Solve the differential equation xy' + y = f(x)varphi(xy)…

Solve the differential equation $xy' + y = f(x)\varphi(xy)$, where $f$ and $\varphi$ are arbitrary (sufficiently smooth) functions.

Recognize the exact left side

Since \(\dfrac{d(xy)}{dx} = y + x\dfrac{dy}{dx}\), the equation \(xy'+y = f(x)\varphi(xy)\) is exactly

\[\frac{d(xy)}{dx} = f(x)\varphi(xy).\]

Substitute \(u=xy\) and separate

With \(u=xy\), \(\dfrac{du}{dx} = f(x)\varphi(u)\), which separates as \(\dfrac{du}{\varphi(u)} = f(x)\,dx\). Integrating,

\[\Phi(u) = F(x) + C, \qquad \Phi(u)=\int\frac{du}{\varphi(u)},\ \ F(x)=\int f(x)\,dx.\]

Verify by differentiating back

Differentiating \(\Phi(xy)=F(x)+C\) gives \(\Phi'(xy)\cdot\frac{d(xy)}{dx}=F'(x)\), i.e. \(\frac{1}{\varphi(xy)}\frac{d(xy)}{dx}=f(x)\), which rearranges to \(xy'+y=f(x)\varphi(xy)\) — the original equation, confirming the derivation.

Special case: \(\varphi\equiv 1\)

Then \(xy=\int f(x)\,dx + C\), so \(y=\frac1x\bigl(\int f(x)dx+C\bigr)\). Verified symbolically for a representative \(f(x)=x\): residual of \(xy'+y-f(x)\varphi(xy)\) is exactly \(0\).

Special case: \(f\equiv 1,\ \varphi(u)=u\)

Then \(\frac{du}{u}=dx \Rightarrow u=Ae^x\), so \(y = \dfrac{Ae^x}{x}\). Verified symbolically: residual is exactly \(0\).

Special case: \(f(x)=x,\ \varphi\equiv1\)

Then \(xy=\frac{x^2}2+C\), so \(y=\dfrac{x}{2}+\dfrac{C}{x}\). Verified symbolically: residual is exactly \(0\).

Special case: \(f(x)=1/x,\ \varphi(u)=u^n\)

Then \(\frac{du}{u^n}=\frac{dx}{x}\). For \(n\ne1\): \((xy)^{1-n} = (1-n)\ln|x| + C_1\), i.e. \(xy=\bigl[(1-n)\ln|x|+C_1\bigr]^{1/(1-n)}\) — verified symbolically for \(n=2,3\), residual exactly \(0\). For \(n=1\): \(\ln|u|=\ln|x|+C \Rightarrow xy=Cx \Rightarrow y=C\), also verified.