The equation of the curve that passes through the point (1,2) and satisfies the differential equation $\frac{d y}{d x} = \frac{- 2 x y}{(x^{2} + 1)}$ is

y (x^{2} + 1) = 4

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y (x^{2} + 1) + 4 = 0

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y (x^{2} - 1) = 4

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Solution

The correct option is A $y (x^{2} + 1) = 4$
$\frac{d y}{d x} = \frac{- 2 x y}{(x^{2} + 1)} \Rightarrow \frac{d y}{y} = - \frac{2 x}{x^{2} + 1} d x$
On integrating, we get
$l o g y = - l o g (1 + x^{2}) + l o g c \Rightarrow y (1 + x^{2}) = c$
Since curve passes through (1, 2), we have
$c = 2 (1 + 1^{2}) \Rightarrow c = 4$
Hence solution is $y (x^{2} + 1) = 4$

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Similar questions

\frac{d y}{d x} = \frac{- 2 x y}{(x^{2} + 1)}

(x^{2} - y^{2}) d x + 2 x y d y = 0

(1, 1)

(x = 1, y = 0)

\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 y} + \frac{y}{x}

(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}

y = y (x)

\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0

y (0) = 1,

y (1)