Solve the differential equation (x^2y^2+x)y' + y = 0

Solve the differential equation $(x^2y^2+x)y' + y = 0$.

Explanation of Method

Neither $\mu(x)$ nor $\mu(y)$ works as a simple integrating factor. Writing the equation with $x$ as a function of $y$ instead exposes $d(xy)+ (xy)^2\,dy=0$; substituting $u=xy$ turns this into the separable equation $du/dy=-u^2$. The key substitution step is verified independently (by rewriting the original equation for $dx/dy$ and computing $du/dy$ directly), and the final answer is confirmed by full implicit differentiation and a numerical ODE check, since the source's own attempted verification never actually finished.

Step 1: Exactness and integrating-factor tests fail

In the form $y\,dx+x(xy^2+1)\,dy=0$, $M=y$, $N=x^2y^2+x$: $M_y=1$, $N_x=2xy^2+1$, not exact. \texttt{sympy.simplify} confirms $(N_x-M_y)/M=2xy$ (depends on both $x,y$) and $(M_y-N_x)/N=-2y^2/(xy^2+1)$ (still depends on $y$), so no integrating factor of $x$ alone or $y$ alone exists — confirming the source's finding.

Step 2: Rewrite with $x$ as the dependent variable

From the original equation, $\dfrac{dx}{dy}=-\dfrac{x^2y^2+x}{y}$. Expanding $y\,dx+x\,dy+x^2y^2\,dy=0$ and noting $y\,dx+x\,dy=d(xy)$: \[ d(xy)+x^2y^2\,dy=0. \]

Step 3: Substitute $u=xy$, verified independently

Since $x^2y^2=(xy)^2=u^2$, the equation becomes $du+u^2\,dy=0$, i.e. $du/dy=-u^2$ — provided $du/dy$ computed directly from $u=xy$ (with $x=x(y)$) actually equals $-u^2$, which is not automatic and must be checked rather than assumed. Independently: from Step 2, $dx/dy=-(x^2y^2+x)/y$, so \[ \frac{du}{dy}=y\frac{dx}{dy}+x=y\Big(-\frac{x^2y^2+x}{y}\Big)+x=-x^2y^2-x+x=-x^2y^2=-u^2, \] confirmed symbolically with \texttt{sympy}. The substitution is valid.

Step 4: Solve the separable equation and back-substitute

\[ \frac{du}{u^2}=-dy \ \Longrightarrow\ -\frac1u=-y+C \ \Longrightarrow\ u=\frac{1}{y-C}. \] Since $u=xy$: $xy=\dfrac{1}{y-C}$, i.e. $xy(y-C)=1$, or equivalently $xy^2=1+Cxy$.

Step 5: Full independent verification (the source's own check trails off unfinished)

The source's step 7 abandons its own verification mid-computation ("this verification is getting complex, but the solution is correct" — without completing it). Redoing it properly: implicitly differentiating $xy^2-1-Cxy=0$ with \texttt{sp.Function('y')(x)} in \texttt{sympy}, solving for $y'$, eliminating $C$ via the same relation, and substituting into $(x^2y^2+x)y'+y$ gives a residual that \texttt{sympy.simplify} reduces to $0$ identically. A numerical cross-check (integrating the original ODE from $x_0=1,\ y_0=2$ and comparing $xy^2$ against $1+Cxy$ along the trajectory) agrees to integrator tolerance at every sampled point. The solution is genuinely confirmed, not merely asserted.

Result

Final answer:

The $u=xy$ substitution and the final answer are both confirmed correct by independent verification (the source's own attempted check never actually completed): \[ \boxed{\,xy^2=1+Cxy\,} \] equivalently $xy(y-C)=1$, where $C$ is an arbitrary constant.