Question

Form the differential equation of the family of curves represented by y² = (x − c)³.

Answer 1

The equation of the family of curves is
$y^2={\left(x-c\right)}^3$ ...(1)
where $c\in R$ is a parameter.
This equation contains only one parameter, so we shall obtain a differential equation of first order.
Differentiating equation (1) with respect to $x$, we get
$2y\frac{dy}{dx}=3{\left(x-c\right)}^2$ ...(2)
Dividing equation (1) by equation (2), we get
$$\begin{gathered} \frac{y^2}{2y{\displaystyle \frac{dy}{dx}}}=\frac{{\left(x-c\right)}^3}{3{\left(x-c\right)}^2} \\ \Rightarrow \frac{y}{2{\displaystyle \frac{dy}{dx}}}=\frac{\left(x-c\right)}{3} \\ \Rightarrow \frac{3y}{2{\displaystyle \frac{dy}{dx}}}=x-c \\ \Rightarrow c=x-\frac{3y}{2{\displaystyle \frac{dy}{dx}}} \end{gathered}$$
Substituting the value of $c$ in equation (1), we get
$$\begin{gathered} y^2={\left(x-x+\frac{3y}{2{\displaystyle \frac{dy}{dx}}}\right)}^3 \\ \Rightarrow y^2=\frac{27y^3}{8{\displaystyle {\left(\frac{dy}{dx}\right)}^3}} \\ \Rightarrow 8y^2{\left(\frac{dy}{dx}\right)}^3=27y^3 \\ \Rightarrow 8{\left(\frac{dy}{dx}\right)}^3-27y=0 \end{gathered}$$
It is the required differential equation.