Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $x(y')^2=y(2y'-1)$.

Step 1: Cast as a Lagrange equation

With $p=y'$, the equation is $xp^2-2yp+y=0$, i.e. $y(2p-1)=xp^2$, so \[y=x\cdot\frac{p^2}{2p-1}=x f(p),\qquad f(p)=\frac{p^2}{2p-1}.\] This is a Lagrange equation (linear in $x$), since $y$ has no additive term independent of $x$.

Step 2: Differentiate to get a linear ODE for $x(p)$

Differentiating $y=xf(p)$ w.r.t. $x$ and using $dy=p\,dx$: $p=f(p)+xf'(p)\dfrac{dp}{dx}$, so \[\frac{dx}{dp}=\frac{f'(p)}{\,p-f(p)\,}\,x.\] Sympy simplifies $p-f(p)=\dfrac{p(p-1)}{2p-1}$ and $f'(p)=\dfrac{2p(p-1)}{(2p-1)^2}$, so \[\frac{f'(p)}{p-f(p)}=\frac{2}{2p-1}\quad\Longrightarrow\quad \frac{dx}{dp}=\frac{2x}{2p-1}.\]

Step 3: Solve the linear (separable) ODE in $x(p)$

Separating and integrating: $\dfrac{dx}{x}=\dfrac{2\,dp}{2p-1}\Rightarrow \ln x=\ln(2p-1)+\text{const}$, so \[x=C(2p-1).\] Back-substituting into $y=xp^2/(2p-1)$ gives the parametric general solution \[x=C(2p-1),\qquad y=Cp^2.\] Sympy confirms $xp^2-2yp+y\equiv 0$ identically for this pair (residual $=0$).

Step 4: Eliminate the parameter $p$

From $x=C(2p-1)$, $p=\dfrac{x+C}{2C}$. Substituting into $y=Cp^2$: \[y=C\left(\frac{x+C}{2C}\right)^2=\frac{(x+C)^2}{4C}.\] Sympy verifies directly in the ORIGINAL equation: with $y=(x+C)^2/(4C)$, $y'=(x+C)/(2C)$, \[x(y')^2-y(2y'-1)=0\] identically (simplifies exactly to $0$) — this is a genuine, elementary, closed-form general solution with one arbitrary constant $C$.

Step 5: $p$-discriminant and singular solutions

From $F=xp^2-2yp+y$: $F_p=2xp-2y=0\Rightarrow p=y/x$; substituting into $F=0$ gives $-y^2/x+y=0\Rightarrow y=0$ or $y=x$. Both satisfy the original equation identically by direct substitution (sympy residual $=0$ for each).

Step 6: Confirm these are genuinely singular (envelope check)

The general-solution family is $(x+C)^2-4Cy=0$. Its envelope satisfies $\partial_C\!\left[(x+C)^2-4Cy\right]=0\Rightarrow 2(x+C)-4y=0\Rightarrow C=2y-x$. Substituting back: $(2y)^2-4(2y-x)y=0\Rightarrow 4y(x-y)=0$, i.e. $y=0$ or $y=x$ — exactly the two candidates found above. So $y=0$ and $y=x$ are true singular solutions (envelopes of the family), not members of it for finite $C$.