An explicit form for this differential equation. So I came across this differential equation:
$$xyy'-y^2=(x+y)^2 e^{-y/x}$$
And I managed to simplify it this way:
First I multiplied by $\frac1{xy}$:
$$y'-\frac{y}{x}=\left(\frac{x}{y}+2+\frac{y}{x}\right) e^{-y/x}$$
Then substituted $v = \frac{y}{x}$ sub to get:
$$\begin{split}
v+xv'-v&=\left(\frac1v+2+v\right) e^{-v}\\
xvv'&=(1+2v+v^2) e^{-v}\\
xvv'&=(v+1)^2 e^{-v}
\end{split}$$
Which is separable. I then proceeded to separate and integrate both sides to get
$$\frac{-e^v}{v+1}\ + e^v=\ln⁡(x)+c$$ or $$\frac{e^v}{v+1}\
=\ln⁡(x)+c$$
Is there a way to get a more explicit equation? Is there something wrong that I did?
 A: I don't see anything wrong with your work, but you can simplify this even further.
You can actually solve the ODE $x \cdot y\left( x \right) \cdot \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} - \left( y\left( x \right) \right)^{2} = \left( x + y\left( x \right) \right)^{2} \cdot e^{-\frac{y\left( x \right)}{x}}$ for $y$ completely explicitly by using the Lambert W-Function $\operatorname{W}_{n}\left( z \right)$ where $\operatorname{W}_{n}\left( z \right) \cdot e^{\operatorname{W}_{n}\left( z \right)} = z$.
First you should substitute $y\left( x \right) ~{:=}~ x \cdot v\left( x \right)$:
$$
\begin{align*}
y\left( x \right) &= x \cdot v\left( x \right)\\
\frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= x \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + v\left( x \right)\\
\end{align*}
$$
So with the substitution you get:
$$
\begin{align*}
x \cdot y\left( x \right) \cdot \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} - \left( y\left( x \right) \right)^{2} &= \left( x + y\left( x \right) \right)^{2} \cdot e^{-\frac{y\left( x \right)}{x}}\\
x \cdot x \cdot v\left( x \right) \cdot x \cdot \left( \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + v\left( x \right) \right) - \left( x \cdot v\left( x \right) \right)^{2} &= \left( x + x \cdot v\left( x \right) \right)^{2} \cdot e^{-\frac{x \cdot v\left( x \right)}{x}}\\
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= \left( x \cdot \left( 1 + v\left( x \right) \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= x^{2} \cdot \left( 1 + v\left( x \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
\end{align*}
$$
Now you can add the terms with the $y\left( x \right)$ from the $x$ and then integrate:
$$
\begin{align*}
x^{3} \cdot v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= x^{2} \cdot \left( 1 + v\left( x \right) \right)^{2} \cdot e^{-v\left( x \right)}\\
\frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}} &= \frac{1}{x}\\
\int \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \int \frac{1}{x}\, \operatorname{d}x\\
\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \ln\left( x \right) + c\\
\end{align*}
$$
Using the rule $\frac{\operatorname{d}\frac{e^{v\left( x \right)}}{v\left( x \right) + 1}}{\operatorname{d}x} = \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}$ we get $\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}} \operatorname{d}x = \int_{0}^{x} \frac{\operatorname{d}\frac{e^{v\left( x \right)}}{v\left( x \right) + 1}}{\operatorname{d}x} \operatorname{d}x = \frac{e^{v\left( x \right)}}{v\left( x \right) + 1}$, so we hold:
$$
\begin{align*}
\int_{0}^{x} \frac{v\left( x \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} \cdot e^{v\left( x \right)}}{\left( 1 + v\left( x \right) \right)^{2}}\, \operatorname{d}x &= \ln\left( x \right) + c\\
\frac{e^{v\left( x \right)}}{v\left( x \right) + 1} &= \ln\left( x \right) + c\\
\left( v\left( x \right) + 1 \right) \cdot e^{-v\left( x \right)} &= \frac{1}{\ln\left( x \right) + c}\\
v\left( x \right) \cdot e^{-v\left( x \right)} + e^{-v\left( x \right)} &= \frac{1}{\ln\left( x \right) + c}\\
\left( -v\left( x \right) + 1 \right) \cdot e^{-v\left( x \right) + 1} &= -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)}\\
-v\left( x \right) + 1 &= \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)\\
v\left( x \right) + 1 &= -\operatorname{W}_{n}\left( -\frac{1}{e \left( \ln\left( x \right) + c \right)} \right)\\
v\left( x \right) &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
\end{align*}
$$
Now we can substitute $v\left( x \right)$ to $y\left( x \right) ~{:=}~ x \cdot v\left( x \right) \Leftrightarrow v\left( x \right) = \frac{y\left( x \right)}{x}$ and get:
$$
\begin{align*}
v\left( x \right) &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
\frac{y\left( x \right)}{x} &= -\operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - 1\\
y\left( x \right) &= -x \cdot \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - x\\
\end{align*}
$$
The probe via Wolfram|Alpha confirms the solution, as you can see here.
However, if you don't accept a concatenation with Lambert W-Function $\operatorname{W}_{n}\left( z \right)$ as the solution, then you could also write the solution in terms of the Fox H-Function $\operatorname{H}_{p,\, q}^{m,\, n}\left( \begin{matrix} a_{1}, &\dots, &a_{p}\\ b_{1}, &\dots, &b_{q}\\\end{matrix} \mid z \right)$ for the branch $n = -1$:$^{\left[ 1. \right]}$
$$
\begin{align*}
y\left( x \right) &= -x \cdot \operatorname{W}_{n}\left( -\frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right) - x\\
y\left( x \right) &= \begin{cases} -\overline{x} \cdot \lim_{\beta \to \alpha^{-}} \left[ \overline{\frac{\alpha^{2} \cdot \left( \left( \alpha - \beta \right) \cdot z \right)^{\frac{\alpha}{\beta}}}{\beta} \cdot \operatorname{H}_{1,\, 2}^{1,\, 1} \left( \begin{matrix} \left( \frac{\alpha + \beta}{\beta},\, \frac{\alpha}{\beta} \right)\\ \left( 0,\, 1 \right),\, \left( -\frac{\alpha}{\beta},\, \frac{\alpha - \beta}{\beta} \right)\\\end{matrix} \mid -\left( -\left( \alpha - \beta \right) \cdot \frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)^{\frac{\alpha}{\beta} - 1} \right) - x} \right],\, \text{for} \left| x \right| < \frac{1}{e \left| \alpha \right|}\\
-\overline{x} \cdot \lim_{\beta \to \alpha^{-}} \left[ \overline{\frac{\alpha^{2} \cdot \left( \left( \alpha - \beta \right) \cdot z \right)^{-\frac{\alpha}{\beta}}}{\beta} \cdot \operatorname{H}_{2,\, 1}^{1,\, 1} \left( \begin{matrix} \left( 1,\, 1 \right),\, \left( \frac{\beta - \alpha}{\beta},\, \frac{\alpha - \beta}{\beta} \right)\\ \left( -\frac{\alpha}{\beta},\, \frac{\alpha}{\beta} \right)\\\end{matrix} \mid -\left( -\left( \alpha - \beta \right) \cdot \frac{1}{e \cdot \left( \ln\left( x \right) + c \right)} \right)^{1 - \frac{\alpha}{\beta}} \right) - x} \right],\, \text{otherwise}\\ \end{cases}\\
\end{align*}
$$
where $\overline{z}$ is the complex conjugate of $z$ and $\alpha = -1$.
