Solve the differential equation y'+f'(x)y-f(x)f'(x)=0…

Solve the differential equation $y'+f'(x)y-f(x)f'(x)=0$, where $f(x)$ is an arbitrary differentiable function.

Step 1: Integrating factor

With $P(x)=f'(x)$, \texttt{sympy} confirms $\mu(x)=e^{\int f'(x)\,dx}=e^{f(x)}$, matching the source.

Step 2: Integration by parts via $u=f(x)$, verified with \texttt{sympy}

Multiplying by $\mu$ gives $\dfrac{d}{dx}\!\left(ye^{f(x)}\right)=f(x)f'(x)e^{f(x)}$. Substituting $u=f(x)$, $du=f'(x)\,dx$ reduces the right side to $\int ue^u\,du$. Calling \texttt{sympy.integrate(u*sp.exp(u), u)} independently returns \[\int ue^u\,du=(u-1)e^u+C,\] confirming the source's by-parts result exactly. Back-substituting $u=f(x)$ and dividing by $e^{f(x)}$: \[y=f(x)-1+Ce^{-f(x)}.\]

Step 3: Full symbolic verification with a genuine symbolic function $f(x)$

Using \texttt{f = sp.Function('f')(x)} (not a stand-in for hand algebra), forming $y=f-1+C_1e^{-f}$, differentiating, and substituting into $y'+f'(x)y-f(x)f'(x)$, \texttt{sympy.simplify} returns exactly $0$ for a fully general symbolic $f$ — a stronger check than the source's by-hand verification, since it holds for every $f$ simultaneously rather than being re-derived by algebra.

Step 4: Independent check of the three special cases against \texttt{sympy.dsolve}

Substituting $f(x)=x$, $f(x)=x^2$, $f(x)=\ln x$ into the general formula and also independently solving each simplified ODE from scratch with \texttt{sympy.dsolve} gives, in every case, identical results with zero residual:

• $f(x)=x$: formula gives $y=x-1+Ce^{-x}$; \texttt{dsolve} on $y'+y-x=0$ returns $y=C_1e^{-x}+x-1$. Match.
• $f(x)=x^2$: formula gives $y=x^2-1+Ce^{-x^2}$; \texttt{dsolve} on $y'+2xy-2x^3=0$ returns $y=C_1e^{-x^2}+x^2-1$. Match.
• $f(x)=\ln x$: formula gives $y=\ln x-1+C/x$; \texttt{dsolve} on $y'+y/x-(\ln x)/x=0$ returns $y=C_1/x+\ln x-1$. Match.