what he said, you should be able to tell us how that changes the conclusions, if it does at all.

You're wrong. It's essentially the same concept, but it's probably best to just focus on the problem at hand than get sidetracked.

I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x).

Same mistake as before. In this problem, you need to differentiate with respect to x and apply the chain rule.

http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf [Broken]

it says to determine wether the fixed points of y'=y^2-1 is stable or not

where y'=y dot=dy/dt

but if we differentiate again we get y''=0? does this mean all points are unstable???

or to determine stability, do we look at the last non 0, ODE

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Mark44

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I don't see how you're getting y'' = 0.also, for the last question on this page

http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf [Broken]

it says to determine wether the fixed points of y'=y^2-1 is stable or not

where y'=y dot=dy/dt

but if we differentiate again we get y''=0? does this mean all points are unstable???

or to determine stability, do we look at the last non 0, ODE

If y' = y

then y'' = 2y*y', by the chain rule.

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vela

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This isn't the approach you want to take. For one thing, you've differentiated incorrectly. Also, you have to keep track of which variable you're talking about. Your equilibrium points are at y=±1. That's completely different than saying t=±1. This complication makes the problem a lot harder.ok, for the question in the opening. the one about stability.

if i differentiate with respect to t. i get

-2ty^2+2t. and t=1 gives

-2y^2+2=0

and y is imaginery. does this mean at t=1, point is unstable?

also at t=-1, i get 2y^2-2==>y^2=1

i got y=+/- 1?? how do i know if it is stable or not?

Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.

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thanks again for response. when differentiating. do i have to do implicit differentiation?This isn't the approach you want to take. For one thing, you've differentiated incorrectly. Also, you have to keep track of which variable you're talking about. Your equilibrium points are at y=±1. That's completely different than saying t=±1. This complication makes the problem a lot harder.

Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.

i differentiated with respect to t, because you said "

- #18

Mark44

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You need to use the chain rule.when differentiating. do i have to do implicit differentiation?

You have dy/dt = yi differentiated with respect to t, because you said "I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with respect to t (not x)."

So, d

- #19

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but if i differentiate w.r.t t i get 0You need to use the chain rule.

You have dy/dt = y^{2}- 1

So, d^{2}y/dt^{2}= d/dt(y^{2}- 1) = d/dy(y^{2}- 1) * dy/dt. Of course, this should be simplified further.

http://www.wolframalpha.com/input/?i=differentiate+with+respect+to+t+dy/dt+=+y2+-+1

this is so confusing. because i have not seen this kind of differentiation. the answer is definately 0. if it isnt, then i have no idea how to get the solution

i put what you said into wolfram

http://www.wolframalpha.com/input/?i=+d/dy(y2+-+1)+*+dy/dt

but how is that findinh y''. this is differentiating the multiply of (^2-1)y', which is just (y')^2. so we are differentiating y'^2!

- #20

Mark44

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No, this is wrong. Wolframalpha is calculating the partial derivative with respect to t, and the assumption is that y and t are independent variables. They are not, since y is a function of t. I already showed you what you needed to do in using the chain rule.but if i differentiate w.r.t t i get 0

http://www.wolframalpha.com/input/?i=differentiate+with+respect+to+t+dy/dt+=+y2+-+1

In any case, and to repeat what vela said, this is NOT THE RIGHT APPROACH.

vela said:Go back and look at HallsofIvy's explanation. Fix the signs and tell us what you get.

this is so confusing. because i have not seen this kind of differentiation. the answer is definately 0. if it isnt, then i have no idea how to get the solution

i put what you said into wolfram

http://www.wolframalpha.com/input/?i=+d/dy(y2+-+1)+*+dy/dt

but how is that findinh y''. this is differentiating the multiply of (^2-1)y', which is just (y')^2. so we are differentiating y'^2!

- #21

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y'=-(t^2- y- 2)(y- 1)(y+ 1)No, this is wrong. Wolframalpha is calculating the partial derivative with respect to t, and the assumption is that y and t are independent variables. They are not, since y is a function of t. I already showed you what you needed to do in using the chain rule.

In any case, and to repeat what vela said, this is NOT THE RIGHT APPROACH.

-1<y<1 for large t, y' is positive

y<-1 for large t,y' is negative

y>1 for large y, y' is negative

<==-1==>1<==

so y=1 is a stable, y=-1 is unstable

but isnt this just the same as my method in post 3, except a small error.

all i have learnt here is that lower values than the fixed point must give a positive gradient. and higher values must be negative gradient

I think of a negative y’ as <== and positive y’ as==>. Then if we have ==><== at either side. Is stable. is this correct?

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- #22

vela

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Right. In your solution, you should probably explain how you determined the sign of y' in the three regions.y'=-(t^2- y- 2)(y- 1)(y+ 1)

-1<y<1 for large t, y' is positive

y<-1 for large t,y' is negative

y>1 for large y, y' is negative

<==-1==>1<==

so y=1 is a stable, y=-1 is unstable

I wouldn't say you made a small error. You had a conceptual error about the difference between a stable and unstable fixed point. Considering the problem was about identifying exactly this characteristic of the fixed points, I'd say it was a major error.but isnt this just the same as my method in post 3, except a small error.

Lower and higher values of what?all i have learnt here is that lower values than the fixed point must give a positive gradient. and higher values must be negative gradient

Yes, it's correct. Can you explain why you think of y'<0 as <== and y'>0 as ==>?I think of a negative y’ as <== and positive y’ as==>. Then if we have ==><== at either side. Is stable. is this correct?

- #23

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no. but i guess i have to explain that in the exam. why is it?Right. In your solution, you should probably explain how you determined the sign of y' in the three regions.

I wouldn't say you made a small error. You had a conceptual error about the difference between a stable and unstable fixed point. Considering the problem was about identifying exactly this characteristic of the fixed points, I'd say it was a major error.

Lower and higher values of what?

Yes, it's correct. Can you explain why you think of y'<0 as <== and y'>0 as ==>?

i just thinking of positive as moving forwards, thus ==>

and negative as moving backwards, thus <==

- #24

vela

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At this point in your education, your instructor probably assumes you understand this concept, so you don't need to explain it on an exam. You should be able to explain it, however. If someone were to ask you why y'>0 means ==> as opposed to <==, you should be able to give a reason better than "it gives me the right answer when I assume that" or "that's what someone else told me."no. but i guess i have to explain that in the exam. why is it?

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ok, i am very grateful for your help. i have my exam in 3 days.At this point in your education, your instructor probably assumes you understand this concept, so you don't need to explain it on an exam. You should be able to explain it, however. If someone were to ask you why y'>0 means ==> as opposed to <==, you should be able to give a reason better than "it gives me the right answer when I assume that" or "that's what someone else told me."

can you please kindly tell me how do i answer part e. wether there is a unique solution through the point (1,2).

i know that to prove uniqueness, we must have y'=f(t,y), and differentiate w.r.t to y. i.e find (f(t,y))'

and then show that (f(t,y))' is continuous.

to show whether or not it has a unique solution at (1,2), should i sub t=1 and y=2 into (f(t,y))', and if i get a unique solution. does that mean there is a unique solution through the point (1,2)??

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