1.2.4 Example 4 \(\left ( y^{\prime }\right ) ^{2}+2\left ( y^{\prime }\right ) ^{3}+y=0\)

Solve (This can also be solved as dAlembert)

\begin{equation} \left ( y^{\prime }\right ) ^{2}+2\left ( y^{\prime }\right ) ^{3}+y=0 \tag {1}\end{equation}

Let \(p=\frac {dy}{dx}\)

\begin{align} y & =p^{2}+2p^{3}\tag {2}\\ y & =f\left ( p\right ) \nonumber \end{align}

Where \(p=\frac {dy}{dx}\). Hence \(dx=\frac {1}{p}dy\). But from the above \(dy=f^{\prime }\left ( p\right ) dp=\left ( 2p+6p^{2}\right ) dp\). Hence

\begin{align*} dx & =\frac {1}{p}\left ( 2p+6p^{2}\right ) dp\\ & =\left ( 2+6p\right ) dp\\ x & =\int \left ( 2+6p\right ) dp\\ & =2p+3p^{2}+c \end{align*}

Solving for \(p\) gives

\[ p=\frac {-1\pm \sqrt {3x+c}}{3}\]

Hence the solution from (2) becomes (for the first root)

\begin{align*} y & =p^{2}+2p^{3}\\ & =\left ( \frac {-1+\sqrt {3x+c}}{3}\right ) ^{2}+2\left ( \frac {-1+\sqrt {3x+c}}{3}\right ) ^{3}\end{align*}

And for the second root

\begin{align*} y & =p^{2}+2p^{3}\\ & =\left ( \frac {-1-\sqrt {3x+c}}{3}\right ) ^{2}+2\left ( \frac {-1-\sqrt {3x+c}}{3}\right ) ^{3}\end{align*}

These methods produce simpler solution if we can solve for \(p\) easily in the above.