given y=(√((1−x)/(1+x))) , then (1−x^2 )(dy/dx)+ky=0. find k


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Question Number 155983 by MathsFan last updated on 06/Oct/21

$g i v e n y = \sqrt{\frac{1 - x}{1 + x}}, t h e n$ $(1 - x^{2}) \frac{d y}{d x} + k y = 0 . f i n d k$
Commented by cortano last updated on 06/Oct/21

$k = 1$
Commented by yeti123 last updated on 07/Oct/21

$(1 - x^{2}) \frac{d y}{d x} + k y = 0$ $(1 - x^{2}) \frac{d}{d x} \sqrt{\frac{1 - x}{1 + x}} + k y = 0$ $(1 - x^{2}) \times \frac{- 1}{(1 + x^{2}) \sqrt{\frac{1 - x}{1 + x}}} + k y = 0$ $(1 - x^{2}) \times \frac{- 1}{(1 + x^{2}) y} + k y = 0$ $- (1 - x^{2}) + {k y}^{2} (1 + x^{2}) = 0$ $k = \frac{1 - x^{2}}{y^{2} (1 + x^{2})}$
	Answered by puissant last updated on 06/Oct/21

	$(1 - x^{2}) \frac{d y}{d x} + k y = 0$ $\Rightarrow (1 - x^{2}) d y = - k y d x \Rightarrow \frac{d y}{k y} = - \frac{d x}{1 - x^{2}}$ $\Rightarrow \frac{1}{k} \int \frac{d y}{y} = \int \frac{d x}{x^{2} - 1} \Rightarrow \frac{1}{k} l n y = - \frac{1}{2} l n (\frac{1 + x}{1 - x})$ $y = \sqrt{\frac{1 - x}{1 + x}} \Rightarrow \frac{1}{k} l n y = \frac{1}{2 k} l n (\frac{1 - x}{1 + x})$ $\Rightarrow \frac{1}{2 k} l n (\frac{1 - x}{1 + x}) = - \frac{1}{2} l n (\frac{1 + x}{1 - x}) = \frac{1}{2} l n (\frac{1 - x}{1 + x})$ $\Rightarrow \frac{1}{2 k} = \frac{1}{2} \Rightarrow ∴∵ k = 1 . . . ◼$
	Commented by MathsFan last updated on 07/Oct/21

	$m e r c i$
	Commented by puissant last updated on 07/Oct/21

	$D^{'} a c c o r d j^{'} e s p e r e q u e t o u s e s t c l a i r e . .$
	Commented by MathsFan last updated on 07/Oct/21

	$o u i m o n s i e u r$
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