Verify that an implicit equation is the solution to the differential equation.

Question

Verify that \begin{equation} a.\;x^3+y^3-3xy=0,\; \mathbb{R}_{x\neq2^{2/3}} \end{equation} is the solution to \begin{equation} b.\;(y^2-x)y' - y+x^2=0 \end{equation} As we know the function g(x) of a. is not easily attainable, therefore we resort to stating one graphically. Such that at $2^{2/3}$ does not exist meaning it makes a jump.

The book I'm reading says that if we implicitly differentiate a. then it agrees with b. and hence the definition of the solution is met.

Definition of solution: Implicit Solution of ODE f(x,y) is the implicit solution of a differential equation: \begin{equation} F(x, y, y',\dots,y^{(n)})=0,\: D \end{equation} on the domain D if it defines a function $g(x)$ in D such that f[x,g(x)]=0 which \begin{equation} F[x, g(x) g(x)',\dots,g(x)^{(n)}]=0 \end{equation}

I don't understand the logic behind the author's assertion and the definition.

Answer 1

You can solve the proposed DE as follows: \begin{align*} (y^{2} - x)y' - y + x^{2} = 0 & \Longleftrightarrow y^{2}y' - xy' - y + x^{2} = 0\\\\ & \Longleftrightarrow y^{2}y' - (xy)' = -x^{2}\\\\ & \Longleftrightarrow \frac{y^{3}}{3} - xy = -\frac{x^{3}}{3} + c_{0}\\\\ & \Longleftrightarrow x^{3} + y^{3} - 3xy = c \end{align*}

which leads exactly to the implicit equation that defines $y$ when one considers $c = 0$.

Hopefully this helps!

Answer 2

Another solution is to rewrite

\begin{equation} \;x^3+y^3-3xy=0 \end{equation}

in the form

$$ (x+y)^3-3x^2y-3xy^2-3xy=0 $$

When you take $\dfrac{d}{dx}$ of both sides you will see that all terms subtract out except

$$ x^2+y^2y^\prime-y-xy^\prime=0 $$