Given differential equation, $y^2(y – xp) = x^4p^2$ where {$p = dy/dx$}
To find the singular solution,
I have extracted the p-discriminant relation which is, $y^3x^2(y + 4x^2) = 0$
From here it is evident that all $x = 0, y = 0$ and $y + 4x^2 = 0$ are the singular solutions when tested by putting back in the differential equation but my text book doesn't mention anything about $x=0$ as singular solution. Am I doing any mistake?
Best Answer
$$y^2\left(y - x\frac{dy}{dx}\right) = x^4\left(\frac{dy}{dx} \right)^2$$ $$y(x)=-4x^2\quad\text{is a singular solution.}$$ $$y(x)=0\quad\text{is a singular solution.}$$ $x(y)=0$ is not solution of the ODE because $\frac{dy}{dx}$ is infinite which doesn't agree with the ODE.
$(x=0, y=0)$ isn't a function but a point. Thus it isn't a solution of the ODE.
The general solution of the ODE is : $$y(x)=\frac{c^2 x}{x-c}$$ The above singular solutions are the envelope of the general solution. $$ $$ IN ADDITION, ANSWER TO THE COMMENT from Hari Prasad (Too long answer to be edited in the comments section) :
The argument raised is :
Let me remind this well-known argument : Given an ODE $$f(x,y,p)=0\tag 1$$ and multiplying by $g(x,y,p)$ gives a different ODE : $$g(x,y,p)f(x,y,p)=0\tag 2$$ which is not equivalent to Eq.$(1)$.The solutions of $(2)$ are the solutions of $(1)$ and the solutions of $(3)$ : $$g(x,y,p)=0\tag 3$$ Thus a solution of $(2)$ is not necessarily a solution of $(1)$. $$ $$ In your argument the equation $$f(x,y,p)=y^2(y-xp)-x^4p^2=0\tag 4$$ is multiplied by $g(x,y,p)=\frac{1}{p^2}$. So the solutions of $$y^2(\frac{y}{p^2}-\frac{x}{p})-x^4=0\tag 5$$ are not only the solutions of $(4)$ but in addition the solutions of $$\frac{1}{p^2}=\left(\frac{dx}{dy}\right)^2=0\tag 6$$ which are $x(y)=c$ and among them $x(y)=0$.
The solution $x=0$ is solution of $(5)$ but not necessarily solution of $(4)$.
Thus your argument is not correct: It doesn't prove that $x(y)=0$ is solution of Eq.$(4)$.