Find the general solution of y y'' + (y')^2 - (2y y')/(x) +…

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

Step 1: Recognize perfect derivatives

Group the terms and recognize two standard derivatives: \[y\,y'' + (y')^2 = \frac{d}{dx}(y\,y'),\qquad \frac{2y\,y'}{x} - \frac{y^2}{x^2} = \frac{d}{dx}\!\left(\frac{y^2}{x}\right).\]

Step 2: Rewrite and integrate once

The equation becomes $\dfrac{d}{dx}\!\left(y\,y' - \dfrac{y^2}{x}\right) = 0$. Integrate: \[y\,y' - \frac{y^2}{x} = C_1. \tag{1}\]

Step 3: A second perfect derivative

Multiply (1) by $2/x^2$: \[\frac{2y\,y'}{x^2} - \frac{2y^2}{x^3} = \frac{2C_1}{x^2}.\]

The left side is $\dfrac{d}{dx}\!\left(\dfrac{y^2}{x^2}\right)$, so: \[\frac{d}{dx}\!\left(\frac{y^2}{x^2}\right) = \frac{2C_1}{x^2}.\]

Step 4: Second integration

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

Step 5: General solution

where $C_1$ and $C_2$ are arbitrary constants.

Step 6: Verification

From $y^2=C_2x^2-2C_1x$: $yy'=C_2x-C_1$, and differentiating again, $(y')^2+yy''=C_2$. Also $\dfrac{2yy'}{x}=2C_2-\dfrac{2C_1}{x}$ and $\dfrac{y^2}{x^2}=C_2-\dfrac{2C_1}{x}$. Check:

\[ yy''+(y')^2-\frac{2yy'}{x}+\frac{y^2}{x^2} = C_2 - 2C_2+\frac{2C_1}{x}+C_2-\frac{2C_1}{x} = 0.\ \checkmark \]