Find general and singular solution of $9p^2(2-p)^2=4(3-y), p=\frac{dy}{dx}$
My attempt:
Let $F(x,y,p)=9p^2(2-p)^2-4(3-y)=0, \frac{\partial F}{\partial p} = 0 \implies p(p-1)(p-2)=0 \implies (y-3)(4y-3)=0$
is required p-discriminant(obtained by putting p=0,1,2 in given equation).
Now $4y=12-9p^2(2-p)^2$
differentiating wrto x,
$4p=-9(2p) \frac{dp}{dx}-9p^2(2)(-1)(2-p)\frac{dp}{dx}$. By simplifing,
$9(p^2-3p+2)dp+dx=0 \implies 6p^3-27p^2+36p+2x=c,$ c is some arbitrary constant. this together with given equation is solution with p as parameter.
But textbook solution is given as $(x+c)^2=y^2(3-y)$. Pls help
Best Answer
If the textbook solution is given as $\quad (x+c)^2=y^2(3-y)\quad$ then : $$2(x+c)dx=(6y-3y^2)dy \quad\to\quad y'=\frac{2(x+c)}{6y-3y^2}$$ Putting these $y$ and $y'$ into the initial equation $\quad 9(y')^2(2-y')^2=4(3-y)\quad$ shows that it doesn't agree.
Thus $\quad (x+c)^2=y^2(3-y)\quad$ is not solution of $\quad 9(y')^2(2-y')^2=4(3-y)$.
So, there is a typo or a mistake in the wording of the question.
Note :
$ (x+c)^2=y^2(3-y)\quad$ is solution of $\quad 9(y')^2(2-y)^2=4(y-3)$
instead of $\quad 9(y')^2(2-y')^2=4(3-y)\quad$ So they are two differences : $y$ instead of $y'$ on the left and the inverse sign on the right.