$y'= y(1 Ã¢ÂˆÂ’ y)$, with $y(0) in [0, 1]$. Then $lim

up vote
-2
down vote

favorite

Let $y(t)$ be a real valued function defined on the real line such that
$y'= y(1 Ã¢ÂˆÂ’ y)$, with $y(0) in [0, 1]$. Then $lim_ttoinfty y(t) = 1$.

The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.

How can i find solution with 3 minutes?

edited Jul 22 at 14:10

Shaun

7,34592972

asked Jul 22 at 13:38

Cloud JR

469414

univ.tifr.res.in/gs2018/Files/GS2017_QP_MTH.pdf. Part A question number 29
â€“Â Cloud JR
Jul 22 at 13:43

1

Here's a MathJax tutorial :)
â€“Â Shaun
Jul 22 at 13:43

add a commentÂ |Â

up vote
-2
down vote

favorite

Let $y(t)$ be a real valued function defined on the real line such that
$y'= y(1 Ã¢ÂˆÂ’ y)$, with $y(0) in [0, 1]$. Then $lim_ttoinfty y(t) = 1$.

The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.

How can i find solution with 3 minutes?

edited Jul 22 at 14:10

Shaun

7,34592972

asked Jul 22 at 13:38

Cloud JR

469414

univ.tifr.res.in/gs2018/Files/GS2017_QP_MTH.pdf. Part A question number 29
â€“Â Cloud JR
Jul 22 at 13:43

1

Here's a MathJax tutorial :)
â€“Â Shaun
Jul 22 at 13:43

add a commentÂ |Â

up vote
-2
down vote

favorite

Let $y(t)$ be a real valued function defined on the real line such that
$y'= y(1 Ã¢ÂˆÂ’ y)$, with $y(0) in [0, 1]$. Then $lim_ttoinfty y(t) = 1$.

The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.

How can i find solution with 3 minutes?

edited Jul 22 at 14:10

Shaun

7,34592972

asked Jul 22 at 13:38

Cloud JR

469414

Let $y(t)$ be a real valued function defined on the real line such that
$y'= y(1 Ã¢ÂˆÂ’ y)$, with $y(0) in [0, 1]$. Then $lim_ttoinfty y(t) = 1$.

The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.

How can i find solution with 3 minutes?

edited Jul 22 at 14:10

Shaun

7,34592972

asked Jul 22 at 13:38

Cloud JR

469414

edited Jul 22 at 14:10

Shaun

7,34592972

edited Jul 22 at 14:10

Shaun

7,34592972

edited Jul 22 at 14:10

Shaun

7,34592972

asked Jul 22 at 13:38

Cloud JR

469414

asked Jul 22 at 13:38

Cloud JR

469414

asked Jul 22 at 13:38

Cloud JR

469414

univ.tifr.res.in/gs2018/Files/GS2017_QP_MTH.pdf. Part A question number 29
â€“Â Cloud JR
Jul 22 at 13:43

1

Here's a MathJax tutorial :)
â€“Â Shaun
Jul 22 at 13:43

add a commentÂ |Â

univ.tifr.res.in/gs2018/Files/GS2017_QP_MTH.pdf. Part A question number 29
â€“Â Cloud JR
Jul 22 at 13:43

1

Here's a MathJax tutorial :)
â€“Â Shaun
Jul 22 at 13:43

univ.tifr.res.in/gs2018/Files/GS2017_QP_MTH.pdf. Part A question number 29
â€“Â Cloud JR
Jul 22 at 13:43

Here's a MathJax tutorial :)
â€“Â Shaun
Jul 22 at 13:43

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
5
down vote

accepted

Take $y(t)=0$. Then $y(0)in[0,1]$ and $y'=0=y(1-y)$. But $lim_ttoinftyy(t)=0neq1$.

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

add a commentÂ |Â

up vote
2
down vote

$$fracdydx = y(1-y) implies dx = fracdyy(1-y)$$

so $$x+C = lnfracy1-y implies Ce^x = fracy1-y$$

and therefore $$y = fracCe^x1+Ce^x$$

The initial condition is

$$y(0) = fracC1+C implies C = fracy(0)1-y(0)$$

so $$y = fracleft(fracy(0)1-y(0)right)e^x1+left(fracy(0)1-y(0)right)e^x$$

If we pick $y(0) = 0$ then $y = 0$ so $lim_ttoinfty y(t) = 0 ne 1$.

answered Jul 22 at 13:46

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
1
down vote

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

1

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

Â |Â
show 1 more comment

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859411%2fy-y1-%25e2%2588%2592-y-with-y0-in-0-1-then-lim-t-to-infty-yt-1%23new-answer', 'question_page');

);

Post as a guest

Name

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
5
down vote

accepted

Take $y(t)=0$. Then $y(0)in[0,1]$ and $y'=0=y(1-y)$. But $lim_ttoinftyy(t)=0neq1$.

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

add a commentÂ |Â

up vote
5
down vote

accepted

Take $y(t)=0$. Then $y(0)in[0,1]$ and $y'=0=y(1-y)$. But $lim_ttoinftyy(t)=0neq1$.

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

add a commentÂ |Â

up vote
5
down vote

accepted

Take $y(t)=0$. Then $y(0)in[0,1]$ and $y'=0=y(1-y)$. But $lim_ttoinftyy(t)=0neq1$.

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

Take $y(t)=0$. Then $y(0)in[0,1]$ and $y'=0=y(1-y)$. But $lim_ttoinftyy(t)=0neq1$.

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

answered Jul 22 at 13:44

JosÃ© Carlos Santos

113k1698177

add a commentÂ |Â

up vote
2
down vote

$$fracdydx = y(1-y) implies dx = fracdyy(1-y)$$

so $$x+C = lnfracy1-y implies Ce^x = fracy1-y$$

and therefore $$y = fracCe^x1+Ce^x$$

The initial condition is

$$y(0) = fracC1+C implies C = fracy(0)1-y(0)$$

so $$y = fracleft(fracy(0)1-y(0)right)e^x1+left(fracy(0)1-y(0)right)e^x$$

If we pick $y(0) = 0$ then $y = 0$ so $lim_ttoinfty y(t) = 0 ne 1$.

answered Jul 22 at 13:46

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
2
down vote

$$fracdydx = y(1-y) implies dx = fracdyy(1-y)$$

so $$x+C = lnfracy1-y implies Ce^x = fracy1-y$$

and therefore $$y = fracCe^x1+Ce^x$$

The initial condition is

$$y(0) = fracC1+C implies C = fracy(0)1-y(0)$$

so $$y = fracleft(fracy(0)1-y(0)right)e^x1+left(fracy(0)1-y(0)right)e^x$$

If we pick $y(0) = 0$ then $y = 0$ so $lim_ttoinfty y(t) = 0 ne 1$.

answered Jul 22 at 13:46

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
2
down vote

$$fracdydx = y(1-y) implies dx = fracdyy(1-y)$$

so $$x+C = lnfracy1-y implies Ce^x = fracy1-y$$

and therefore $$y = fracCe^x1+Ce^x$$

The initial condition is

$$y(0) = fracC1+C implies C = fracy(0)1-y(0)$$

so $$y = fracleft(fracy(0)1-y(0)right)e^x1+left(fracy(0)1-y(0)right)e^x$$

If we pick $y(0) = 0$ then $y = 0$ so $lim_ttoinfty y(t) = 0 ne 1$.

answered Jul 22 at 13:46

mechanodroid

22.2k52041

$$fracdydx = y(1-y) implies dx = fracdyy(1-y)$$

so $$x+C = lnfracy1-y implies Ce^x = fracy1-y$$

and therefore $$y = fracCe^x1+Ce^x$$

The initial condition is

$$y(0) = fracC1+C implies C = fracy(0)1-y(0)$$

so $$y = fracleft(fracy(0)1-y(0)right)e^x1+left(fracy(0)1-y(0)right)e^x$$

If we pick $y(0) = 0$ then $y = 0$ so $lim_ttoinfty y(t) = 0 ne 1$.

answered Jul 22 at 13:46

mechanodroid

22.2k52041

answered Jul 22 at 13:46

mechanodroid

22.2k52041

answered Jul 22 at 13:46

mechanodroid

22.2k52041

answered Jul 22 at 13:46

mechanodroid

22.2k52041

add a commentÂ |Â

up vote
1
down vote

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

1

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

Â |Â
show 1 more comment

up vote
1
down vote

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

1

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

Â |Â
show 1 more comment

up vote
1
down vote

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 13:50

Mohammad Riazi-Kermani

27.5k41852

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

1

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

Â |Â
show 1 more comment

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

1

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

How can we claim y(0)$in[0,1]$ then $y'=y(1-y)>0$?
â€“Â Cloud JR
Jul 22 at 14:05

@CloudJR I had it if $y(0) in (0,1)$ not$ [0,1]$
â€“Â Mohammad Riazi-Kermani
Jul 22 at 14:29

Okay but still how?
â€“Â Cloud JR
Jul 22 at 15:19

@CloudJR positive times positive makes positive
â€“Â Mohammad Riazi-Kermani
Jul 22 at 16:00

y=1 is an equilibrium point. If you start below y=1 no can never pass it.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 19:43

Â |Â
show 1 more comment

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

ukmuiik