4.9.1.1.2 Example 2 \(y^{\prime \prime }+ay\left ( y^{\prime }\right ) +by^{3}=0\)

\begin{equation} y^{\prime \prime }+ay\left ( y^{\prime }\right ) +by^{3}=0 \tag {1}\end{equation}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=pp^{\prime }\). Hence the ode becomes

\begin{equation} pp^{\prime }+ayp+by^{3}=0 \tag {2}\end{equation}

Which is now a first order ode.

\begin{equation} p^{\prime }=-ay+b\frac {y^{3}}{p} \tag {3}\end{equation}

Solving for \(p\) gives

\[ \frac {1}{4\sqrt {a^{2}+8b}}\left ( \ln \left ( -by^{4}+ay^{2}p+2p^{2}\right ) \sqrt {a^{2}+8b}+2a\operatorname {arctanh}\left ( \frac {ax^{2}+4p}{y^{2}\sqrt {a^{2}+8b}}\right ) \right ) =c_{1}\]

Then \(y\) is found by solving \(y^{\prime }=p\), another first order ode.

\[ \frac {1}{4\sqrt {a^{2}+8b}}\left ( \ln \left ( -by^{4}+ay^{2}y^{\prime }+2\left ( y^{\prime }\right ) ^{2}\right ) \sqrt {a^{2}+8b}+2a\operatorname {arctanh}\left ( \frac {ax^{2}+4y^{\prime }}{y^{2}\sqrt {a^{2}+8b}}\right ) \right ) =c_{1}\]

But this second one could not solve. Actually ode (3) is homogeneous, class G and should use formula given in Kamke’s book, p. 19. but I have yet to implement this.