I am trying to solve the following equation and find a real-valued function $y\left(x\right)$ which satisfies
$$ \exp{y\left(x\right)} + y\left(x\right) + 1 = 0, \hspace{0.2cm}\forall x>0 $$
The trivial solution $y\left(x\right) = x_0$ where $x_0$ is the root of $$ \exp{x} + x + 1 = 0 $$ is probably one solution. Is there any way to find a non-trivial one? Any ideas?
On
No matter that you have marked it as $y(x)$, the solution will be the same as
$$e^x+x+1=0$$
In order $y(x)$ to be a function $x$ must show up elsewhere not just within $y(x)$. Otherwise $y(x)$ has no different meaning than any other symbols you could use $H,u,p,g(x),d...$
Therefore there is no other solution than within reals
$$y(x)=-W(\frac1{e})-1$$
where $W(x)$ is Lambert W-function
or if you take a continuation of W-function it would be
$$y(x)=-W_n(\frac1{e})-1$$
but either way it can be just a constant.
There is certainly at least no differentiable solution. If we differentiate both sides, we get $y'(x)\exp(y(x)) + y'(x) = 0$, or $y'(x)(\exp(y(x)) + 1 = 0)$. Since $\exp(y(x))+1$ is never zero, this implies that $y'(x) = 0$, so that $y(x)$ is a constant, and hence $y(x) = x_0$ where $x_0$ is as you specified.
This reasoning shows that there cannot be a solution that is differentiable on any interval of $\mathbb{R}^+$. Maybe there's crazy function that does work, but there probably isn't a solution that you can write down nicely.