Advanced topic in Odes 2

[ra_forever8]

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]

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?

 

[ra_forever8]

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 don't know how to calculate the stability.

 

[ra_forever8]

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 stablekindly someone please check my answer for the fixed points and normal stabilityfor the modified euler scheme, i will try it after i got the above answer correct.

 

[blue_raver22]

Your answer is correct so far. To find the fixed points of the modified Euler Scheme, we set dy/dt = 0 and solve for y.

dy/dt = y(1-ky) = 0
y = 0 or 1/k

So the fixed points are y = 0 and y = 1/k.

To investigate their stability, we can plug these values into the original equation and see if they are stable or unstable.

For y = 0:
dy/dt = 0(1-0) = 0
This means that the solution will remain at y = 0, making it a stable fixed point.

For y = 1/k:
dy/dt = (1/k)(1-1) = 0
This also means that the solution will remain at y = 1/k, making it a stable fixed point.

So both fixed points are stable.