Find the general solution of y y'' = y'(y' + 1) , by…

Find the general solution of $y\,y'' = y'(y' + 1)$, by transforming both sides into perfect derivatives.

Step 1: Recognize perfect derivatives

Rewrite as $yy''-(y')^2=y'$ and divide by $y^2$: $\frac{yy''-(y')^2}{y^2} = \frac{y'}{y^2}.$

Recognize $\dfrac{yy''-(y')^2}{y^2} = \dfrac{d}{dx}\!\left(\dfrac{y'}{y}\right)$ and $\dfrac{y'}{y^2} = -\dfrac{d}{dx}\!\left(\dfrac1y\right)$, so: $\frac{d}{dx}\!\left(\frac{y'+1}{y}\right) = 0.$

Step 2: First integration

\[ \frac{y'+1}{y} = C_1 \implies y' = C_1\,y - 1. \]

Step 3: Second integration

Separate variables: $\dfrac{dy}{C_1y-1}=dx$. For $C_1\neq0$, integrate: \[C_1 y - 1 = A\,e^{C_1 x} \implies y = \frac{1}{C_1} + \frac{A}{C_1}e^{C_1x}.\]

Writing the two free constants as $a = A/C_1$ and $b=C_1$:

Step 4: General solution

where $a$ and $b \neq 0$ are arbitrary constants. (The case $C_1=0$, $y=-x+C$, is a limiting case: it satisfies $yy''=0=y'(y'+1)=(-1)(0)$.)

Step 5: Verification

Let $y=ae^{bx}+1/b$. Then $y'=abe^{bx}$, $y''=ab^2e^{bx}$. Compute: $yy'' = \left(ae^{bx}+\frac1b\right)ab^2e^{bx} = a^2b^2e^{2bx}+abe^{bx}$, $y'(y'+1) = abe^{bx}(abe^{bx}+1) = a^2b^2e^{2bx}+abe^{bx}$.

The two sides are equal. $\checkmark$