Find degree of differential equation of family of curves described as $y=c(x-c)^2$

Question

Find degree of differential equation of family of curves described as $$y=c(x-c)^2$$

I can see that the order is $1$ but am not able to identify the degree.

Answer 1

Given $y=c(x-c)^2$ $$\dfrac{dy}{dx}=2c(x-c).....(1)$$$$x-c=\dfrac{1}{2c}\left(\dfrac{dy}{dx}\right).....(2)$$$$y=c\left(\dfrac{1}{2c}\dfrac{dy}{dx}\right)^2$$$$y=c\left(\dfrac{1}{4c^2}\left(\dfrac{dy}{dx}\right)^2\right)$$$$c=\dfrac{1}{4y}\left(\dfrac{dy}{dx}\right)^2.....(3)$$Now plug in $(3)$ in $(1)$ we get,$$\dfrac{dy}{dx}=2\left(\dfrac{1}{4y}\left(\dfrac{dy}{dx}\right)^2\right)\left(x-\dfrac{1}{4y}\left(\dfrac{dy}{dx}\right)^2\right)$$$$1=\dfrac{1}{2y}\left(\dfrac{dy}{dx}\right)\left(x-\dfrac{1}{4y}\left(\dfrac{dy}{dx}\right)^2\right)$$$$1+\dfrac{1}{8y^2}\left(\dfrac{dy}{dx}\right)^3=\dfrac{x}{2y}\left(\dfrac{dy}{dx}\right)$$$$8y^2+\left(\dfrac{dy}{dx}\right)^3=4xy\dfrac{dy}{dx}$$ Therefore, $$\mbox{order }=1$$ $$\mbox{degree }=3$$

Answer 2

This is not an answer to the OP question. This a comment too long to be edited in the comments section.

In fact, the answer to the OP question was already given by Key Flex, who already found the ODE : $$8y^2+\left(\dfrac{dy}{dx}\right)^3=4xy\dfrac{dy}{dx} \tag 1$$ My comment is an answer to a question raised in comments by Claude Leib0vici who I welcome.

The question is : How to solve the ODE $(1)$ ?

Let $y(x)=u(x)^2\quad;\quad y'=2uu'$ $$8u^4+(2uu')^3=4xu^2(2uu')$$ $$u+(u')^3=xu'\tag 2$$ $$u'+3(u')^2u''=u'+xu''$$ $$(3(u')^2-x)u''=0$$ We get two families of solutions :

From $3(u')^2-x=0$ the first family $u=\frac{2}{3\sqrt{3}}x^{3/2}+C$ must be rejected because putting it into Eq.$(2)$ shows that it doesn't agree. This is a non convenient solution introduced by the differentiation of $(2)$.

From $u''=0$ the second family $u=ax+b$ is put into Eq.$(2)$ and agrees if $(ax+b)+a^3=xa$ which implies $b=-a^3$. Thus : $$u=ax-a^3$$ $$y=(ax-a^3)^2=a^2(x-a^2)$$ Let $a^2=c$ $$y=c(x-c)^2$$ Note that $a$ is not necessarily real, but can be complex. So, the case of negative $c$ is not forgotten.