Solve y'=e^(xy'/y) using the parameter method
Solve $y'=e^{xy'/y}$ using the parameter method.
Explanation of Method
Parameter substitution $p=y'$, logarithm, reduction to a separable Lagrange-type ODE, all steps independently re-verified with \texttt{sympy}.
Step 1: Introduce the parameter and take logs
Let $p=y'$. Taking $\ln$ of both sides of $p=e^{xp/y}$ gives $\ln p=xp/y$, hence
\[ y=\frac{xp}{\ln p} \](requiring $p>0$ and $p\neq1$).
Step 2: Differentiate and impose $dy/dx=p$ (sympy)
Treating $p$ as a function of $x$, \texttt{sympy} differentiates $y=xp/\ln p$ and sets the result equal to $p$. Simplifying and solving for $p'(x)$ gives exactly
\[ \frac{dp}{dx}=\frac{p\ln p}{x} \]which rearranges to the separable form $p\ln p\,dx=x\,dp$, i.e. $\dfrac{dx}{x}=\dfrac{dp}{p\ln p}$ — matching the source's Step 2–3 exactly.
Step 3: Integrate (sympy) and back-substitute
\texttt{sympy.integrate} confirms $\int\frac{dp}{p\ln p}=\ln|\ln p|$ and $\int\frac{dx}{x}=\ln|x|$, and \texttt{sympy.dsolve} applied directly to $dp/dx=p\ln p/x$ returns $p(x)=e^{C_1x}$, equivalent (relabelling $C=1/C_1$) to the source's $x=C\ln p$. Substituting $\ln p=x/C$ into $y=xp/\ln p$ gives $y=Cp$, i.e. $p=y/C$; combining with $x=C\ln p=C\ln(y/C)$ and solving for $y$ gives
\[ y=Ce^{x/C} \]Step 4: Final verification (sympy)
Substituting $y=Ce^{x/C}$ (so $y'=e^{x/C}$) into $y'-e^{xy'/y}$ and simplifying with \texttt{sympy} returns $0$ identically — the source's Step 7 verification checkmark is correct. The intermediate identity $y=xp/\ln p$ with $p=e^{x/C}$ was checked separately and also holds identically.
Result
Final answer:
Verified — no correction needed. The explicit general solution is
\[ y=Ce^{x/C},\qquad C\neq0 \]confirmed to satisfy $y'=e^{xy'/y}$ identically by direct \texttt{sympy} substitution. Domain/sign remarks: $p=y'=e^{x/C}>0$ always, consistent with the requirement $p>0$ from Step 1; the auxiliary relation $y=xp/\ln p$ has a removable $0/0$ singularity at $p=1$ (i.e. $x=0$), but the final closed form $y=Ce^{x/C}$ is perfectly regular there ($y(0)=C$). As $C\to0$, the family degenerates non-uniformly (the limit depends on the sign of $x$ rather than tending to a single limiting curve), so $C=0$ is excluded.