Given that $$8x(y')^3 + 12y(y')^2 - 9y^5 = 0$$ Find all solutions and study singular solutions.
My solution
After noting that $y' = 0$ is solution only when $y = 0$ we subtitute $y' = \frac{1}{x'}$ and then $x' = p$. After that we get $$\frac{8x}{p^3} + \frac{12y}{p^2} - 9y^5 = 0 \implies 8x + 12yp - 9y^5p^3 = 0 \tag{1}$$ Then we take differentials of both sides and note that $dx = pdy$. Now we have $$8pdy + 12ydp + 12pdy - 45y^4p^3dy - 27y^5p^2dp = 0 \implies$$ $$\implies (20p - 45y^4p^3)dy + (12y - 27y^5p^2)dp = 0 \implies$$ $$\implies 5p(4 - 9y^4p^2)dy + 3y(4 - 9y^4p^2)dp = 0 \tag{2}$$ From here we take solution $4 - 9y^4p^2 = 0 \implies p = \pm\frac{2}{3y^2}$ and substitute in $(1)$: $$8x \pm (\frac{8}{y} - \frac{8}{3y}) = 0 \implies y = \pm\frac{2}{3x}$$ Now we continue solving $(2)$ divided by $4 - 9y^4p^2$. We have $$5pdy + 3ydp = 0 \implies y^5p^3 = C_1 \implies p = \frac{C_2}{\sqrt[3]{y^5}}$$ Substitute this in $(1)$ $$8x + \frac{12C_2}{\sqrt[3]{y^2}} - 9C_2^3 = 0$$ Now we make a substitution $C = \frac{3}{2}C_2$ and get $$8x + \frac{8C}{\sqrt[3]{y^2}} - \frac{8C^3}{3} = 0 \implies 3x + \frac{3C}{\sqrt[3]{y^2}} - C^3 = 0 \implies $$ $$\implies y = \pm \left(\frac{3C}{C^3 - 3x}\right)^\frac{3}{2}$$
And here the first problem appears: solution in the book is $y = \pm \left(\frac{3C}{C - 3x}\right)^\frac{3}{2}$, but I can't replace $C^3$ to $C$ because then $C$ in numerator turns into $\sqrt[3]{C}$. After all, solutions we have are:
$\quad \large 1) y = 0$
$\quad \large 2) y = \pm\frac{2}{3x}$
$\quad \large 3) y = \pm \left(\frac{3C}{C^3 - 3x}\right)^\frac{3}{2}$
Then we study singular solutions. Firstly, $$\frac{\partial F}{\partial y'} = 24x(y')^2 + 24yy' = 0 \implies xy' + y = 0 \implies$$ $$\implies xy = C$$ Solutions of this kind that we have are $$xy = \pm\frac{2}{3}$$ Let's try to find intersections and tangent lines of our solutions with other solutions. $$y_1(x_0) = y_2(x_0) \implies \pm\frac{2}{3x} = \pm \left(\frac{3C}{C^3 - 3x}\right)^\frac{3}{2} \tag{3}$$ This equation seems to be hard to solve, so let's just leave it as it is and look for tangent lines' condition $$y_1'(x_0) = y_2'(x_0) \implies -\frac{2}{3x^2} = \left(\frac{3C}{C^3 - 3x}\right)^\frac{3}{2}\cdot \frac{9}{2(C^3 - 3x)} = \frac{2}{3x}\frac{9}{2(C^3 - 3x)} \tag{4}$$ The last part of the equation follows from $(3)$. After some manipulations we obtain $$6x - 2C^3 = 9x \implies x = -\frac{2C^3}{3}$$ So maybe we'll now think that singular solutions really exist from our computations, but they don't. If we put our solution to $(3)$ we'll get $$\mp\frac{1}{C^3} = \pm\frac{1}{C^3}$$ which obviously does not hold for any $C\in \mathbb{R}$. So I concluded that there are no singular solutions as such. But the book, again, says that there are two singular solutions $y = \pm\frac{2}{3x}$
I have 2 questions:
$\quad 1)$ Are the solutions found correct?
$\quad 2)$ Do singular solutions exist? Maybe I've made a mistake somewhere and everything went wrong.
In principle it is looking good. One can start discussing composite solutions at the point (2) where you factorize $$ 0=(5p+3yp')(4−9y^4p^2)=(5p+3yp')(2-3y^2p)(2+3y^2p) $$ This means that on solutions one can mark out segments according to which factor is zero. The factor can change at points where two factors are zero. Obviously, the last two factors never are simultaneously zero.
Another way to simplify the situation is to take equation (1) $$ 0=8x+12py-9(py^{5/3})^3 $$ and compare it with the Clairaut equation $$ v(u)=v'(u)u+f(v'(u)), $$ where comparing the form of the last terms suggests the transformation $u=y^{-2/3}$, $x=v(u)=v(y^{-2/3})$, so that $p=x'=-\tfrac23y^{-5/3}v'(y^{-2/3})$ and $$ 0=8v(u)-8v'(u)u+\frac83v'(u)^3\iff v(u)=v'(u)u-\frac13v'(u)^3 $$ is indeed already in Clairaut form. This now has a well-known solution structure, first the family of linear solutions $$ v(u)=Cu-\frac13C^3\iff x=\frac{C}{y^{2/3}}-\frac{C^3}3 $$ and then the singular solution that results from $0=u+f'(v'(u))=u-v'(u)^2$, which inserted into the original Clairaut equation gives $$ v(u)=v'(u)u-\frac13v'(u)u=\pm\frac23u^{3/2}\implies x=\pm\frac2{3y}. $$ A curve $v=Cu+f(C)$ from the regular family meets the singular curve where $u=-f'(C)$, here $y^{-2/3}=C^2\iff y=\pm C^{-3}$, $x=\frac23C^3$. The sign of $y$ decides the branch of the singular curve, it is possible that you mixed the wrong sign choices there. The slopes of the tangents at these points are \begin{align} x=\frac{C}{y^{2/3}}-\frac{C^3}3&\implies x'=-\frac23Cy^{-5/3}=\mp\frac23C^6,\\ x=\pm\frac{2}{3y}&\implies x'=\mp\frac2{3y^2}=\mp\frac23C^6, \end{align} so that indeed, as constructed, the tangents at intersections are the same.