Advanced topic in Odes 
For the equation $$\dfrac{dy}{dt} = y ( 1 - ky ),$$ 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?

$$\begin{align}
\dfrac{dy}{dt} &= y ( 1 - ky ) \\
\dfrac{dy}{dt} &= y - ky^2 \\
\dfrac{dy}{dt} - y &= - ky^2 \end{align}$$
Let $v = y^{-1}$, so $y = 1/v$.
$$\begin{align}
\dfrac{dy}{dt} &= \frac{-1}{v^2} \cdot \dfrac{dv}{dt} \\
\frac{-1}{v^2} \dfrac{dv}{dt} - \frac{1}{v} &= ( - k ) \left(\frac{1}{v}\right)^2 \\
\dfrac{dv}{dt} + v &= k \\
e^t \dfrac{dv}{dt} + e^t v &= ke^t \\
e^t v &= ke^t + C \\
v &= k + Ce^{-t} \\
y &= \frac{1}{k + Ce^{-t}} 
\end{align}$$
Can someone please help me after this. 
 A: We are given:
$$\dfrac{dy}{dt} = y (1-k y)$$
To find the fixed points, we set $\dfrac{dy}{dt} = 0$ and find the roots, which yields:
$$y (1-k y) = 0 \implies y = 0, y = \dfrac{1}{k}$$
An additional point of interest is $k = 0$, in which case $y = 0$ is the only fixed point.
To investigate the stability, we would look at a direction field plot. I am going to give three samples as that shows the stability for all general cases with $k = -1, 0, 1$.
k = -1

Notice that we have $y = 0$ as an unstable point and $y = -1$ as a stable point. Generally $y = 0$ will be unstable and $y = -k, k \gt 0~$ is stable.
k = 0

Notice that $y = 0$ is unstable.
k = 1

Notice that the stability argument is flipped for the $k = -1$ case, but this same effect is also true for a general value of $k$.
Here is an animated version of the direction field:

Now that you have this as a sample, you can repeat for the modified Euler argument.
A: 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: $f'(y)= 1- 2ky$
$$f'(0)= 1 >0  \mbox{ unstable}$$
$$f'(1/k) = -1<0 \mbox{ stable} $$
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.
