Solve the differential equation y' = y^2 f(xy) , where f is…
Solve the differential equation $y' = y^2 f(xy)$, where $f$ is an arbitrary (sufficiently smooth) function.
Explanation of Method
The right-hand side depends on \(y\) through the product \(xy\), which suggests the substitution \(u=xy\). This turns the equation into a separable one in \(u\) and \(x\), reducing the problem to a single quadrature that depends on the specific form of \(f\).
Substitute \(u=xy\)
Let \(u=xy\), so \(\dfrac{du}{dx} = y + x\dfrac{dy}{dx}\). Using the ODE, \(x y' = xy^2f(xy) = uy\,f(u)\), hence
\[\frac{du}{dx} = y + uy\,f(u) = y\bigl[1+uf(u)\bigr].\]
Since \(y=u/x\), this becomes
\[\frac{du}{dx} = \frac{u\bigl[1+uf(u)\bigr]}{x}.\]
Separate and integrate
Separating variables,
\[\frac{du}{u\bigl[1+uf(u)\bigr]} = \frac{dx}{x},\]
so the general (implicit) solution is
\[\int \frac{du}{u\bigl[1+uf(u)\bigr]} = \ln|x| + C, \qquad u = xy.\]
Verify by differentiating back
Differentiating \(\int \frac{du}{u(1+uf(u))} = \ln|x|+C\) with respect to \(x\) gives \(\dfrac{1}{u(1+uf(u))}\dfrac{du}{dx} = \dfrac1x\), i.e. \(\dfrac{du}{dx}=\dfrac{u(1+uf(u))}{x}\). Substituting \(u=xy\), \(\frac{du}{dx}=y+xy'\), this rearranges exactly to \(y'=y^2f(xy)\), confirming the derivation (verified symbolically with sympy).
Special case: \(f(xy)=k\) (constant)
Then \(y'=ky^2\), which separates directly: \(-1/y = kx+C\), so \(y=-\dfrac{1}{kx+C}\). Direct substitution back into \(y'=ky^2\) confirms this exactly (residual \(=0\)).
Special case: \(f(xy)=1/(xy)\)
Then \(y' = y^2\cdot\frac{1}{xy} = y/x\), a homogeneous equation. With \(y=vx\): \(v+xv' = v \Rightarrow xv'=0 \Rightarrow v=C\), giving \(y=Cx\). Direct substitution back into \(y'=y^2f(xy)\) confirms this exactly.
Special case: \(f(xy)=(xy)^n\)
Then \(y'=y^2(xy)^n\), and with \(u=xy\) the general formula gives \(\displaystyle\int\frac{du}{u(1+u^{n+1})}=\ln|x|+C\). Setting \(n=0\) recovers the constant case; \(n=-1\) recovers \(f=1/(xy)\).
Result
Final answer:
The general implicit solution, obtained via \(u=xy\), is:
\[\boxed{\int \frac{du}{u\bigl[1+u\,f(u)\bigr]} = \ln|x| + C, \qquad u = xy}\]
with special cases: for \(f\equiv k\), \(y=-\dfrac{1}{kx+C}\); for \(f(xy)=1/(xy)\), \(y=Cx\); for \(f(xy)=(xy)^n\), \(\displaystyle\int\frac{du}{u(1+u^{n+1})}=\ln|x|+C\). All cases verified by direct substitution — the source's derivation and results are correct as given.