Question

The graphs of members of the one-parameter family $x^3+y^3=3cxy$ are called folia of Descartes. Verify that this family is an implicit solution of the first-order differential

equation

$\frac{dy}{dx}=\frac{y}{x}\frac{(y^3-2x^3)}{(2y^3-x^3)}$

Short answer

The wanted equation using implicit differentiation of $x^3+y^3=3cxy$is showed.

Step-by-step solution

Implicit Solution of an ODE.

A relation $G(x,y)=0$is said to be an implicit solution of an ordinary differential equation on an interval$I$ , provided that there exists at least one function $\varphi$that satisfies the relation as well as the differential equation on $I$.

Implicit differentiation

By implicit differentiation of$x^3+y^3=3cxy$, you can obtain the following,

$\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(y^3\right)=\frac{d}{dx}\left(3cxy\right)$

We get,

$3x^2+3y^2\frac{dy}{dx}=3c\left(y+x\frac{dy}{dx}\right)$

Divide3on both sides,

$x^2+y^2\frac{dy}{dx}=cy+cx\frac{dy}{dx}$

Rearrange the terms,

$y^2\frac{dy}{dx}-cx\frac{dy}{dx}=cy-x^2$

Taking out the common term,

$\frac{dy}{dx}(y^2-cx)=cy-x^2$

$\frac{dy}{dx}=\frac{cy-x^2}{y^2-cx}$......(1)

Step 3:Manipulate the expressions.

You can see that this isn’t exactly what we are looking for, but let’s manipulate the expression and use the following equations that follow from$x^3+y^3=3cxy$,

$$\begin{gathered} y^3=3cxy-x^3......\left(2\right) \\ {x}^3=3cxy-y^3......\left(3\right) \end{gathered}$$

From (1), it follows,

$\frac{dy}{dx}=\frac{cy-x^2}{y^2-cx}$

Multiply and divide by$3xy$,

$$\begin{gathered} \frac{dy}{dx}=\frac{3xy}{3xy}\frac{cy-x^2}{y^2-cx} \\ \frac{dy}{dx}=\frac{y}{x}\frac{3x(cy-x^2)}{3y(y^2-cx)} \\ \frac{dy}{dx}=\frac{y}{x}\frac{(3cxy-3x^3)}{(3y^3-3cxy)} \end{gathered}$$

So,

$\frac{dy}{dx}=\frac{y}{x}\frac{(3cxy-x^3-2x^3)}{(2y^3+y^3-3cxy)}$

Use the equations (2) and (3),

$\frac{dy}{dx}=\frac{y}{x}\frac{(y^3-2x^3)}{(2y^3-x^3)}$

Hence, the statement is established.

Show the wanted equation using implicit differentiation of$x^3+y^3=3cxy$.