# 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}

Question
Integrals
Give the correct answer and solve the given equation $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}},{y}{\left({2}\right)}={2}$$

2020-11-03
Given that
$$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}},$$
$$\displaystyle{y}{\left({2}\right)}={2}$$
$$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}}$$
$$\displaystyle\Rightarrow\frac{{\left.{d}{y}\right.}}{{{y}^{2}-{1}}}=\frac{{\left.{d}{x}\right.}}{{{x}^{2}-{1}}}$$
$$\displaystyle\Rightarrow\int\frac{{\left.{d}{y}\right.}}{{{y}^{2}-{1}}}=\int\frac{{\left.{d}{x}\right.}}{{{x}^{2}-{1}}}$$
$$\displaystyle\Rightarrow\frac{ \ln{{\left({\left|{y}-{1}\right|}\right)}}}{{{\left|{y}+{1}\right|}}}=\frac{ \ln{{\left({\left|{x}-{1}\right|}\right)}}}{{{\left|{x}+{1}\right|}}}+ \ln{{c}}.$$
Now putting $$\displaystyle{y}{\left({2}\right)}={2}$$ we have
$$\displaystyle\frac{ \ln{{1}}}{{3}}=\frac{ \ln{{1}}}{{3}}+ \ln{{c}}$$
$$\displaystyle\Rightarrow\frac{1}{{3}}={c}\frac{1}{{3}}$$
$$\displaystyle\Rightarrow{c}={1}$$
Therefore we have
$$\displaystyle\Rightarrow\frac{ \ln{{\left({\left|{y}-{1}\right|}\right)}}}{{{y}+{1}}}=\frac{ \ln{{\left({\left|{x}-{1}\right|}\right)}}}{{{\left|{x}+{1}\right|}}}+ \ln{{1}}$$
$$\displaystyle\Rightarrow\frac{ \ln{{\left({\left|{y}-{1}\right|}\right)}}}{{{y}+{1}}}=\frac{ \ln{{\left({\left|{x}-{1}\right|}\right)}}}{{{\left|{x}+{1}\right|}}}$$
$$\displaystyle\Rightarrow\frac{{{\left|{y}-{1}\right|}}}{{{y}+{1}}}=\frac{{{\left|{x}-{1}\right|}}}{{{\left|{x}+{1}\right|}}}$$

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