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प्रश्न
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
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उत्तर
The equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1............(1)\]
where \[a\text{ and }b\] are parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating equation (1) with respect to x, we get
\[\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} =0..........(2)\]
Differentiating equation (2) with respect to x, we get
\[\frac{2}{a^2} - \frac{2}{b^2}\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = 0\]
\[ \Rightarrow \frac{1}{a^2} = \frac{1}{b^2}\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right]\]
\[ \Rightarrow \frac{b^2}{a^2} = \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] \left......( 3 \right)\]
Now, from equation (2), we get
\[\frac{2x}{a^2} = \frac{2y}{b^2}\frac{dy}{dx}\]
\[ \Rightarrow \frac{b^2}{a^2} = \frac{y}{x}\frac{dy}{dx} ........(4)\]
From (3) and (4), we get
\[\frac{y}{x}\frac{dy}{dx} = \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right]\]
\[ \Rightarrow y\frac{dy}{dx} = xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 \]
\[ \Rightarrow xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 - y\frac{dy}{dx} = 0\]
It is the required differential equation.
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