MHB Fixed Points and Stability Analysis of dy/dt = y(1-ky) and Modified Euler Scheme

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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?
 
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grandy said:
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?
 
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.
 
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.
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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