Find the general solution of x y'' - y' = x^2 y y'
Find the general solution of $x\,y'' - y' = x^2\,y\,y'$.
Explanation of Method
Both sides of the equation are exact derivatives; one integration reduces it to a separable first-order equation in y.
Step 1: Both sides are perfect derivatives
Divide by $x^2$: $\frac{y''}{x} - \frac{y'}{x^2} = y\,y'.$
The left side is $\dfrac{d}{dx}\!\left(\dfrac{y'}{x}\right) = \dfrac{y''x-y'}{x^2}$, and the right side is $\dfrac{d}{dx}\!\left(\dfrac{y^2}{2}\right)$, so
\[ \frac{d}{dx}\!\left(\frac{y'}{x}\right) = \frac{d}{dx}\!\left(\frac{y^2}{2}\right). \]
Step 2: First integration
Integrating once gives a first-order equation that is already separable: \[\frac{y'}{x} = \frac{y^2}{2}+C_1 \qquad\Longrightarrow\qquad \frac{dy}{y^2+2C_1} = \frac{x\,dx}{2}. \tag{1}\]
Step 3: Integrate by cases on the sign of $C_1$
The antiderivative of the left side of (1) takes a different elementary form depending on the sign of $C_1$:
Case A · C₁ = a² > 0.
\[ \frac{1}{a\sqrt2}\arctan\frac{y}{a\sqrt2} = \frac{x^2}{4}+C_2. \]
Case B · C₁ = 0.
\[ -\frac{1}{y} = \frac{x^2}{4}+\frac{C_2}{4} \qquad\Longrightarrow\qquad y=-\frac{4}{x^2+C_2}. \]
Case C · C₁ = −a² < 0 (write k = a√2).
\[ \frac{1}{2k}\ln\left|\frac{y-k}{y+k}\right| = \frac{x^2}{4}+C_2 \qquad\Longrightarrow\qquad y = k\,\frac{1+Ae^{kx^2/2}}{1-Ae^{kx^2/2}}. \]
Step 4: General solution
The three cases exhaust the possible signs of the integration constant $C_1$, with $a$, $k$, $A$, $C_2$ arbitrary constants.
Step 5: Verification (Case B)
Let $y=-\dfrac{4}{x^2+C_2}$. Then $y'=\dfrac{8x}{(x^2+C_2)^2}$, and
\[ y'' = \frac{8(x^2+C_2)^2-8x\cdot2(x^2+C_2)\cdot2x}{(x^2+C_2)^4} = \frac{8(C_2-3x^2)}{(x^2+C_2)^3}. \]
Left side: \[xy''-y' = \frac{8x(C_2-3x^2)}{(x^2+C_2)^3}-\frac{8x}{(x^2+C_2)^2} = \frac{8x\big[(C_2-3x^2)-(x^2+C_2)\big]}{(x^2+C_2)^3} = \frac{-32x^3}{(x^2+C_2)^3}.\]
Right side: \[x^2yy' = x^2\!\left(-\frac{4}{x^2+C_2}\right)\!\frac{8x}{(x^2+C_2)^2} = \frac{-32x^3}{(x^2+C_2)^3}.\]
The two sides agree, so $xy''-y'=x^2yy'$. $\checkmark$
Result
Final answer:
\[\boxed{\,y=-\frac{4}{x^2+C_2}\ \ (C_1{=}0);\qquad y = k\,\frac{1+Ae^{kx^2/2}}{1-Ae^{kx^2/2}}\ \ (C_1{=}-\tfrac{k^2}{2}{<}0)\,}\]
Remark
Remark.
Equation (1) can also be read as a Bernoulli/Riccati equation in $y$, but no substitution is needed — it is already separable in the form shown, and integrating it directly is the shortest path to the answer.