Equation of curve through point(1,0) which satisfies the differential equation $(1 + y^{2}) d x - x y d y = 0$ , is

x^{2} + y^{2} = 1

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x^{2} - y^{2} = 1

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2 x^{2} + y^{2} = 2

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Solution

The correct option is B $x^{2} - y^{2} = 1$
We have $\frac{d x}{x} = \frac{y d y}{1 + y^{2}}$
Integrating, we get $l o g | x | = \frac{1}{2} l o g (1 + y^{2}) + l o g c$
or $| x | = c \sqrt{(1 + y^{2})}$
But it passes through (1, 0), so we get c =1
$∴$ Solution is $x^{2} = y^{2} + 1$ or $x^{2} - y^{2} = 1$ .

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(1, 0)

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