# Give the correct answer and solve the given equation displaystylefrac{{left.{d}{y}right.}}{{left.{d}{x}right.}}=frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}},{y}{left({2}right)}={2}

Give the correct answer and solve the given equation $\frac{dy}{dx}=\frac{{y}^{2}-1}{{x}^{2}-1},y\left(2\right)=2$
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Lacey-May Snyder
Given that
$\frac{dy}{dx}=\frac{{y}^{2}-1}{{x}^{2}-1},$
$y\left(2\right)=2$
$\frac{dy}{dx}=\frac{{y}^{2}-1}{{x}^{2}-1}$
$⇒\frac{dy}{{y}^{2}-1}=\frac{dx}{{x}^{2}-1}$
$⇒\int \frac{dy}{{y}^{2}-1}=\int \frac{dx}{{x}^{2}-1}$
$⇒\frac{\mathrm{ln}\left(|y-1|\right)}{|y+1|}=\frac{\mathrm{ln}\left(|x-1|\right)}{|x+1|}+\mathrm{ln}c.$
Now putting $y\left(2\right)=2$ we have
$\frac{\mathrm{ln}1}{3}=\frac{\mathrm{ln}1}{3}+\mathrm{ln}c$
$⇒\frac{1}{3}=c\frac{1}{3}$
$⇒c=1$
Therefore we have
$⇒\frac{\mathrm{ln}\left(|y-1|\right)}{y+1}=\frac{\mathrm{ln}\left(|x-1|\right)}{|x+1|}+\mathrm{ln}1$
$⇒\frac{\mathrm{ln}\left(|y-1|\right)}{y+1}=\frac{\mathrm{ln}\left(|x-1|\right)}{|x+1|}$
$⇒\frac{|y-1|}{y+1}=\frac{|x-1|}{|x+1|}$