Prove that $y = \frac{c - x}{1 + c x}$ is the solution of differential equation $(1 + x^{2}) \frac{d y}{d x} + (1 + y^{2}) = 0$

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Solution

$y = \frac{c - x}{1 + c x}$
Differentiate w.r.t.x,
$\frac{d y}{d x} = \frac{- 1 (1 + c x) - c (c - x)}{(1 + c x)^{2}}$
$\frac{d y}{d x} = \frac{- 1 - c x - c^{2} + c x}{(1 + c x)^{2}}$
$= \frac{- (1 + c^{2})}{(1 + c x)^{2}}$ ....... $(2)$
From $(1)$ ,
$y (1 + c x) = c - x$
$\Rightarrow c = \frac{x + y}{1 - x y}$
Substituting value of $c$ in $(2)$ ,
$\frac{d y}{d x} = \frac{- [1 + {(\frac{x + y}{1 - x y})}^{2}]}{{[1 + \frac{(x + y) x}{1 - x y}]}^{2}}$
$= \frac{1 [(1 - x y)^{2} + x^{2} + y^{2} + 2 x y]}{(1 - x y + x^{2} + x y)^{2}}$
$= \frac{- (1 + x^{2} y^{2} + x^{2} + y^{2})}{(1 + x^{2})^{2}}$
$= \frac{- (1 + x^{2}) (1 + y^{2})}{(1 + x^{2})^{2}}$
$= \frac{- (1 + y^{2})}{1 + x^{2}}$
$\Rightarrow \frac{d y}{d x} (1 + x^{2}) = - (1 + y^{2})$
$\Rightarrow (1 + x^{2}) \frac{d y}{d x} + (1 + y^{2}) = 0$

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