Find the general integral of the homogeneous equation…
Find the general integral of the homogeneous equation $1+e^{x/y}+e^{x/y}\left(1-\dfrac{x}{y}\right)y'=0$.
Explanation of Method
Since the equation depends on $x/y$, use $x=vy$ (equivalently work with $\dfrac{dx}{dy}=1/y'$) to reduce it to a separable equation in $v$ and $y$, recognizing the antiderivative $\int\frac{1+e^v}{e^v+v}\,dv=\ln|e^v+v|$.
Step 1: Substitute $x=vy$
With $v=x/y$, $y'=\dfrac{-(1+e^v)}{e^v(1-v)}$, and since $x=vy\Rightarrow\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}=\dfrac{1}{y'}$, substituting and simplifying gives
\[y\frac{dv}{dy}=-\frac{e^v+v}{1+e^v}.\]
Step 2: Separate and integrate
\[\frac{1+e^v}{e^v+v}\,dv=-\frac{dy}{y}.\]
Recognizing that $\dfrac{d}{dv}(e^v+v)=1+e^v$, the left side integrates to $\ln|e^v+v|$, so
\[\ln|e^v+v|=-\ln|y|+C \quad\Longrightarrow\quad y(e^v+v)=C.\]
Step 3: Back-substitute $v=x/y$
\[\boxed{\,x+ye^{x/y}=C.}\]
Result
Final answer:
\[\boxed{\,x+ye^{x/y}=C.}\]