A question in Ordinary differential equations Find the modified equation when the implicit Euler method is applied to $y'= f(y)$. If $f(y)=λy$, where $λ$ is negative. what is the effect on the amplication factor?
$y ' = λ y$ 
$\Rightarrow \frac{dy}{dx} = λ y$ 
$\Rightarrow dy / y = λ dx $
$\Rightarrow \ln y = λx + C$ 
$\Rightarrow y = Ae^{\lambda x}$,
where the constant factor does not depend on $\lambda$. 
This is what I have done so far. Can anyone help me what to do next to answer the question?

The implicit Euler scheme is given by:
$$y_{n+1} = y_n + h f(y_{n+1}, t_{n+1})$$
For $f(y)=\lambda~y$, we have:
$$y_{n+1} = y_n + h f(y_{n+1}, t_{n+1}) = y_n + h~\lambda~y_{n+1}$$
Solving this for $y_{n+1}$ (in general, this is not possible), we arrive at:
$$\tag 1 y_{n+1} = \dfrac{y_n}{1 - h ~\lambda}$$
From $(1)$, we can see that if $|1 - h ~ \lambda| \ge 1$, the solution is decaying (stable). Compare this to the actual solution of $y(x) = c e^{\lambda~x}$.
If we have $\lambda$ being negative, we would have:
$$\tag 2 y_{n+1} = \dfrac{y_n}{1 + h ~\lambda}$$
Compare this to the actual solution of $y(x) = c e^{-\lambda~x}$. What conclusion can you draw?
I would recommend actually generating numerical data using this scheme and seeing what is going on since we have the closed form solution to compare against.
Try it for $\lambda = \pm 1$ and see what happens to stability.
