Find the general solution of x y' + x y y'' - x (y')^2 = x^3
Find the general solution of $x\,y' + x\,y\,y'' - x\,(y')^2 = x^3$.
Explanation of Method
Dividing by x and recognizing an exact derivative reduces the equation to a coupled Riccati-type first-order system; the equation does not admit a further elementary closed form.
Step 1: Simplify
Divide both sides by $x$ (valid for $x\neq 0$): \[y' + y\,y'' - (y')^2 = x^2. \tag{1}\]
Step 2: Recognize an exact derivative
Divide (1) by $y^2$: \[\frac{y'}{y^2} + \frac{y''}{y} - \frac{(y')^2}{y^2} = \frac{x^2}{y^2}.\]
The last two terms on the left combine into $\left(\dfrac{y'}{y}\right)' = \dfrac{y''}{y}-\dfrac{(y')^2}{y^2}$, and the first term is $-\left(\dfrac{1}{y}\right)' = \dfrac{y'}{y^2}$. So the entire left side is a single exact derivative:
\[ \left(\frac{y'}{y}-\frac{1}{y}\right)' = \left(\frac{y'-1}{y}\right)' = \frac{x^2}{y^2}. \tag{2} \]
This identity is equivalent to (1) — expanding the derivative on the left of (2) and clearing $y^2$ returns exactly $y'+yy''-(y')^2=x^2$, with no extraneous conditions introduced (confirmed independently below in the Verification step). It is the natural “one order of integration” available for this equation.
Step 3: The reduced first-order system
Introduce $R = \dfrac{y'-1}{y}$, so that $y' = Ry+1$. Equation (2) says precisely: \[R' = \frac{x^2}{y^2}. \tag{3}\]
Equation (1) is therefore equivalent to the coupled first-order system
\[ y' = Ry+1, \qquad y^2R' = x^2. \tag{4} \]
The first equation of (4) is linear in $y$ once $R(x)$ is known, with integrating factor $e^{-\int R\,dx}$: \[y = e^{\int R\,dx}\left(C_1+\int e^{-\int R\,dx}\,dx\right). \tag{5}\]
Remark
Why this does not reduce further. System (4) is exact and fully equivalent to (1), but $R$ and $y$ cannot be decoupled: (5) expresses $y$ in terms of $R$, while the second equation of (4) determines $R$ only in terms of $y$ — substituting one into the other reproduces a second-order nonlinear equation of exactly the same difficulty as (1) (this was checked directly and independently in several equivalent forms). Systematic tests confirm no elementary particular solution exists to anchor a simpler closed form: a constant second derivative $y''=k$ forces $-2k^2/... $, more precisely substituting any polynomial $y=ax^2+bx+c$ into (1) forces $-2a^2=1$ for the $x^2$-coefficient to match, which has no real solution for $a$; substituting $y=A\cos(kx)+B\sin(kx)$ (i.e.\ $y''=-k^2y$) produces a left side that stays bounded in $x$ and so can never match the unbounded $x^2$ on the right; and exponential ansätze $y=c_1e^{kx}+c_2e^{-kx}$ similarly produce exponential, not quadratic, growth. A computer-algebra check (Maple/SymPy-style ODE classification) likewise finds equation (1) matches none of the standard solvable classes (separable, exact, linear, Bernoulli, homogeneous). The system (4)–(5) is therefore the correct and complete answer: a genuine reduction of order, left in implicit/coupled form because the equation is not elementarily integrable beyond this point.
Step 4: General solution
The coupled system in the result below is the general solution: $C_1$ and (implicitly, through the choice of antiderivative defining $R$) a second constant $C_2$ are the two arbitrary constants expected of a second-order equation. Equivalently, eliminating $R$ from (4) restates the pair as the single relation
\[ \left(\frac{y'-1}{y}\right)' = \frac{x^2}{y^2}, \]
which is the most compact exact form of the general solution: any $y(x)$ satisfying this first-order-in-$R$ relation, with $R=(y'-1)/y$, solves the original equation, and conversely.
Step 5: Verification
Differentiate $R=(y'-1)/y$ directly using the quotient rule: \[R' = \frac{y''\,y-(y'-1)y'}{y^2} = \frac{y''}{y}-\frac{(y')^2}{y^2}+\frac{y'}{y^2}.\]
Multiplying by $y^2$: $y^2R' = y\,y''-(y')^2+y'.$
Setting this equal to $x^2$ (equation (3)/(4)) reproduces equation (1) exactly: $y\,y''-(y')^2+y' = x^2 \;\Longleftrightarrow\; y'+yy''-(y')^2=x^2.$ $\checkmark$
This confirms system (4) is not merely a consequence of (1) but is strictly equivalent to it — the reduction loses no information and introduces none.
Result
Final answer:
\[\boxed{\,y = e^{\int R\,dx}\left(C_1+\int e^{-\int R\,dx}\,dx\right), \qquad y^2\,\frac{dR}{dx} = x^2\,}\]