Find the general integral of the third-order equation y' = x…
Find the general integral of the third-order equation $y' = x y'' - \frac{(y'')^2}{2}$.
Explanation of Method
Setting $p=y'$, $q=y''$ reveals a Clairaut equation in $(x,p)$, yielding both a two-parameter general solution and a singular solution that is its envelope.
Step 1: Recognizing the Clairaut structure
Set $p=y'$ and $q=y''=\dfrac{dp}{dx}$. The equation becomes
\[ p = xq - \frac{q^2}{2}, \]
which is a Clairaut equation in the pair $(x,p)$, with $q=dp/dx$ playing the role of the "slope" variable. It has exactly the form $p = x q - \varphi(q)$ with $\varphi(q)=q^2/2$.
Step 2: Differentiating with respect to $x$
Differentiate $p=xq-q^2/2$ with respect to $x$, using $dp/dx=q$: \[\frac{dp}{dx} = q + x\frac{dq}{dx} - q\frac{dq}{dx} \;\Longrightarrow\; q = q + (x-q)\frac{dq}{dx} \;\Longrightarrow\; (x-q)\frac{dq}{dx} = 0.\]
As with any Clairaut equation, this factored condition splits the problem into two independent cases.
Step 3: Case I — the general (two-parameter) solution
Case I: General Solution.
Take $dq/dx=0$, i.e. $q=C_1$, a constant. Substituting this constant value back into $p=xq-q^2/2$: $p = C_1x - \frac{C_1^2}{2}, \qquad \text{i.e.} \qquad y' = C_1x - \frac{C_1^2}{2}.$
Integrating once more in $x$:
$y = \frac{C_1$.
This is the general integral: a two-parameter family of parabolas, one parameter ($C_1$) coming from the Clairaut structure and the other ($C_3$) from the final integration.
Step 4: Case II — the singular solution (envelope)
Case II: Singular Solution.
Take instead $x-q=0$, i.e. $q=x$, so that $y''=x$. Substitute $q=x$ back into $p=xq-q^2/2$: \[p = x\cdot x - \frac{x^2}{2} = \frac{x^2}{2}, \qquad \text{i.e.} \qquad y' = \frac{x^2}{2}.\]
Integrating:
$y = \frac{x^3$.
Unlike Case I, this branch carries only a single free constant $C$ — it is the singular solution of the Clairaut equation.
Step 5: Verification
Case II directly: with $y'=x^2/2$ and $y''=x$, check $y'=xy''-\tfrac12(y'')^2$: \[xy'' - \frac{(y'')^2}{2} = x\cdot x - \frac{x^2}{2} = \frac{x^2}{2} = y'.\ \checkmark\]
Case I: with $y'=C_1x-C_1^2/2$ and $y''=C_1$, the same substitution gives $xy''-\tfrac12(y'')^2=xC_1-C_1^2/2=y'$ identically, for every $C_1$ and $C_3. $\checkmark$
Result
Final answer:
\[\boxed{\,y = \frac{C_1}{2}x^2 - \frac{C_1^2}{2}x + C_3\,}\]
Final answer:
\[\boxed{\,y = \frac{x^3}{6} + C\,}\]
Remark
Geometric remark. The general-solution family of Case I is tangent to the singular solution of Case II — it is the envelope of the family. Comparing the two slope conditions $p=C_1x-C_1^2/2$ and $p=x^2/2$ at fixed $C_1$:
\[ \frac{x^2}{2} - C_1x + \frac{C_1^2}{2} = \frac{(x-C_1)^2}{2} = 0 \]
has the double root $x=C_1$, confirming that each member of the Case I family touches the Case II curve tangentially at $x=C_1$, which is exactly the envelope property expected of the singular solution of a Clairaut equation.