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Family of Curves

Obtain the di...

Question

Obtain the differential equation of the family of circles $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ ; where g , f & c are arbitary constants.

2 y^{'} (y^{''})^{2} + y^{'''} [1 - (y^{'})^{3}] = 0

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3 y^{'} (y^{''})^{2} - y^{'''} [1 + (y^{'})^{2}] = 0

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3 y^{'} (y^{''})^{2} + y^{'''} [1 - (y^{'})^{2}] = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 y^{'} (y^{''})^{2} - y^{'''} [1 + (y^{'})^{3}] = 0

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Solution

The correct option is A $3 y^{'} (y^{''})^{2} - y^{'''} [1 + (y^{'})^{2}] = 0$
Differentiation, we get,
$\Rightarrow 2 x + 2 y y^{'} + 2 g + 2 f y^{'} = 0$
Again differentiating,
$\Rightarrow 1 + (y^{'})^{2} + y y^{''} + f y^{''} = 0$ ........(i)
Again differentiating,
$\Rightarrow 2 (y^{'}) y^{''} + y^{'} y^{''} + y y^{'''} + f y^{'''} = 0$
$\Rightarrow 3 y^{'} y^{''} + y^{'''} [y - \frac{1}{y^{''}} - \frac{(y^{'})^{2}}{y^{''}} - y] = 0$ (from (i))
$\Rightarrow 3 y^{'} (y^{''})^{2} - y^{'''} [1 + (y^{'})^{2}] = 0$

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