Solve $e^x cdot 2y’ = y^2 + y’^2$

**My Attempt**

Let $p = y’$ then $2e^x cdot p=y^2+p^2$. No differentiate both sides to get:

$$2e^x cdot p + 2e^x frac{dp}{dx} = 2yp + 2pfrac{dp}{dx}$$

$$frac{dp}{dx}(2e^x-2p) = 2yp-2e^xp rightarrow frac{dp}{dx}(e^x-p) = yp-e^xp$$

I can not easily identify this form so I make the substitution $y = ux$ and $y’ = u’x + u$

$$(u’x+u)(e^x-ux) = yux-e^xux$$

Later manipulation does not really get me to an easier form, where should I go from here?