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

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

Step 1: Regroup using a product rule

Since $\dfrac{d}{dx}(yy') = (y')^2+yy''$, the first two terms combine exactly into $x\,\dfrac{d}{dx}(yy')$, so the equation can be rewritten with no approximation as

\[ x\cdot\frac{d}{dx}(yy') - yy' = 0. \]

Step 2: Substitute $z=yy'$

With $z=yy'$, the equation becomes $xz'-z=0$, which is separable (for $z\neq0,\ x\neq0$): \[\frac{dz}{z} = \frac{dx}{x} \implies \ln|z| = \ln|x|+\ln|C_1| \implies z = C_1x.\]

Step 3: Integrate back to $y$

Since $z=yy'=\dfrac12\dfrac{d}{dx}(y^2)$, the equation $z=C_1x$ becomes

\[ \frac12\frac{d}{dx}\bigl(y^2\bigr) = C_1x \implies \frac{d}{dx}\bigl(y^2\bigr) = 2C_1x \implies y^2 = C_1x^2+C_2, \]

integrating once more and renaming the constant of integration $C_2$.

Step 4: General integral

$y^2 = C_1x^2 + C_2$.

Step 5: Verification

Differentiating $y^2=C_1x^2+C_2$ twice: \[2yy' = 2C_1x \implies yy' = C_1x, \qquad\text{and again:}\qquad (y')^2+yy'' = C_1.\]

Substituting into the original expression: $x\bigl(yy''+(y')^2\bigr) - yy' = xC_1 - C_1x = 0.$ $\checkmark$

(The choice $C_1=0$ recovers the constant solutions $y=\pm\sqrt{C_2}$, which trivially satisfy the equation and are correctly included as a special case of the general integral.)