# Advanced topic in Odes 2

#### grandy

##### Member
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 ) ]

#### Prove It

##### Well-known member
MHB Math Helper
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 ) ]

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
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
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.