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प्रश्न
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
उत्तर
The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 . . . . . . . . \left( 1 \right)\]
Here, a and b are parameters.
Since this equation contains two parameters, so we get a second order differential equation.
Differentiating (1) with respect to x, we get
\[\frac{2x}{a^2} - \frac{2y}{b^2}y' = 0 . . . . . . . . \left( 2 \right)\]
Differentiating (2) with respect to x, we get
\[\frac{2}{a^2} - \frac{2}{b^2}\left[ yy'' + \left( y' \right)^2 \right] = 0\]
\[ \Rightarrow \frac{1}{a^2} = \frac{1}{b^2}\left[ yy'' + \left( y' \right)^2 \right]\]
\[ \Rightarrow \frac{b^2}{a^2} = \left[ yy'' + \left( y' \right)^2 \right] . . . . . . . . (3)\]
From (2), we get
\[\frac{2x}{a^2} = \frac{2y}{b^2}y'\]
\[ \Rightarrow \frac{b^2}{a^2} = \frac{y}{x}y' . . . . . . . . (4)\]
From (3) and (4), we get
\[\frac{y}{x}y' = \left[ yy'' + \left( y' \right)^2 \right]\]
\[ \Rightarrow yy' = xyy'' + x \left( y' \right)^2 \]
\[\text{Hence, }xyy'' + x \left( y' \right)^2 - yy' = 0\text{ is the required differential equation.}\]
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