Find the general and singular solutions of…
Find the general and singular solutions of $4(1-y)=(3y-2)^2(y')^2$.
Explanation of Method
Separate variables and integrate (verified independently via implicit differentiation of the result) for the general solution; then test the $p$-discriminant candidate rigorously by checking whether the separated integral converges there, which decides — more reliably than the algebraically degenerate C-discriminant — whether it is a genuine envelope.
Step 1: Separate variables
$y'=\pm\dfrac{2\sqrt{1-y}}{3y-2}$ (real only for $y\le1$), so $\dfrac{(3y-2)\,dy}{2\sqrt{1-y}}=\pm dx$.
Step 2: Integrate and verify
Symbolic integration of $\dfrac{3y-2}{2\sqrt{1-y}}$ gives $-y\sqrt{1-y}$ (up to sign/constant, confirmed by symbolic integration), so the general solution is \[y\sqrt{1-y}=x+C.\] Independent verification: implicitly differentiating $y\sqrt{1-y}-x-C=0$ gives $y'=\dfrac{-2\sqrt{1-y}}{3y-2}$, and substituting into $4(1-y)-(3y-2)^2(y')^2$ gives residual $\equiv0$ symbolically. The source's integration was correct.
Step 3: $p$-discriminant candidate
$F=4(1-y)-(3y-2)^2p^2$, $F_p=-2(3y-2)^2p=0\Rightarrow p=0$ or $y=2/3$. Testing $y=2/3$ in $F=0$: $4(1-2/3)-0=4/3\neq0$, so $y=2/3$ is spurious (discarded — it is merely a vertical-tangent turning point on individual solution curves, where $y'\to\infty$, not a separate solution). Testing $p=0$ in $F=0$: $4(1-y)=0\Rightarrow y=1$. Check: $y\equiv1\Rightarrow y'=0$, and $4(0)-(1)^2(0)^2=0$ — satisfied.
Step 4: The naive C-discriminant is degenerate here (as in Problem 151)
$G=y\sqrt{1-y}-x-C$, $\partial G/\partial C=-1$, never zero. This purely-additive-$C$ family makes the algebraic elimination fail mechanically — but (unlike Problem 151) this failure does not settle the question; it only means this particular test is uninformative here and a direct check is needed.
Step 5: Decisive convergence/tangency test
At $y=1$: $y\sqrt{1-y}=0$, so every member reaches $y=1$ at the finite point $x=-C$ — not merely in a limit. Confirming this is a genuine (not asymptotic) contact: $\displaystyle\int_{0}^{1}\frac{dy}{2\sqrt{1-y}}=1$ (finite, symbolically confirmed) — a finite amount of $x$-travel reaches $y=1$, in sharp contrast to Problem 151's divergent integral. Moreover the slope matches: from Step 2, $y'\to \dfrac{-2\sqrt0}{1}=0$ as $y\to1$, exactly the slope of the candidate horizontal line $y=1$. So every family member genuinely touches $y=1$ at a finite point with matching tangent — the defining property of an envelope.
Result
Final answer:
\[\boxed{\,y\sqrt{1-y}=x+C\,}\qquad(y\le1)\]
Verdict on $y=1$: although the algebraic C-discriminant elimination degenerates (same superficial symptom as Problem 151), the correct decisive test — convergence of the separated integral at the boundary value — comes out opposite to Problem 151: the integral is finite, so every solution curve actually reaches $y=1$ at a finite $x=-C$ with matching zero slope. $y=1$ is therefore a genuine singular solution (envelope) of the family, distinct in kind from Problem 151's $y=0$. The source's final conclusion is right, but its justification ("still singular via boundary argument") is hand-wavy; the rigorous justification is the convergent-integral / finite-tangency argument given above.