Find the fixed points of the implicit Euler scheme \begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1}) \end{equation} when applied to the differential equation $y'=y(1-y)$ and investigate their stability?
=> implicit Euler scheme \begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1}) \end{equation}
$y'=y(1-y)$ \begin{equation} y_{n+1}=y_{n}+hy_{n+1}(1-y_{n+1}) \end{equation} \begin{equation} y_{n+1}=y_{n}+hy_{n+1}-hy^2_{n+1} \end{equation}
For fixed points $y_{n+1}=y_{n}$ \begin{equation} y_{n}=y_{n}+hy_{n}-hy^2_{n} \end{equation} $y_{n}=0$ or $1$
I got problem with stability but this is what I have done
$y_{n}= \alpha +\epsilon^n$, $y_{n+1}= \alpha +\epsilon^{n+1}$, \begin{equation} \alpha +\epsilon^{n+1}= \alpha +\epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation} \begin{equation} \epsilon^{n+1}= \epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation} When $y_{n}=0=\alpha$ \begin{equation} \epsilon^{n+1}= \epsilon^n + h \epsilon^{n+1}(1-\epsilon^{n+1}) \end{equation}
I don't what to say or do after that to determine the stability.
When $y_{n}=1=\alpha$ \begin{equation} \epsilon^{n+1}= \epsilon^n - h \epsilon^{n+1}(1+\epsilon^{n+1}) \end{equation}
same again what can say about with my answer to investigate the stability.
Starting from $$ y_{n+1}=y_{n}+hy_{n+1}(1-y_{n+1}) $$ multiply with $4h$ and rearrange $$ (2hy_{n+1})^2+2(2hy_{n+1})(1-h)=4hy_n \\ (2hy_{n+1}+(1-h))^2=(1-h)^2+4hy_n $$ to get \begin{align} y_{n+1}&=\frac1{2h}\left(-(1-h)+\sqrt{(1-h)^2+4hy_n}\right) \\ &=1+\frac1{2h}\left(-(1+h)+\sqrt{(1+h)^2-4h+4hy_n}\right) \\[0.8em] &=\begin{cases} \dfrac{2y_n}{(1-h)+\sqrt{(1-h)^2+4hy_n}} &\approx \dfrac{y_n}{1-h} &&\text{ for } y_n\approx 0 \\[0.3em] 1+\dfrac{2(y_n-1)}{(1+h)+\sqrt{(1+h)^2+4h(y_n-1)}} &\approx 1+\dfrac{y_n-1}{1+h} &&\text{ for } y_n\approx 1 \end{cases} \end{align} In this direct form, the reasoning about stability should be a little easier.
Or even more directly: Stability analysis is about the linearization, so just remove the term that are quadratic in $ϵ^{n+1}$ to get magnification factors of $(1-h)^{-1}$ and $(1+h)^{-1}$, indicating one stable and one unstable solution.