How to solve $y'=e^{\frac{xy'}y}$? How to solve the following equation?
$$y'=e^{\frac{xy'}y}$$
We must find a common solution.
 A: $$y'=e^{xy'/y}$$
$$y\ln(y')=xy'$$


*

*Substitute $y=e^{Ax+C}$ to get
$$e^{Ax+C}\ln(Ae^{Ax+C})=xAe^{Ax+C}$$
$$\ln(Ae^{Ax+C})=Ax$$
$$Ae^{Ax}e^C=e^{Ax}$$
$$Ae^C=1$$
$$A=e^{-C}$$
So the solution is $$y=e^{e^{-C}x+C}$$

*Substitute $y=-e^{Ax+C}$ and follow steps from above to get the solution $$y=-e^{-e^{-C}x+C}$$
Be aware that by substitutions we are not guaranteed to find all solutions.
A: $$y'=e^{\frac{xy'}{y}}$$
Let : $u=\frac{xy'}{y}$ so, $y'=\frac{yu}{x}$
$$\frac{yu}{x}=e^u$$
$$y=x\frac{e^u}{u}$$
$$y'=\frac{e^u}{u}+x\frac{e^u}{u}u'-x\frac{e^u}{u^2}u'=e^u$$
This allows to eliminate $e^u$
$$\frac{1}{u}+x\frac{1}{u}u'-x\frac{1}{u^2}u'=1$$
$$x(u-1)u'=u^2-u=(u-1)u$$
A particular solution is $u=1$ and the trivial solution $y=x\frac{e^1}{1}=ex$
General case $u\neq 1$ :
$$xu'=u$$
$$u=cx$$
Then, the general solution :
$$y=x\frac{e^u}{u}=x\frac{e^{cx}}{cx}$$
$$y=\frac{e^{cx}}{c}$$
A: $$y'(x)=e^{\frac{xy'(x)}{y(x)}}\Longleftrightarrow$$

Let $y(x)=xv(x)$, which gives $y'(x)=v(x)+xv'(x)$:

$$xv'(x)+v(x)=e^{\frac{xv'(x)+v(x)}{v(x)}}\Longleftrightarrow$$
$$xv'(x)+v(x)=e^{1+\frac{xv'(x)}{v(x)}}\Longleftrightarrow$$
$$v'(x)=\frac{-v(x)-\text{W}\left(-\frac{1}{v(x)}\right)v(x)}{x}\Longleftrightarrow$$
$$\frac{v'(x)}{-v(x)-\text{W}\left(-\frac{1}{v(x)}\right)v(x)}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{v'(x)}{-v(x)-\text{W}\left(-\frac{1}{v(x)}\right)v(x)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$\ln\left|\text{W}\left(-\frac{1}{v(x)}\right)\right|=\ln\left|x\right|+\text{C}\Longleftrightarrow$$
$$v(x)=-\frac{e^{-e^{\text{C}}x-\text{C}}}{x}\Longleftrightarrow$$
$$y(x)=-e^{-e^{\text{C}}x-\text{C}}$$
