Show that the differential equation y' = ((x^2+y^2-α)y+λ…

Show that the differential equation \[y' = \frac{(x^2+y^2-\alpha)y+\lambda x}{(x^2+y^2-k)x-\beta y}\] where \(\alpha,\beta,k,\lambda\) are parameters, can be reduced to separable form by the substitution \(x=\rho\cos\varphi\), \(y=\rho\sin\varphi\).

Set up the substitution

With \(x=\rho\cos\varphi,\ y=\rho\sin\varphi\) (so \(x^2+y^2=\rho^2\)) and the standard differentials, the equation \(y'=N/D\) with \[N=(x^2+y^2-\alpha)y+\lambda x, \qquad D=(x^2+y^2-k)x-\beta y\] becomes, after substituting and cancelling one overall factor of \(\rho\) from numerator and denominator, \[\frac{\sin\varphi\,d\rho+\rho\cos\varphi\,d\varphi}{\cos\varphi\,d\rho-\rho\sin\varphi\,d\varphi} = \frac{(\rho^2-\alpha)\sin\varphi+\lambda\cos\varphi}{(\rho^2-k)\cos\varphi-\beta\sin\varphi}.\]

Cross-multiply and collect $d\rho$, $d\varphi$ coefficients

Cross-multiplying and expanding fully with a computer algebra system (expanding all products of trigonometric and polynomial terms symbolically, rather than collecting by hand) gives an equation of the form \(F(\varphi)\,d\rho + G(\rho,\varphi)\,d\varphi = 0\), where the individual coefficients \[F(\varphi) = (\alpha-k)\sin\varphi\cos\varphi - \beta\sin^2\varphi - \lambda\cos^2\varphi,\] \[G(\rho,\varphi) = \rho\big[\rho^2 - k\cos^2\varphi - \alpha\sin^2\varphi + (\lambda-\beta)\sin\varphi\cos\varphi\big]\] were confirmed to exactly match the source's expressions for the \(d\rho\)- and \(d\varphi\)-coefficients individually — this part of the source's algebra is correct.

Sign error found in combining the two coefficients

The source's final "separate variables" step (and its boxed summary) asserts \[F(\varphi)\,d\rho = G(\rho,\varphi)\,d\varphi \quad \text{(source's claim — INCORRECT)}.\] Redoing the full cross-multiplication symbolically from \(N\,dx-D\,dy=0\) (rather than collecting "Left side" and "Right side" terms by hand as the source did) shows this is off by a sign: the correct combined relation is \[F(\varphi)\,d\rho + G(\rho,\varphi)\,d\varphi = 0, \quad\text{i.e.}\quad F(\varphi)\,d\rho = -\,G(\rho,\varphi)\,d\varphi.\] The source's own "collect \(d\rho\) terms" and "collect \(d\varphi\) terms" sub-steps are each individually correct (both equal \(F(\varphi)\) and \(G(\rho,\varphi)\) respectively as differences "Left \u2212 Right"), but since the original cross-multiplied equation reads Left \(=\) Right, moving all terms to one side necessarily gives (Left\(-\)Right)\(_\rho\, d\rho\) \(+\) (Left\(-\)Right)\(_\varphi\,d\varphi = 0\) — a sum equal to zero, not an equation setting the two collected pieces equal to each other. The source dropped this sign when it wrote \(F\,d\rho = G\,d\varphi\) in its final steps.

Independent confirmation via back-substitution

To settle this conclusively, the corrected relation \(d\rho/d\varphi = -G/F\) was substituted back into \(dy/dx = \big(\cos\varphi\,\tfrac{d\rho}{d\varphi}\sin\varphi\cdots\big)\) — concretely, computing \(\frac{dx}{d\varphi}=\cos\varphi\,\rho'-\rho\sin\varphi\) and \(\frac{dy}{d\varphi}=\sin\varphi\,\rho'+\rho\cos\varphi\) with \(\rho'=-G/F\), and forming \(dy/dx\), then simplifying symbolically with a computer algebra system: this reproduces the original right-hand side \(N/D\) exactly (residual \(0\)). Repeating the same check using the source's uncorrected sign (\(\rho'=+G/F\)) produces a nonzero residual — i.e. it does not reproduce the original equation. This confirms definitively that a minus sign is required.

Correct separated (variables-separable) form

\[\boxed{\big[(\alpha-k)\sin\varphi\cos\varphi-\beta\sin^2\varphi-\lambda\cos^2\varphi\big]\,d\rho \;=\; -\,\rho\big[\rho^2-k\cos^2\varphi-\alpha\sin^2\varphi+(\lambda-\beta)\sin\varphi\cos\varphi\big]\,d\varphi}\] equivalently, dividing through, \[\frac{d\rho}{\rho\big[\rho^2-k\cos^2\varphi-\alpha\sin^2\varphi+(\lambda-\beta)\sin\varphi\cos\varphi\big]} = \frac{-d\varphi}{(\alpha-k)\sin\varphi\cos\varphi-\beta\sin^2\varphi-\lambda\cos^2\varphi},\] which is separated in \(\rho\) and \(\varphi\) (a function of \(\rho\) alone times \(d\rho\) equals a function of \(\varphi\) alone times \(d\varphi\)), as required — with the sign corrected relative to the source.

Special case check ($\lambda=\beta,\ \alpha=k$)

When \(\lambda=\beta\) and \(\alpha=k\), \(F(\varphi)\equiv0\) is no longer forced trivially — but \(G(\rho,\varphi)=\rho(\rho^2-k)\), so the (corrected) relation gives \(0=-\rho(\rho^2-k)\,d\varphi\), consistent with the circular solutions \(\rho=\sqrt{k}=\text{const}\) noted by the source in this degenerate case.