Solving the following non-separable differential equation: $y'=\ln(x+y)$ I'm getting stuck trying to separate, and thus solve this differential equation. Here $y' := \frac{\mathrm dy}{\mathrm dx}$

$$y'=\ln(x+y)$$

Is this differential separable using elementary functions? WolframAlpha is mentioning something about the Exponential integral ($\text{E}_i$). We haven't been taught about this $\text{E}_i$ term...
 A: The differential equation cannot be solved in terms of a finite number of elementary functions. In this answer, we do not restrict ourselves to elementary functions.

One can reduce this to a separable ODE by substituting:
$$v=x+y \implies \frac{dv}{dx}=1+\frac{dy}{dx}$$
This gives:
$$\frac{dv}{dx}=\ln(v)+1$$
One can separate both sides and integrate:
$$\int \frac{1}{\ln(v)+1}~dv=\int dx$$
The left hand side cannot be integrated in terms of elementary functions as you mentioned. However, it can be evaluated using the Exponential Integral $\operatorname*{Ei}(x)$.

The trick here is to substitute:
$$u=\ln(v)+1 \implies v=e^{u-1} \implies dv=e^{u-1}~du$$
This gives:
$$\int \frac{e^{u-1}}{u}~du=\int dx \tag{1}$$
It follows from the definition that:
$$\int \frac{e^u}{u}~du=\operatorname*{Ei}(u)+C$$
Putting $(1)$ in that form gives:
$$\frac{1}{e}\int \frac{e^u}{u}~du=\int dx$$
$$\frac{\operatorname*{Ei}(u)}{e}=x+c \tag{2}$$
All that remains to do now is to substitute back to obtain an implicit solution for $y(x)$.
A: If you take a new variable $z = x + y$, the equation becomes $z' = 1 + \ln(z)$.
This is separable, but the integration is not elementary: that's where the $Ei$ comes in.
