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**Ackbeet** Hmm. Strange. I don't get quite the same integration you do. Using your substitution, I get

$\displaystyle \left(2\,\dfrac{y}{x}\,e^{y/x}+1\right)y'=2+2\,\dfrac{y}{x}\implies$

$\displaystyle (2ue^{u}+1)(xu'+u)=2+2u\implies$

$\displaystyle xu'+u=\dfrac{2+2u}{2ue^{u}+1}\implies$

$\displaystyle xu'=\dfrac{2+2u}{2ue^{u}+1}-u=\dfrac{2+2u-2u^{2}e^{u}-u}{2ue^{u}+1}=\dfrac{2+u-2u^{2}e^{u}}{2ue^{u}+1}\implies$

$\displaystyle \dfrac{(2ue^{u}+1)\,du}{2+u-2u^{2}e^{u}}=\dfrac{dx}{x}.$

The integral on the left is intractable. Have I made a mistake somewhere?

I wonder if the opposite substitution, that of $\displaystyle vy=x,$ would give you a more do-able integral. *Working it out...*

[EDIT]: Nope, that integral is just as intractable.

It looks to me like the best you can do is reduce the DE to quadratures.