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
\[2y\frac{dy}{dx} - 2a\frac{dy}{dx} + 2x = 0\]
\[ \Rightarrow 2y\frac{dy}{dx} + 2x = 2a\frac{dy}{dx}\]
\[ \Rightarrow y + \frac{x}{\frac{dy}{dx}} = a\]
Substituting the value of a in equation (2), we get
\[y^2 - 2\left( y + \frac{x}{\frac{dy}{dx}} \right)y + x^2 = \left( y + \frac{x}{\frac{dy}{dx}} \right)^2 \]
\[ \Rightarrow \frac{y^2 \frac{dy}{dx} - 2\left( y\frac{dy}{dx} + x \right)y + x^2 \frac{dy}{dx}}{\frac{dy}{dx}} = \frac{\left( y\frac{dy}{dx} + x \right)^2}{\left( \frac{dy}{dx} \right)^2}\]
\[ \Rightarrow y^2 \left( \frac{dy}{dx} \right)^2 - 2 y^2 \left( \frac{dy}{dx} \right)^2 - 2xy\left( \frac{dy}{dx} \right) + x^2 \left( \frac{dy}{dx} \right)^2 = y^2 \left( \frac{dy}{dx} \right)^2 + 2xy\left( \frac{dy}{dx} \right) + x^2 \]
\[ \Rightarrow \left( x^2 - 2 y^2 \right) \left( \frac{dy}{dx} \right)^2 - 4xy\left( \frac{dy}{dx} \right) - x^2 = 0 \]
It is the required differential equation.