Question

Form the differential equation corresponding to y² − 2ay + x² = a² by eliminating a.

Answer 1

The equation of the family of curves is
$y^2-2ay+x^2=a^2$ ...(1)
where $a$ is a parameter.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to $x$, we get
$$\begin{gathered} 2y\frac{dy}{dx}-2a\frac{dy}{dx}+2x=0 \\ \Rightarrow 2y\frac{dy}{dx}+2x=2a\frac{dy}{dx} \\ \Rightarrow y+\frac{x}{{\displaystyle \frac{dy}{dx}}}=a \end{gathered}$$
Substituting the value of $a$ in equation (2), we get
$$\begin{gathered} y^2-2\left(y+\frac{x}{{\displaystyle \frac{dy}{dx}}}\right)y+x^2={\left(y+\frac{x}{{\displaystyle \frac{dy}{dx}}}\right)}^2 \\ \Rightarrow \frac{y^2{\displaystyle \frac{dy}{dx}}-2\left(y{\displaystyle \frac{dy}{dx}}+x\right)y+x^2{\displaystyle \frac{dy}{dx}}}{{\displaystyle \frac{dy}{dx}}}=\frac{{\left(y{\displaystyle \frac{dy}{dx}}+x\right)}^2}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2} \\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2-2y^2{\left(\frac{dy}{dx}\right)}^2-2xy\left(\frac{dy}{dx}\right)+x^2{\left(\frac{dy}{dx}\right)}^2=y^2{\left(\frac{dy}{dx}\right)}^2+2xy\left(\frac{dy}{dx}\right)+x^2 \\ \Rightarrow \left(x^2-2y^2\right){\left(\frac{dy}{dx}\right)}^2-4xy\left(\frac{dy}{dx}\right)-x^2=0 \\ \mathrm{It}\mathrm{is}\mathrm{the}\text{required differential equation.} \end{gathered}$$