Find the general integral of the Bernoulli equation…

Find the general integral of the Bernoulli equation $y'+2xy=2x^3y^3$.

Explanation of Method

With $n=3$, substitute $v=y^{-2}$ (dividing by $y^3$ first) to reduce to a linear equation for $v(x)$.

Step 1: Substitute $v=y^{-2}$

Dividing by $y^3$: $\dfrac{y'}{y^3}+\dfrac{2x}{y^2}=2x^3$. With $v=y^{-2}$, $v'=-2y^{-3}y'$, so $\dfrac{y'}{y^3}=-\dfrac{v'}{2}$:

\[-\frac{v'}{2}+2xv=2x^3\quad\Longrightarrow\quad v'-4xv=-4x^3.\]

Step 2: Solve the linear equation

With $\mu(x)=e^{-2x^2}$: $\dfrac{d}{dx}(e^{-2x^2}v)=-4x^3e^{-2x^2}$. Integrating by parts (with $u=-2x^2$) gives $e^{-2x^2}v=e^{-2x^2}\!\left(x^2+\tfrac12\right)+C$, so $v=x^2+\tfrac12+Ce^{2x^2}$.

Result

Final answer:

\[\boxed{\,\frac{1}{y^2}=x^2+\frac12+Ce^{2x^2},\qquad C\in\mathbb{R}}\]