Solve the differential equation (2xy^3+y)y' + 2y^2 - 4 = 0

Solve the differential equation $(2xy^3+y)y' + 2y^2 - 4 = 0$.

Explanation of Method

Written as $(4-2y^2)\,dx-(2xy^3+y)\,dy=0$, the equation is not exact, but $(N_x-M_y)/M$ reduces to $y$, a function of $y$ alone, giving an integrating factor $\mu(y)=e^{y^2/2}$. The potential function is then found by integration and the final relation confirmed by implicit differentiation.

Step 1: Exactness test and integrating factor

With $M=4-2y^2$, $N=-(2xy^3+y)$: $M_y=-4y$, $N_x=-2y^3$, not exact. \texttt{sympy.simplify} confirms \[ \frac{N_x-M_y}{M}=\frac{-2y^3+4y}{4-2y^2}=\frac{2y(2-y^2)}{2(2-y^2)}=y, \] a function of $y$ alone, so $\mu(y)=e^{y^2/2}$.

Step 2: Multiply through and confirm exactness

$\tilde M=(4-2y^2)e^{y^2/2}$, $\tilde N=-(2xy^3+y)e^{y^2/2}$. \texttt{sympy} confirms $\tilde M_y=\tilde N_x=-2y^3e^{y^2/2}$, so the equation is now exact.

Step 3: Integrate $\tilde M$ with respect to $x$

\[ F(x,y)=\int(4-2y^2)e^{y^2/2}\,dx=x(4-2y^2)e^{y^2/2}+g(y). \]

Step 4: Determine $g(y)$

\texttt{sympy} computes $\partial_y\big[x(4-2y^2)e^{y^2/2}\big]=-2xy^3e^{y^2/2}$, so requiring $F_y=\tilde N$ gives $g'(y)=-ye^{y^2/2}$. Substituting $u=y^2/2$, \texttt{sympy.integrate} confirms \[ g(y)=-\int ye^{y^2/2}\,dy=-e^{y^2/2}. \]

Step 5: Assemble and simplify the solution

\[ F(x,y)=e^{y^2/2}\big[x(4-2y^2)-1\big]=C_1. \] Dividing by $e^{y^2/2}$: \[ 4x-2xy^2-1=Ce^{-y^2/2}. \]

Step 6: Independent verification

Implicitly differentiating $4x-2xy^2-1-Ce^{-y^2/2}=0$ with \texttt{sympy}, solving for $y'$, eliminating $C$ using the same relation, and substituting into $(2xy^3+y)y'+2y^2-4$ gives a residual that \texttt{sympy.simplify} reduces to $0$ identically.

Result

Final answer:

The source's derivation and final relation are confirmed correct by independent \texttt{sympy} verification: \[ \boxed{\,4x-2xy^2-1=Ce^{-y^2/2}\,} \] where $C$ is an arbitrary constant (equivalently $(4x-2xy^2-1)e^{y^2/2}=C$).