4.1.2.1.2 Example 2

\begin{align*} 3\beta y^{\prime \prime }+yy^{\prime } & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =3\beta y^{\prime \prime }+yy^{\prime }\end{align*}

Applying the test

\begin{equation} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) =0 \tag {1}\end{equation}

Therefore

\begin{align*} \frac {\partial F}{\partial y} & =y^{\prime }\\ \frac {\partial F}{\partial y^{\prime }} & =y\\ \frac {\partial F}{\partial y^{\prime \prime }} & =3\beta \end{align*}

Hence (1) becomes

\begin{align*} \left ( y^{\prime }\right ) -\frac {d}{dx}\left ( y\right ) +\frac {d^{2}}{dx^{2}}\left ( 3\beta \right ) & =0\\ y^{\prime }-y^{\prime } & =0\\ 0 & =0 \end{align*}

Therefore this exact. Therefore we see that \(\left ( \frac {y^{2}}{2}+3\beta y^{\prime }\right ) ^{\prime }=3\beta y^{\prime \prime }+yy^{\prime }=0\). Which implies the ode can be written as

\begin{align*} \int \left ( \frac {y^{2}}{2}+3\beta y^{\prime }\right ) ^{\prime }dx & =0\\ \frac {y^{2}}{2}+3\beta y^{\prime } & =c \end{align*}

Solving this first order ode gives the solution

\[ y=\tanh \left ( \frac {1}{6r}\sqrt {c_{1}}\left ( c_{2}+x\right ) \sqrt {2}\right ) \sqrt {2}\sqrt {c_{1}}\]