Find the general and singular solutions (if they exist) of…
Find the general and singular solutions (if they exist) of $(y')^2-4y^3=0$.
Explanation of Method
Separate variables to get the general solution, then use the $p$-discriminant to find a candidate singular solution and test it rigorously: a genuine singular solution must be tangent to the family at an actual finite point, which is decided cleanly by checking whether the separated integral converges at the candidate value.
Step 1: Separate and integrate
$(y')^2=4y^3 \Rightarrow y'=\pm 2y^{3/2}$ (for $y>0$). Separating, $\dfrac{dy}{2y^{3/2}}=\pm dx$, so $-y^{-1/2}=\pm x + C_0$, giving $y^{-1/2}=\mp x - C_0$. Squaring and relabeling the constant, $y=\dfrac{1}{(x+C)^2}$.
Step 2: Verify the general solution by substitution
With $y=(x+C)^{-2}$, $y'=-2(x+C)^{-3}$, so $(y')^2=4(x+C)^{-6}=4y^3$. Residual $\equiv 0$ (confirmed symbolically). $\boxed{y=\dfrac{1}{(x+C)^2}}$ is verified.
Step 3: $p$-discriminant candidate
$F=p^2-4y^3$, $F_p=2p=0\Rightarrow p=0$; substituting into $F=0$ gives $y^3=0$, i.e. $y=0$. Direct check: $y\equiv0\Rightarrow y'=0$, and $0-0=0$, so $y=0$ does satisfy the ODE.
Step 4: Does the family ever reach $y=0$?
$y=(x+C)^{-2}>0$ for every finite $x,C$ — no finite $C$ produces $y\equiv0$, and no member of the family passes through any point with $y=0$ at finite $x$; $y=0$ is only the asymptotic limit as $x+C\to\pm\infty$. The C–discriminant confirms this algebraically: writing $G=y(x+C)^2-1=0$, $\partial G/\partial C=2y(x+C)=0$ forces $y=0$ or $x=-C$, but substituting either back into $G=0$ gives $-1=0$ — a contradiction, so the elimination produces no curve at all.
Step 5: Decisive convergence test
The rigorous way to settle whether a $p$-discriminant boundary value is actually attained at finite $x$ (hence a genuine envelope) or only approached asymptotically is to check convergence of the separated integral at that value: $\displaystyle\int_0^{1}\frac{dy}{2y^{3/2}}$ diverges ($=\infty$, confirmed symbolically). An infinite amount of $x$-travel is required to reach $y=0$, so $y=0$ is never actually reached by any solution curve at finite $x$ — it is a horizontal asymptote shared by the whole family, not a point of tangential contact.
Result
Final answer:
\[\boxed{\,y=\dfrac{1}{(x+C)^2}\,}\]
Verdict on $y=0$: it solves the ODE and lies on the $p$-discriminant locus, but it fails the C–discriminant/envelope test (the elimination yields no consistent equation) and the separated integral diverges at $y=0$, meaning no family member ever actually touches $y=0$ at a finite point — it is only the limiting curve as $x+C\to\infty$. By the strict, classical definition (an envelope must be tangent to each member at a genuine finite contact point), $y=0$ is therefore not a genuine singular solution; the source's "limiting envelope" language is imprecise. The honest description is: $y=0$ is a spurious/extraneous branch of the $p$-discriminant locus — an asymptotic solution of the family, not a singular solution.