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Advanced topic in Odes 2

grandy

Member
Dec 26, 2012
73
For the equation dy/dt =y(1-ky), where k is a constant,find the fixed points and investigate their stability.
what are the fixed points of the modified Euler Scheme applied to this equation and what is their stability?
=>

dy / dt = y ( 1 - ky )
dy / dt = y - ky^2
dy / dt - y = - ky^2
Let v = y^( - 1 )
y = 1 / v
dy / dt = ( - 1 / v^2 ) * dv / dt

( - 1 / v^2)dv / dt - ( 1 / v ) = ( - k ) ( 1 / v)^2
(dv / dt ) + v = k
e^(t ) dv / dt + ( e^t )v = ke^t
(e^t )v = ke^t + C
v = k + Ce^(- t )

y = 1 / [ k + Ce^( - t ) ]


Can someone please help me after this. did I answer the question correctly? Is my answer incomplete?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
For the equation dy/dt =y(1-ky), where k is a constant,find the fixed points and investigate their stability.
what are the fixed points of the modified Euler Scheme applied to this equation and what is their stability?
=>

dy / dt = y ( 1 - ky )
dy / dt = y - ky^2
dy / dt - y = - ky^2
Let v = y^( - 1 )
y = 1 / v
dy / dt = ( - 1 / v^2 ) * dv / dt

( - 1 / v^2)dv / dt - ( 1 / v ) = ( - k ) ( 1 / v)^2
(dv / dt ) + v = k
e^(t ) dv / dt + ( e^t )v = ke^t
(e^t )v = ke^t + C
v = k + Ce^(- t )

y = 1 / [ k + Ce^( - t ) ]


Can someone please help me after this. did I answer the question correctly? Is my answer incomplete?
I believe you have solved the DE correctly (although I would have just separated the variables). Have you found the critical points yet? Have you checked their stability?
 

grandy

Member
Dec 26, 2012
73
To find the fixed points, we set dy/dt=0 and find the roots, which yields:

y ( 1 - ky ) =0
that gives
y=0,y=1/k
An additional point of interest is k=0, in which case y=0 is the only fixed point.
is there any fixed point, that i need to calculate?
To investigate the stability, i would look at a direction field plot. but dont know how to calculate the stability.
 

grandy

Member
Dec 26, 2012
73
i found out the fixed points, by setting dy/dt=0 and find the roots, which yields:


y ( 1 - ky ) =0
that gives
y=0,y=1/k

To investigate the stability:
f'(y)= 1- 2ky
f'(0)= 1 >0 unstable
f'(1/k) = -1<0 stable


kindly someone please check my answer for the fixed points and normal stability


for the modified euler scheme, i will try it after i got the above answer correct.