4.9.1.1.9 Example 9 \(y\left ( y^{\prime \prime }\right ) ^{3}+y^{3}y^{\prime }=0\)

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

First we note that this factors to

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

Hence \(y=0\) is solution. Now we just need to solve

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

Since the above is missing \(x\) then then make everything as \(\frac {du}{dy}\) using the substitution \(u=y^{\prime },y^{\prime \prime }=u\frac {du}{dy},y^{\prime \prime \prime }=u^{2}\frac {d^{2}u}{dy^{2}}+u\left ( \frac {du}{dy}\right ) ^{2}\) and so on. The above becomes

\begin{align*} \left ( u\frac {du}{dy}\right ) ^{3}+y^{2}u & =0\\ u^{3}\left ( \frac {du}{dy}\right ) ^{3}+y^{2}u & =0\\ u\left ( u^{2}\left ( \frac {du}{dy}\right ) ^{3}+y^{2}\right ) & =0 \end{align*}

Hence one solution is \(u=0\) or \(y^{\prime }=0\) or \(y=c_{1}\). So now we just need to solve

\[ u^{2}\left ( \frac {du}{dy}\right ) ^{3}+y^{2}=0 \]

Taking all three roots gives

\begin{align*} \frac {du}{dy} & =\left ( \frac {-y^{2}}{u^{2}}\right ) ^{\frac {1}{3}}\\ & =-\frac {\left ( -y^{2}u\right ) ^{\frac {1}{3}}}{2u}-\frac {i\sqrt {3}}{2u}\left ( -y^{2}u\right ) ^{\frac {1}{3}}\\ & =-\frac {\left ( -y^{2}u\right ) ^{\frac {1}{3}}}{2u}+\frac {i\sqrt {3}}{2u}\left ( -y^{2}u\right ) ^{\frac {1}{3}}\end{align*}

Let look at first root.

\begin{align*} \frac {du}{dy} & =\left ( \frac {-y^{2}}{u^{2}}\right ) ^{\frac {1}{3}}\\ & =\left ( \frac {-y^{2}u}{u^{3}}\right ) ^{\frac {1}{3}}\\ & =\frac {\left ( -y^{2}u\right ) ^{\frac {1}{3}}}{u}\end{align*}

This is homogeneous, class A. Let \(u=vy\). Then \(\frac {du}{dy}=y\frac {dv}{dy}+v\). The above becomes

\begin{align*} y\frac {dv}{dy}+v & =\frac {\left ( -y^{2}vy\right ) ^{\frac {1}{3}}}{vy}\\ & =\frac {-yv^{\frac {1}{3}}}{vy}\\ & =-v^{\frac {-2}{3}}\\ y\frac {dv}{dy} & =\left ( -v^{\frac {-2}{3}}-v\right ) \\ \frac {dv}{v^{\frac {-2}{3}}+v} & =-\frac {1}{y}dy \end{align*}

Which is separable. Integrating gives

\begin{align*} \frac {3}{5}\ln \left ( 1+v^{\frac {5}{3}}\right ) & =-\ln y+c_{1}\\ \ln \left ( 1+v^{\frac {5}{3}}\right ) & =-\frac {5}{3}\ln y+c_{2}\\ & =\ln y^{-\frac {5}{3}}+c_{2}\end{align*}

Hence

\begin{align*} 1+v^{\frac {5}{3}} & =c_{3}y^{-\frac {5}{3}}\\ v^{\frac {5}{3}} & =c_{3}y^{-\frac {5}{3}}-1\\ v & =\left ( c_{3}y^{-\frac {5}{3}}-1\right ) ^{\frac {3}{5}}\end{align*}

Hence

\begin{align*} u & =vy\\ & =y\left ( c_{3}y^{-\frac {5}{3}}-1\right ) ^{\frac {3}{5}}\end{align*}

But \(u=y^{\prime }\), therefore

\[ y^{\prime }=y\left ( c_{3}y^{-\frac {5}{3}}-1\right ) ^{\frac {3}{5}}\]

This is first order quadrature. Solving it gives

\[ y=x-\int ^{y\left ( x\right ) }\frac {1}{a\left ( \frac {-a^{\frac {5}{3}}+c_{1}}{a^{\frac {5}{3}}}\right ) ^{\frac {3}{5}}}da+c_{2}\]

The above is solution of the first root. We can solve the other two roots.