3.3.4.12 Example 12 ${\left(y^{\prime }\right)}^4+4y{\left(y^{\prime }\right)}^3+6y^2{\left(y^{\prime }\right)}^2-(1-4y^3)y^{\prime }-(3-y^3)y=0$

$${\left(y^{\prime }\right)}^4+4y{\left(y^{\prime }\right)}^3+6y^2{\left(y^{\prime }\right)}^2-(1-4y^3)y^{\prime }-(3-y^3)y=0$$

With IC

$$y\left(x_0\right)=y_0$$

Since $x$ is missing then this is of the form $y^{\prime }=f\left(y\right)$ we just need to solve for $y^{\prime }$. The solution is in terms of RootOf

$$y^{\prime }=\mathrm{RootOf}(\mathrm{\_}{Z}^4+4y\mathrm{\_}{Z}^3+6y^2\mathrm{\_}{Z}^2-(1-4y^3)\mathrm{\_}Z-(3-y^3)y)$$

Integrating gives

$$\begin{array}{rl}\int \frac{dy}{\mathrm{RootOf}(\mathrm{\_}{Z}^4+4y\mathrm{\_}{Z}^3+6y^2\mathrm{\_}{Z}^2-(1-4y^3)\mathrm{\_}Z-(3-y^3)y)}& =\int dx\\ {\int }^{y\left(x\right)}\frac{d\tau }{\mathrm{RootOf}(\mathrm{\_}{Z}^4+4\tau \mathrm{\_}{Z}^3+6{\tau }^2\mathrm{\_}{Z}^2-(1-4{\tau }^3)\mathrm{\_}Z-(3-{\tau }^3)\tau )}& =x+c\end{array}$$

Applying IC the above becomes

$$\begin{array}{c}{\int }_0^{y_0}\frac{d\tau }{\mathrm{RootOf}(\mathrm{\_}{Z}^4+4\tau \mathrm{\_}{Z}^3+6{\tau }^2\mathrm{\_}{Z}^2-(1-4{\tau }^3)\mathrm{\_}Z-(3-{\tau }^3)\tau )}\hfill \\ \hfill +{\int }_{y_0}^{y\left(x\right)}\frac{d\tau }{\mathrm{RootOf}(\mathrm{\_}{Z}^4+4\tau \mathrm{\_}{Z}^3+6{\tau }^2\mathrm{\_}{Z}^2-(1-4{\tau }^3)\mathrm{\_}Z-(3-{\tau }^3)\tau )}=x-x_0\end{array}$$