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
asked 2020-11-02
Give the correct answer and solve the given equation \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}^{2}-{1}}}{{{x}^{2}-{1}}},{y}{\left({2}\right)}={2}\)

Answers (1)

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
\(\displaystyle\Rightarrow\frac{ \ln{{\left({\left|{y}-{1}\right|}\right)}}}{{{y}+{1}}}=\frac{ \ln{{\left({\left|{x}-{1}\right|}\right)}}}{{{\left|{x}+{1}\right|}}}+ \ln{{1}}\)
\(\displaystyle\Rightarrow\frac{ \ln{{\left({\left|{y}-{1}\right|}\right)}}}{{{y}+{1}}}=\frac{ \ln{{\left({\left|{x}-{1}\right|}\right)}}}{{{\left|{x}+{1}\right|}}}\)
\(\displaystyle\Rightarrow\frac{{{\left|{y}-{1}\right|}}}{{{y}+{1}}}=\frac{{{\left|{x}-{1}\right|}}}{{{\left|{x}+{1}\right|}}}\)
0

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