y = log{(√x) +(1/((√x) ))}^2 show that x(x+1)^2 y_2 + (x+1)^2 y


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Question Number 52820 by 33 last updated on 13/Jan/19
$y = log{(√x) +(1/((√x) ))}^2 show that x(x+1)^2 y_2 + (x+1)^2 y_1 = 2$
$y = l o g {\sqrt{x} + \frac{1}{\sqrt{x}}}^{2}$ $s h o w t h a t$ $x {(x + 1)}^{2} y_{2} + {(x + 1)}^{2} y_{1} = 2$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19

	$y = l n (x + 2 + \frac{1}{x}) = l n (\frac{x^{2} + 2 x + 1}{x})$ $y = 2 l n (x + 1) - l n x$ $y_{1} = \frac{2}{x + 1} - \frac{1}{x} = \frac{2 x - x - 1}{x^{2} + x} = \frac{x - 1}{x^{2} + x}$ $y_{2} = \frac{x^{2} + x (1) - (x - 1) (2 x + 1)}{x^{2} {(x + 1)}^{2}}$ $y_{2} = \frac{x^{2} + x - 2 x^{2} - x + 2 x + 1}{x^{2} {(x + 1)}^{2}} = \frac{- x^{2} + 2 x + 1}{x^{2} {(x + 1)}^{2}}$ $x {(x + 1)}^{2} y_{2} = - x + 2 + \frac{1}{x}$ ${(x + 1)}^{2} y_{1} = (x + 1) (x + 1) \times \frac{x - 1}{x (x + 1)} = \frac{x^{2} - 1}{x} = x - \frac{1}{x}$ $s o x {(x + 1)}^{2} y_{2} + {(x + 1)}^{2} y_{1}$ $= - x + 2 + \frac{1}{x} + x - \frac{1}{x}$ $= 2 h e n c e p r o v e d$
	Commented by 33 last updated on 14/Jan/19

	$t h a n k y o u s o m u c h$
	Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jan/19

	$m o s t w e l c o m e . . .$
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