Solve the differential equation 2x^2yy' + y^2 - 2x^3 - x^2 = 0

Solve the differential equation $2x^2yy' + y^2 - 2x^3 - x^2 = 0$.

Explanation of Method

Written in differential form $(2x^3+x^2-y^2)\,dx-2x^2y\,dy=0$, the equation is not exact, but admits an integrating factor $\mu(x)=e^{-1/x}/x^2$ depending on $x$ alone. Multiplying through by $\mu$ makes the equation exact, and the potential function $F(x,y)$ is recovered by integration (\texttt{sympy.integrate}, cross-checked by hand) and then confirmed by implicit differentiation, reproducing the original equation exactly.

Step 1: Differential form and exactness test

Rearranging: $(2x^3+x^2-y^2)\,dx-2x^2y\,dy=0$, so $M=2x^3+x^2-y^2$, $N=-2x^2y$. Then $M_y=-2y$ and $N_x=-4xy$, which are unequal, so the equation is not exact.

Step 2: Find the integrating factor $\mu(x)$

\texttt{sympy.simplify} confirms \[ \frac{M_y-N_x}{N}=\frac{-2y+4xy}{-2x^2y}=\frac{1-2x}{x^2}, \] a function of $x$ alone, so $\mu(x)$ exists with $\mu'/\mu=(1-2x)/x^2=1/x^2-2/x$. Integrating, $\ln|\mu|=-1/x-2\ln|x|$, so \[ \mu(x)=\frac{e^{-1/x}}{x^2}. \]

Step 3: Multiply through and confirm exactness

Multiplying $M,N$ by $\mu$ gives \[ \tilde M=\mu M=e^{-1/x}\Big(2x+1-\frac{y^2}{x^2}\Big),\qquad \tilde N=\mu N=-2ye^{-1/x}. \] \texttt{sympy} confirms $\tilde M_y-\tilde N_x=0$ identically, so the equation is now exact.

Step 4: Find the potential function $F(x,y)$

Integrating $\tilde N$ with respect to $y$ (holding $x$ fixed) is the easier route: \[ F(x,y)=\int \tilde N\,dy = -y^2e^{-1/x}+h(x). \] Requiring $F_x=\tilde M$ gives \[ -y^2\cdot\frac{e^{-1/x}}{x^2}+h'(x)=e^{-1/x}(2x+1)-\frac{y^2e^{-1/x}}{x^2} \ \Longrightarrow\ h'(x)=(2x+1)e^{-1/x}. \] \texttt{sympy.integrate((2*x+1)*sp.exp(-1/x), x)} returns $h(x)=x^2e^{-1/x}$ (independently checked by hand: $\frac{d}{dx}[x^2e^{-1/x}]=2xe^{-1/x}+x^2\cdot \frac{1}{x^2}e^{-1/x}=(2x+1)e^{-1/x}$, matching $h'(x)$ exactly). Hence \[ F(x,y)=-y^2e^{-1/x}+x^2e^{-1/x}=(x^2-y^2)e^{-1/x}. \]

Step 5: General solution and independent verification

The implicit solution is $(x^2-y^2)e^{-1/x}=C$. Differentiating implicitly: \[ (2x-2yy')e^{-1/x}+(x^2-y^2)e^{-1/x}\cdot\frac{1}{x^2}=0. \] Dividing by $e^{-1/x}$ and multiplying by $x^2$: \[ 2x^3-2x^2yy'+x^2-y^2=0\ \Longrightarrow\ 2x^2yy'+y^2-2x^3-x^2=0, \] exactly the original equation. \texttt{sympy} (implicit differentiation via \texttt{sp.Function('y')(x)}, eliminating $C$) confirms the residual is identically $0$, and a numerical ODE integration cross-check agrees to within integrator tolerance. This is a genuine, verified closed-form solution.

Result

Final answer:

The source's own derivation of the integrating factor $\mu=e^{-1/x}/x^2$ (its Step 3) was correct, but it then abandoned that path and cycled through three unverified, mutually contradictory guesses before giving up. Carrying the exact-equation method through to completion gives the genuine closed-form solution, confirmed by implicit differentiation to reproduce the original equation exactly: \[ \boxed{\,(x^2-y^2)e^{-1/x}=C\,} \] equivalently $x^2-y^2=Ce^{1/x}$, where $C$ is an arbitrary constant.