Solve the differential equation y=(y'-tfrac12√(x))^3
Solve the differential equation $y=\left(y'-\tfrac12\sqrt{x}\right)^3$.
Explanation of Method
Writing \(p=y'\), this has the form \(y=f(x,p)\) with \(f\) cubic in \(p\). The naive \(p\)-discriminant (\(\partial f/\partial p=0\)) is checked directly against the original equation rather than assumed valid. The genuine reduction comes from setting \(u=p-\tfrac12\sqrt x\), then \(s=\sqrt x\), then \(w=u/s\): each substitution is chosen to exploit the equation's built-in scaling, and the last one turns it into a separable, exactly integrable equation.
Set up the parametric equation
With \(p=y'\), \(y=(p-\tfrac12\sqrt x)^3\). Differentiating in \(x\): \[p=3\left(p-\tfrac12\sqrt x\right)^2\left(\frac{dp}{dx}-\frac{1}{4\sqrt x}\right).\]
Check the naive singular solution — it is spurious
Eliminating \(p\) between \(y=(p-\tfrac12\sqrt x)^3\) and \(\partial y/\partial p=3(p-\tfrac12\sqrt x)^2=0\) gives \(p=\tfrac12\sqrt x\), hence \(y=0\). But substituting \(y=0,\,y'=0\) into the original equation gives \(0=(0-\tfrac12\sqrt x)^3=-\tfrac18x^{3/2}\), which holds only at the single point \(x=0\), not identically. So \(y=0\) is not a genuine singular solution — the naive \(p\)-discriminant here is spurious, and the source's own step 7 shows this but then incorrectly boxes \(y=0\) anyway. This point is discarded.
Introduce \(u=p-\tfrac12\sqrt x\) and \(s=\sqrt x\)
Since \(y=u^3\) and \(dy=p\,dx=(u+\tfrac12\sqrt x)\,dx\), while also \(dy=3u^2\,du\), we get \(3u^2\,du=(u+\tfrac12 s)\,dx\) with \(x=s^2\), \(dx=2s\,ds\). This yields, treating \(u\) as a function of \(s\): \[3u^2\frac{du}{ds}=2us+s^2.\] Every term is homogeneous of the same weight in \((u,s)\) once \(du/ds\) is treated as weight 0, so a further substitution should separate it.
Substitute \(w=u/s\): the equation separates
With \(u=ws\), \(u'=w+sw'\), the equation becomes (after dividing by \(s^2\)): \[3sw^2\frac{dw}{ds}=1+2w-3w^3.\] This is separable: \[\frac{3w^2\,dw}{1+2w-3w^3}=\frac{ds}{s}.\]
Detect the embedded exact solution \(w\equiv1\)
The right side of the numerator factors as \(1+2w-3w^3=(1-w)(3w^2+3w+1)\), so \(w\equiv1\) is an equilibrium of the reduced equation. Tracing back: \(w=1\Rightarrow u=s\Rightarrow p=u+\tfrac12 s=\tfrac32 s\), and \(y=u^3=s^3=x^{3/2}\). Direct substitution confirms \[y=x^{3/2}\ \Longrightarrow\ y-\left(y'-\tfrac12\sqrt x\right)^3\equiv0\] for all \(x\ge0\) (checked symbolically). This is the genuine special/envelope solution replacing the source's incorrect \(y=0\).
Integrate the separable equation (general solution)
Using \(1+2w-3w^3=(1-w)(3w^2+3w+1)\) and partial fractions, \[\int\frac{3w^2\,dw}{1+2w-3w^3}=-3\ln|w-1|-2\ln(3w^2+3w+1)+2\sqrt3\arctan\!\big(\sqrt3(2w+1)\big)+\text{const.}\] Setting this equal to \(7\ln s+\text{const}\) (from \(\int ds/s\), scaled by 7) and writing \(s=\sqrt x\), \(w=y^{1/3}/\sqrt x\), gives the implicit general solution below. This was verified both symbolically (the antiderivative's derivative reproduces the integrand) and numerically (integrating the reduced ODE and checking \(dy/dx=p\) and the original equation hold along the resulting curve, to machine precision).
Result
Final answer:
Corrected. The source's boxed "singular solution" \(y=0\) is spurious (contradicted by its own verification step) and its general solution was never actually found. The correct results are:
Genuine special solution (replacing \(y=0\)):
\[\boxed{\,y=x^{3/2}\,}\quad(x\ge0)\]
General solution, in closed implicit form with \(w=y^{1/3}/\sqrt{x}\):
\[\boxed{\,-3\ln|w-1|-2\ln(3w^2+3w+1)+2\sqrt3\arctan\!\big(\sqrt3(2w+1)\big)-\tfrac72\ln x=C\,}\]
This is a genuine closed-form (transcendental) relation obtained via the reduction \(u=y'-\tfrac12\sqrt x\), \(s=\sqrt x\), \(w=u/s\), which turns the original cubic equation into a separable ODE — contrary to the source's claim that the problem "requires numerical methods."