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

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

Step 1: Recognize the perfect derivative

Divide by $(y')^2$ ($y'\neq0$): \[\frac{y'''}{y'} - \frac{(y'')^2}{(y')^2} = \frac{d}{dx}\!\left(\frac{y''}{y'}\right) = 0.\]

Step 2: First integration

\[ \frac{y''}{y'} = C_1. \]

Step 3: Second integration

The equation $y''=C_1y'$ is separable: $\dfrac{d(y')}{y'}=C_1\,dx$, so $y' = A\,e^{C_1x}$.

Step 4: Third integration

For $C_1 \neq 0$, integrate $y'=Ae^{C_1x}$: $y = \frac{A}{C_1}e^{C_1x}+C_2.$

Relabelling $A/C_1 \to C_1$ (the linear case $C_1=0$ is the limit $y=ax+b$):

Step 5: General solution

where $C_1$, $C_2$, $C_3$ are arbitrary constants (the linear solution $y=ax+b$ is the limiting case $C_2\to0$).

Step 6: Verification

Let $y=C_1e^{C_2x}+C_3$. Then $y'=C_1C_2e^{C_2x}$, $y''=C_1C_2^2e^{C_2x}$, $y'''=C_1C_2^3e^{C_2x}$. Check: $y'y''' - (y'')^2 = C_1^2C_2^4e^{2C_2x} - C_1^2C_2^4e^{2C_2x} = 0.$ $\checkmark$