Solve $y=(2y^4+2x)y'$ using Laplace Transform

Question

My attempt: \begin{align*} \mathscr{L}\{y&=(2y^4+2x)y'\}\\ Y&=2\mathscr{L}\{y^4y'\}+2\mathscr{L}\{xy'\}\\ \end{align*} Consider \begin{align*} \mathscr{L}\{y'\}&=sY-Y(0)\\ \int_0^\infty y'e^{-sx}dx&=sY-Y(0)\\ \int_0^\infty \frac{\partial}{\partial s}(y'e^{-sx})dx&=\dfrac{d}{ds}(sY-Y(0))\\ \mathscr{L}\{xy'\}&=-(x\frac{dY}{ds}+Y) \end{align*} I can't think of a way to evaluate $\mathscr{L}\{y^4y'\}$ and to continue solving differential equation.

Answer 1

$$y=(2y^4+2x)y'$$ Rewrite the differential equation as: $$x'y=2y^4+2x$$ Then you can apply the Laplace transform. $$X'(s)+2X(s)=-\dfrac {2 \times 4!}{s^5}$$