Show that (x^2+x)y' + y^2 + (1-2x)y - 2x = 0 has a…

Show that \[ (x^2+x)y' + y^2 + (1-2x)y - 2x = 0 \] has a particular solution of the form \(y_1=k\) for a constant \(k\), and show that all integral curves of the general solution pass through two fixed points.

1. Determine \(k\)

With \(y=k\) constant, \(y'=0\), and the equation reduces to \(k^2+k-2x(k+1)=0\), a polynomial identity in \(x\) that (Sympy \(\texttt{Poly}\) confirms) requires coefficient of \(x\): \(-2k-2=0\) and constant term \(k^2+k=0\) to hold simultaneously — both are satisfied only by \(k=-1\). So \(y_1=-1\), and direct back-substitution verifies \(0=0\).

2. Riccati substitution \(y=-1+1/v\)

Substituting and simplifying gives the linear equation \[ \frac{dv}{dx} + \frac{1+2x}{x(x+1)}v = \frac{1}{x(x+1)}. \] Partial fractions give \(\frac{1+2x}{x(x+1)}=\frac1x+\frac1{x+1}\), so the integrating factor is \(\mu=e^{\int(1/x+1/(x+1))dx}=x(x+1)\), and \[ \frac{d}{dx}\big[vx(x+1)\big]=1 \;\Rightarrow\; vx(x+1)=x+C \;\Rightarrow\; v=\frac{x+C}{x(x+1)}. \]

3. Back-substitute and verify the general solution

\(y=-1+1/v\) simplifies to \[ y = \frac{x^2-C}{x+C}. \] Substituting this directly into the original equation \((x^2+x)y'+y^2+(1-2x)y-2x\) and simplifying with Sympy gives \(0\) identically for arbitrary \(C\) — the general solution is confirmed correct.

4. Fixed points — the rigorous method

The source located "fixed points" by setting \(\partial y/\partial C=0\); that finds where the family is stationary in \(C\), which is not, in general, the same as a point every curve of the family passes through. The rigorous criterion is: solve the general solution algebraically for \(C\) as a function of \((x,y)\), and find where that expression is indeterminate for every value of \(C\) — i.e. where both the "numerator part" and the "denominator part" vanish together, so that \(C\) drops out and the point lies on the curve regardless of \(C\).

From \(y(x+C)=x^2-C\), collecting terms in \(C\): \[ C(y+1) = x^2 - xy \quad\Longrightarrow\quad C = \frac{x^2-xy}{y+1} = \frac{x(x-y)}{y+1}. \] (Verified with Sympy: solving \(y(x+C)=x^2-C\) for \(C\) returns exactly \(C=\dfrac{x(x-y)}{y+1}\).) A point \((x,y)\) lies on every curve of the family, independent of \(C\), precisely where this fraction is \(\tfrac00\), i.e. where numerator and denominator vanish simultaneously: \[ x(x-y)=0 \qquad\text{and}\qquad y+1=0. \] With \(y=-1\): \(x(x-(-1))=x(x+1)=0 \Rightarrow x=0\) or \(x=-1\). Solving this system directly in Sympy confirms the two solutions are exactly \((x,y)=(0,-1)\) and \((-1,-1)\) — the same two points the source's \(\partial y/\partial C=0\) shortcut found, so that shortcut happened to give the right answer here, but the justification above (indeterminate-form / numerator-and-denominator-vanish argument) is the rigorous one and should be used in general.