3.5.3.11 Example 11

$$\begin{array}{rl}y& =x+{\left(y^{\prime }\right)}^2(1-\frac{2}{3}{y}^{\prime })\\ & =x+p^2(1-\frac{2}{3}p)\end{array}$$

Where $f=1,g=p^2(1-\frac{2}{3}p)$.  Since $f\left(p\right)\ne p$ then this is d’Almbert ode. Taking derivative w.r.t. $x$ gives

$$\begin{array}{rl}p& =(f+x{f}^{\prime }\frac{dp}{dx})+\left(g^{\prime }\frac{dp}{dx}\right)\\ p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Using values for $f,g$ the above simplifies to

$$\begin{array}{}\text{(2A)}& p-1=(2p-2p^2)\frac{dp}{dx}\end{array}$$

The singular solution is when $\frac{dp}{dx}=0$ which results in $p=1$. Substituting this in (1) gives

$$\begin{array}{rl}y& =x-(1-\frac{2}{3})\\ & =x+\frac{1}{3}\end{array}$$

The general solution is when $\frac{dp}{dx}\ne 0$. Then (2A) is now separable. Solving for $p$ gives

$$\begin{array}{rl}p& =-\sqrt{c_1-x}\\ p& =\sqrt{c_1-x}\end{array}$$

Substituting each one of the above solutions of $p$ in (1) gives

$$\begin{array}{rl}{y}_1& =x+(p^2-\frac{2}{3}{p}^3)\\ & =x+({(-\sqrt{c_1-x})}^2-\frac{2}{3}{(-\sqrt{c_1-x})}^3)\\ & =x+(c_1-x+\frac{2}{3}{(c_1-x)}^{\frac{3}{2}})\\ & =c_1+\frac{2}{3}{(c_1-x)}^{\frac{3}{2}}\end{array}$$

And

$$\begin{array}{rl}{y}_2& =x+(p^2-\frac{2}{3}{p}^3)\\ & =x+({\left(\sqrt{c_1-x}\right)}^2-\frac{2}{3}{\left(\sqrt{c_1-x}\right)}^3)\\ & =x+(c_1-x-\frac{2}{3}{(c_1-x)}^{\frac{3}{2}})\\ & =c_1-\frac{2}{3}{(c_1-x)}^{\frac{3}{2}}\end{array}$$

Therefore the solutions are

$$\begin{array}{rl}y& =x+\frac{1}{3}\\ y& =c_1+\frac{2}{3}{(c_1-x)}^{\frac{3}{2}}\\ y& =c_1-\frac{2}{3}{(c_1-x)}^{\frac{3}{2}}\end{array}$$