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

Verify that $$$$a.\;x^3+y^3-3xy=0,\; \mathbb{R}_{x\neq2^{2/3}}$$$$ is the solution to $$$$b.\;(y^2-x)y' - y+x^2=0$$$$ As we know the function g(x) of a. is not easily attainable, therefore we resort to stating one graphically. Such that at $$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: $$$$F(x, y, y',\dots,y^{(n)})=0,\: D$$$$ on the domain D if it defines a function $$g(x)$$ in D such that f[x,g(x)]=0 which $$$$F[x, g(x) g(x)',\dots,g(x)^{(n)}]=0$$$$

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

• This is not a function. It is just a relation. Jan 31, 2022 at 2:59
• 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. Jan 31, 2022 at 3:07
• @JackFrosher okay. Thanks for the explanation. Jan 31, 2022 at 4:37

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!

Another solution is to rewrite

$$$$\;x^3+y^3-3xy=0$$$$

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

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

• How does this manipulation indicate a logical satisfaction of the definition. Jan 31, 2022 at 3:48
• 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. Jan 31, 2022 at 3:56
• Oh, ok got it. Thank You. Jan 31, 2022 at 4:28