$y' = \exp(-\frac{y}{x}) + \frac{y}{x}$ Could you help me to solve this differential equation:
$$y' = \exp\left(-\frac{y}{x}\right) + \frac{y}{x}$$
 A: Hint: Try the change of coordinates $u=\frac{y}{x}$.
A: Following the hint of user71352 we substitute
$$y(x) = x\cdot v(x), \qquad y'(x) = v(x) + x\cdot v'(x)$$
back  and we get
\begin{align*}
v(x) + xv'(x) &=e^{-v(x)} + v(x) \\
xv'(x) &= e^{-v(x)} \\
v'(x) &= \dfrac{e^{-v(x)}}{x} \\
e^{v(x)}v'(x)&= \dfrac{1}{x} \\
\int e^{v(x)}v'(x) \, dx&= \int\dfrac{1}{x}\, dx \\
e^{v(x)} &= \log(x) + C \\
v(x) &= \log(\log(x)+C) \implies y(x) = x\log(\log(x)+C)
\end{align*}
where $C$ is an arbitrary constant
A: $$y' = \exp\left(-\frac{y}{x}\right) + \frac{y}{x}$$
$$x \frac{xy'-y}{x^2} = \exp\left(-\frac{y}{x}\right)$$
$$x \left(\frac{y}{x}\right)' = \exp\left(-\frac{y}{x}\right)$$
$$x u' = e^{-u}$$
$$e^{u} u' = \frac{1}{x}$$
$$(e^{u})' = \frac{1}{x}$$
$$e^{u}-e^{u_0}=\ln x-\ln x_0$$
A: Given the transcendental term $\exp(-\frac{y}{x})$, we are pushed to consider a change of variable $z=\frac yx$, i.e. $y=zx$, and
$$z'x+z=\exp(-z)+z.$$
Then we get a separable equation which we can integrate straight away
$$z'\exp(z)=\frac1x,$$
$$\exp(z)-\exp(z_0)=\ln|x|-\ln|x_0|,$$
$$z=\ln(\ln|x|-\ln|x_0|+\exp(-z_0)),$$
$$\frac yx=\ln(\ln|x|-\ln|x_0|+\exp(-\frac{y_0}{x_0})),$$
and
$$y=x\ln(\ln\left|\frac x{x_0}\right|+\exp(-\frac{y_0}{x_0})).$$
The function is odd and undefined when $|x|\le|x_0|\exp(-\exp(-\frac{y_0}{x_0}))$. It has two vertical asymptotes, and due to the very slow growth of the function $\ln(\ln x)$, it is quasi-linear for larger $x$.
