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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. ![Text](https://stackoverflow.com/[![image.jpg][1]][1])

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.

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  • $\begingroup$ This is not a function. It is just a relation. $\endgroup$ Jan 31, 2022 at 2:59
  • $\begingroup$ An implicit function is also a term for an implicit equation. It's not an actual function. I've edited the title to make that clear. $\endgroup$ Jan 31, 2022 at 3:07
  • $\begingroup$ @JackFrosher okay. Thanks for the explanation. $\endgroup$ Jan 31, 2022 at 4:37

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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!

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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

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

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  • $\begingroup$ How does this manipulation indicate a logical satisfaction of the definition. $\endgroup$ Jan 31, 2022 at 3:48
  • $\begingroup$ The idea behind the definition is to expand the concept of 'solution' as it applies to a differential equation. Generally speaking, a solution of an algebraic equation is of the form $x=\{some number set\}$. The definition is merely acknowledging that the solution of a DE needn't be of the form $y=f(x)$ but can be of the form $g(x,y)=0$. At least that is my take on the exercise. $\endgroup$ Jan 31, 2022 at 3:56
  • $\begingroup$ Oh, ok got it. Thank You. $\endgroup$ Jan 31, 2022 at 4:28

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