Find the general solution of y' y''' - 2(y'')^2 = 0 , by…

Find the general solution of $y'\,y''' - 2(y'')^2 = 0$, by transforming both sides into perfect derivatives.

Step 1: Recognize the perfect derivative

Divide by $(y')^2$ and let $p = y''/y'$. Using $\dfrac{d}{dx}\!\left(\dfrac{y''}{y'}\right) = \dfrac{y'''}{y'}-\dfrac{(y'')^2}{(y')^2}$, the equation $\dfrac{y'''}{y'}-2\dfrac{(y'')^2}{(y')^2}=0$ becomes:

\[ p' - p^2 = 0. \tag{$*$} \]

Step 2: First integration — solve for $p$

Equation ($*$) is separable, $\dfrac{dp}{p^2}=dx$: \[-\frac{1}{p} = x - a \implies \frac{y''}{y'} = p = \frac{1}{a-x}.\]

Step 3: Second integration

The equation $y''=\dfrac{y'}{a-x}$ is separable: \[\frac{d(y')}{y'} = \frac{dx}{a-x} \implies y' = \frac{b}{a-x}.\]

Step 4: Third integration

Integrate $y'=\dfrac{b}{a-x}$: $y = -b\ln|a-x| + c.$

Relabel $-b\to C_1$, $a\to C_2$, $c\to C_3$:

Step 5: General solution

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

Step 6: Verification

Let $y=C_1\ln|x-C_2|+C_3$. Then $y'=\dfrac{C_1}{x-C_2}$, $y''=-\dfrac{C_1}{(x-C_2)^2}$, $y'''=\dfrac{2C_1}{(x-C_2)^3}$. Check: $y'y''' = \frac{2C_1^2}{(x-C_2)^4} = 2(y'')^2.$ $\checkmark$