non-linear to linear differential equation, Can $xz'=z^2e^{1/z}$ be solved?

Question

$(E) :\qquad x^2y'=y^2e^{\frac{x}{y}}+xy$

Attempt :

Let $z:=\dfrac{y}{x}$ thus $z'=\dfrac{xy'-y}{x^2}\iff x^2z'=xy'-y\iff y'=xz'+\dfrac{y}{x}\iff y'=xz'+z$

We deduce that $x^2(xz'+z)=(xz)^2e^{1/z}+x^2z\iff x^3z'=x^2z^2e^{1/z}\iff xz'=z^2e^{1/z}$

So $\displaystyle \dfrac{z'}{z^2e^{1/z}}=\dfrac{1}{x}\iff \int \dfrac{1}{s^2e^{1/s}}\cdot ds=\int\dfrac{1}{x}\cdot dx$, can we go further?

Answer 1

Observe that $$ \frac{\rm d}{{\rm d}x}e^{-x/y}=e^{-x/y}\frac{xy'-y}{y^2}. $$ The original equation is equivalent to $$ x\frac{xy'-y}{y^2}=e^{x/y}\iff\frac{\rm d}{{\rm d}x}e^{-x/y}=\frac{1}{x}, $$ So straightforwardly, $$ y=-\frac{x}{\log\left(\log x+C\right)} $$ for some constant $C$.

Here $x$ and $y$ are supposed to be real.

Answer 2

$$\displaystyle \dfrac{z'}{z^2e^{1/z}}=\dfrac{1}{x}\iff \int \dfrac{1}{s^2e^{1/s}}\cdot ds=\int\dfrac{1}{x}\cdot dx$$

can we go further?

You were so close to the solution...

Yes just substitute $u=-1/s \implies ds=s^2du$ and integrate

I got the same result as yours...

$$ x^2y'=y^2e^{\frac{x}{y}}+xy$$ $$ x(xy'-y)=y^2e^{\frac{x}{y}}$$ $$ \frac {(xy'-y)}{x^2}=y^2e^{\frac{x}{y}}\frac 1 {x^3}$$ $$z'=\frac {e^{1/z}}{x}z^2$$ $$\int\frac {e^{-1/z}dz}{z^2}=\int \frac {dx}{x}$$ $$\int\frac {e^{-1/z}dz}{z^2}=\ln x + K$$ substitute $u=-1/z \implies dz=z^2du$ $$\int e^{u}du=\ln x + K$$ $$e^{-x/y}=\ln x + K$$ $$\boxed{y(x)=-\frac x {\ln (\ln x + K)}}$$