Steady State Solution Non-Linear ODE I'm working through some problems studying for a numerical methods course, but I'm stuck on how to answer the following question analytically. It says to find the steady state solution of the following equation:
$\frac{dy}{dx} = -ay + e^{-y}$ where $y(0)=0$. It says that the steady state solution is when $\frac{dy}{dx} = 0$. It's also claimed that the steady state solution is a fixed point of the system.
I'm stuck on where to start with this. If I immediately try to solve $0 = -ay + e^{-y}$, I can't isolate y. Then I tried to start with the homogenous equation, but that devolved also.
What am I missing?
 A: Firstly, I will expand on my comment.
If $a\neq 0$, then $0=-ay+e^{-y}$, which can be rearranged to $ye^y=\dfrac{1}{a}$. This has the "analytic" solution using the LambertW function $y=LambertW(1/a)$.
If $a=0$, then the original ODE is $y'=e^{-y}$, which has solution $y(x)=\log(x+c)$. This does not have a steady state.
Now, given that you are in a numerical analysis class, you can also solve this equation using numerical methods. Hence, a solution could be obtained using Newton's method. Given $f(y)=e^{-y}-ay$, the iteration would be
$$
y_{n+1}=y_n-\frac{e^{-y_n}-ay_n}{-e^{-y_n}-a}=y_n+\dfrac{1+ay_ne^{y_n}}{1+ae^{y_n}}.
$$
This will break down when $1+ae^y=0$.
Hence, if $a>0$, The iteration will not break down and is guaranteed to find a solution.
If $a<0$, it is possible that the expression is satisfied. Now, appealing to the geometric interpretation of the problem, we see that there is a solution provided that the graphs of $l_1=ay$ and $l_2=e^{-y}$ intersect. The least magnitude negative value for $a$ must then be the value where these two graphs are tangent. This leads to $a=-e^{-p}$ where $y=p$ is the point of tangency. But, we also require the line $l_1=ay$ to be satisfied at $(p,l_2(p))$ Hence, $e^{-p}=-e^{-p}p\implies p=-1$. Thus, the least magnitude negative value for $a$ is $a=-e$. The method will then also not breakdown if $a\leq-e$. This result is confirmed as the $LambertW(x)$ function is defined only when $x\geq-1/e$, which corresponds to $a\leq-e$ when $a<0$.
Conclusion
A solution will exist provided $a>0$ or $a\leq-e$. This is confirmed by this implicit plot of the solution.

