How to determine the inflection point of non-equilibrium solution? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Differential Equation ,Finding solution by sketching the graphDetermining the stability of an equilibrium point of a system of non-linear odes.Equilibrium points of the ODE $y'=sin y−fracy2$.Show that the equilibrium point (r, 0) = (1,0) is not stable, even though all nearby solution tend to it (eventually).How to determine the nature of this equilibrium point?Check equilibrium point for stabilityIs Equilibrium Solution Always Constant?Equilibrium and general solution of this polynomial non linear ODEExistence and Uniqueness of Equilibrium Points in Non-Linear Dynamical Systemsdetermine the stability of an equilibrium point(x,0).
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How to determine the inflection point of non-equilibrium solution?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Differential Equation ,Finding solution by sketching the graphDetermining the stability of an equilibrium point of a system of non-linear odes.Equilibrium points of the ODE $y'=sin y−fracy2$.Show that the equilibrium point (r, 0) = (1,0) is not stable, even though all nearby solution tend to it (eventually).How to determine the nature of this equilibrium point?Check equilibrium point for stabilityIs Equilibrium Solution Always Constant?Equilibrium and general solution of this polynomial non linear ODEExistence and Uniqueness of Equilibrium Points in Non-Linear Dynamical Systemsdetermine the stability of an equilibrium point(x,0).
$begingroup$
$$
fracdPdt = 0.2P(1-fracP1000)
$$
I am asked to find out the common inflection point of each non-equilibrium solution. How to do that? (I know the equilibrium points are $P$ = $0$, and $P$ = $1000$)
ordinary-differential-equations
asked Feb 20 '17 at 19:14
user214969user214969
835
$endgroup$
add a comment |
$begingroup$
$$
fracdPdt = 0.2P(1-fracP1000)
$$
I am asked to find out the common inflection point of each non-equilibrium solution. How to do that? (I know the equilibrium points are $P$ = $0$, and $P$ = $1000$)
ordinary-differential-equations
asked Feb 20 '17 at 19:14
user214969user214969
835
$endgroup$
add a comment |
$begingroup$
$$
fracdPdt = 0.2P(1-fracP1000)
$$
I am asked to find out the common inflection point of each non-equilibrium solution. How to do that? (I know the equilibrium points are $P$ = $0$, and $P$ = $1000$)
ordinary-differential-equations
asked Feb 20 '17 at 19:14
user214969user214969
835
$endgroup$
$$
fracdPdt = 0.2P(1-fracP1000)
$$
I am asked to find out the common inflection point of each non-equilibrium solution. How to do that? (I know the equilibrium points are $P$ = $0$, and $P$ = $1000$)
ordinary-differential-equations
ordinary-differential-equations
asked Feb 20 '17 at 19:14
user214969user214969
835
asked Feb 20 '17 at 19:14
user214969user214969
835
edited Feb 20 '17 at 20:46
user214969
asked Feb 20 '17 at 19:14
user214969user214969
835
asked Feb 20 '17 at 19:14
user214969user214969
835
asked Feb 20 '17 at 19:14
user214969user214969
835
835
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The second derivative is
$$
fracd^2Pdt^2=0.2left(1-2fracP1000right),fracdPdt
$$
which you can now easily set to zero.
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
$endgroup$
add a comment |
$begingroup$
first derivative ( let $ M=1000)$
$$ P-P^2/M$$
second derivative vanishes at inflection point
$$ 1- 2 P/M =0,, P=M/2 = 500 $$
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
$endgroup$
add a comment |
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$begingroup$
The second derivative is
$$
fracd^2Pdt^2=0.2left(1-2fracP1000right),fracdPdt
$$
which you can now easily set to zero.
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
$endgroup$
add a comment |
$begingroup$
The second derivative is
$$
fracd^2Pdt^2=0.2left(1-2fracP1000right),fracdPdt
$$
which you can now easily set to zero.
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
$endgroup$
add a comment |
$begingroup$
The second derivative is
$$
fracd^2Pdt^2=0.2left(1-2fracP1000right),fracdPdt
$$
which you can now easily set to zero.
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
$endgroup$
The second derivative is
$$
fracd^2Pdt^2=0.2left(1-2fracP1000right),fracdPdt
$$
which you can now easily set to zero.
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
answered Feb 20 '17 at 20:50
LutzLLutzL
60.8k42157
60.8k42157
add a comment |
add a comment |
$begingroup$
first derivative ( let $ M=1000)$
$$ P-P^2/M$$
second derivative vanishes at inflection point
$$ 1- 2 P/M =0,, P=M/2 = 500 $$
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
$endgroup$
add a comment |
$begingroup$
first derivative ( let $ M=1000)$
$$ P-P^2/M$$
second derivative vanishes at inflection point
$$ 1- 2 P/M =0,, P=M/2 = 500 $$
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
$endgroup$
add a comment |
$begingroup$
first derivative ( let $ M=1000)$
$$ P-P^2/M$$
second derivative vanishes at inflection point
$$ 1- 2 P/M =0,, P=M/2 = 500 $$
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
$endgroup$
first derivative ( let $ M=1000)$
$$ P-P^2/M$$
second derivative vanishes at inflection point
$$ 1- 2 P/M =0,, P=M/2 = 500 $$
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
answered Feb 20 '17 at 21:08
NarasimhamNarasimham
21.3k62258
21.3k62258
add a comment |
add a comment |
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