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

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

Step 1: Divide by $y^2$

Assuming $y\neq 0$, divide the whole equation by $y^2$ to express everything in terms of the ratios $y''/y$ and $y'/y$: \[x\,\frac{y''}{y} + \frac{y'}{y} - x\left(\frac{y'}{y}\right)^{\!2} = 0.\]

Step 2: The key insight: a logarithmic substitution

The combination $y'/y$ is exactly the derivative of $\ln y$, which motivates setting $z=\ln y$. Then

\[ z' = \frac{y'}{y}, \qquad z'' = \frac{y''}{y} - \left(\frac{y'}{y}\right)^{\!2} \implies \frac{y''}{y} = z'' + z'^2. \]

Step 3: Substitute

Replacing $y''/y$ and $y'/y$ in the Step 1 equation: \[x(z''+z'^2) + z' - xz'^2 = 0 \implies xz'' + xz'^2 + z' - xz'^2 = 0 \implies xz'' + z' = 0.\]

The $xz'^2$ terms cancel completely, leaving a linear equation in $z$.

Step 4: Recognize a perfect derivative

The left side is exactly $\dfrac{d}{dx}(xz')$, since $\dfrac{d}{dx}(xz') = xz''+z'$. So

\[ \frac{d}{dx}(xz') = 0. \]

Step 5: Integrate twice

First integration: $xz' = C_1$, so $z' = C_1/x$. Second integration: $z = C_1\ln|x| + C_2.$

Step 6: Return to $y$

Since $z=\ln y$, \[y = e^z = e^{C_1\ln|x|+C_2} = e^{C_2}|x|^{C_1}.\] $y = A\,x^{C_1$.

where $A=e^{C_2}$ and $C_1$ are arbitrary constants (equivalently $\ln y = C_1\ln x + C_2$).

Step 7: Verification

With $y=Ax^{C_1}$: $y'=AC_1x^{C_1-1}$, $y''=AC_1(C_1-1)x^{C_1-2}$. Substituting into $xyy''-xy'^2+yy'$: \[x\cdot Ax^{C_1}\cdot AC_1(C_1-1)x^{C_1-2} - x\cdot A^2C_1^2x^{2C_1-2} + Ax^{C_1}\cdot AC_1x^{C_1-1}\]

\[ = A^2C_1x^{2C_1-1}\bigl[(C_1-1) - C_1 + 1\bigr] = A^2C_1x^{2C_1-1}\cdot 0 = 0.\ \checkmark \]