Find the general integral of the third-order equation…
Find the general integral of the third-order equation $3y'(y'')^2 - \bigl(1+(y')^2\bigr)y''' = 0$.
Explanation of Method
A double reduction of order — first $q(p)$, then $p(x)$ — shows this third-order equation characterizes curves of constant curvature: circles, and degenerately, lines.
Step 1: Reduction of order: $q$ as a function of $p$
Set $p=y'$ and $q=y''$. Where $q\neq0$, treat $q$ as a function of $p$, so that $y'''=\dfrac{dq}{dx}=\dfrac{dq}{dp}\dfrac{dp}{dx}=q\dfrac{dq}{dp}$. Substituting into the equation:
\[ 3pq^2 - (1+p^2)\,q\frac{dq}{dp} = 0. \]
Dividing by $q$ (excluding the trivial branch $q\equiv0$): $\frac{dq}{dp} = \frac{3pq}{1+p^2}$, a separable first-order equation for $q(p)$.
Step 2: Solving for $q(p)$
Separating variables: $\frac{dq}{q} = \frac{3p\,dp}{1+p^2}.$
Integrating: \[\ln|q| = \frac32\ln(1+p^2) + \ln|A| \;\Longrightarrow\; q = A(1+p^2)^{3/2}.\]
Step 3: Solving for $p(x)$
Since $q=dp/dx$: $\frac{dp}{dx} = A(1+p^2)^{3/2}.$
Separating variables: $\frac{dp}{(1+p^2)^{3/2}} = A\,dx.$
The standard integral $\displaystyle\int\frac{dp}{(1+p^2)^{3/2}} = \frac{p}{\sqrt{1+p^2}}$ (via $p=\tan\theta$) gives
\[ \frac{p}{\sqrt{1+p^2}} = Ax + B. \]
Step 4: Inverting for $p$
Let $\xi = Ax+B$. Squaring, $p^2/(1+p^2)=\xi^2$, so $p^2(1-\xi^2)=\xi^2$ and
\[ p = \frac{\xi}{\sqrt{1-\xi^2}} = \frac{Ax+B}{\sqrt{1-(Ax+B)^2}}. \]
Step 5: Integrating $p=y'$ to recover $y$
Since $p=y'$, substitute $u=Ax+B$ ($du=A\,dx$): \[y = \int \frac{u}{\sqrt{1-u^2}}\cdot\frac{du}{A} = -\frac{1}{A}\sqrt{1-u^2} + C = -\frac1A\sqrt{1-(Ax+B)^2} + C.\]
Rearranging, \[A(y-C) = -\sqrt{1-(Ax+B)^2} \;\Longrightarrow\; A^2(y-C)^2 = 1-(Ax+B)^2.\]
Step 6: General integral
$(Ax+B)^2 + A^2(y-C)^2 = 1$, where $A$, $B$, $C$ are arbitrary constants — a three-parameter family of circles of radius $1/|A|$ centered at $\left(-\dfrac{B}{A},\,C\right)$.
Degenerate case $A=0$.
The branch $q\equiv0$ excluded in Step 1 corresponds to $y''=0$, i.e. straight lines $y=\alpha x+\beta$. These trivially satisfy the original equation, since $y''=y'''=0$ there. This is exactly the $A\to0$ limit of the circle family (infinite radius).
Step 7: Verification and geometric interpretation
Implicit differentiation of $(Ax+B)^2+A^2(y-C)^2=1$ recovers $p=y'=-\dfrac{Ax+B}{A(y-C)}$, and a further differentiation reproduces the original third-order equation $3y'(y'')^2-(1+(y')^2)y'''=0$ identically. [Confirmed by full symbolic computer-algebra substitution.]
Result
Final answer:
\[\boxed{\,(Ax+B)^2 + A^2(y-C)^2 = 1\,}\]
Remark
Remark. The equation $3y'(y'')^2-(1+(y')^2)y'''=0$ characterizes curves of constant curvature: differentiating the curvature formula $\kappa=y''/(1+(y')^2)^{3/2}$ with respect to $x$ and setting $d\kappa/dx=0$ produces exactly this equation. Circles (and, degenerately, lines) are precisely the plane curves of constant curvature, which is why they emerge as the general integral.