Find the general solution of y'' - x y''' + (y''')^2 = 0

Find the general solution of $y'' - x\,y''' + (y''')^2 = 0$.

Step 1: Substitution

Set $q = y''$, so $q' = y'''$. The equation becomes: \[q - x\,q' + (q')^2 = 0. \tag{1}\]

This is a Clairaut equation in $q(x)$.

Step 2: Clairaut equation

Rearranged, (1) reads: $q = x\,q' - (q')^2$.

the standard Clairaut form $q = x\,q' + f(q')$ with $f(t) = -t^2$.

Step 3: General solution of the Clairaut equation

For a Clairaut equation, the general solution is obtained by setting $q' = C_1$ (an arbitrary constant): $q = C_1 x + f(C_1) = C_1 x - C_1^2.$

Thus: \[y'' = C_1 x - C_1^2. \tag{2}\]

Step 4: Integrate to find $y'$

\[ y' = \frac{C_1}{2}x^2 - C_1^2 x + C_2. \tag{3} \]

Step 5: Integrate to find $y$

\[ y = \frac{C_1}{6}x^3 - \frac{C_1^2}{2}x^2 + C_2 x + C_3. \]

Step 6: General solution

where $C_1$, $C_2$, $C_3$ are arbitrary constants.

Step 7: Singular solution

The singular solution of the Clairaut equation is found from $x + f'(q') = 0$: $x - 2q' = 0 \implies q' = \frac{x}{2}.$

Substitute back into the Clairaut equation: \[q = x\cdot\frac{x}{2} - \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}.\] So $y'' = \dfrac{x^2}{4}$. Integrate twice: $y' = \frac{x^3}{12} + C_2,\qquad y = \frac{x^4}{48} + C_2 x + C_3.$

This gives the singular solution family: $y = \frac{x^4}{48} + C_2 x + C_3.$

Step 8: Verification of general solution

With $y'' = C_1 x - C_1^2$ and $y''' = C_1$: $y'' - x\,y''' + (y''')^2 = (C_1 x - C_1^2) - x\cdot C_1 + C_1^2 = 0.$ $\checkmark$