3.3.4.12 Example 12

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

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

\[ y^{\prime }=\operatorname {RootOf}\left ( \_Z^{4}+4y\_Z^{3}+6y^{2}\_Z^{2}-\left ( 1-4y^{3}\right ) \_Z-\left ( 3-y^{3}\right ) y\right ) \]

Integrating gives

\begin{align*} \int \frac {dy}{\operatorname {RootOf}\left ( \_Z^{4}+4y\_Z^{3}+6y^{2}\_Z^{2}-\left ( 1-4y^{3}\right ) \_Z-\left ( 3-y^{3}\right ) y\right ) } & =\int dx\\ \int ^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \_Z^{4}+4\tau \_Z^{3}+6\tau ^{2}\_Z^{2}-\left ( 1-4\tau ^{3}\right ) \_Z-\left ( 3-\tau ^{3}\right ) \tau \right ) } & =x+c \end{align*}

Applying IC the above becomes

\begin{multline*} \int _{0}^{y_{0}}\frac {d\tau }{\operatorname {RootOf}\left ( \_Z^{4}+4\tau \_Z^{3}+6\tau ^{2}\_Z^{2}-\left ( 1-4\tau ^{3}\right ) \_Z-\left ( 3-\tau ^{3}\right ) \tau \right ) }\\ +\int _{y_{0}}^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \_Z^{4}+4\tau \_Z^{3}+6\tau ^{2}\_Z^{2}-\left ( 1-4\tau ^{3}\right ) \_Z-\left ( 3-\tau ^{3}\right ) \tau \right ) }=x-x_{0}\end{multline*}