3.5.3.11 Example 11
\begin{align*} y & =x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \\ & =x+p^{2}\left ( 1-\frac {2}{3}p\right ) \end{align*}
Where \(f=1,g=p^{2}\left ( 1-\frac {2}{3}p\right ) \). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} p-1=\left ( 2p-2p^{2}\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is when \(\frac {dp}{dx}=0\) which results in \(p=1\).
Substituting this in (1) gives
\begin{align*} y & =x-\left ( 1-\frac {2}{3}\right ) \\ & =x+\frac {1}{3}\end{align*}
The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2A) is now separable. Solving for \(p\) gives
\begin{align*} p & =-\sqrt {c_{1}-x}\\ p & =\sqrt {c_{1}-x}\end{align*}
Substituting each one of the above solutions of \(p\) in (1) gives
\begin{align*} y_{1} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( -\sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( -\sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}
And
\begin{align*} y_{2} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( \sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( \sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}
Therefore the solutions are
\begin{align*} y & =x+\frac {1}{3}\\ y & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\\ y & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}