{y}{\left({x}^{2}-{1}\right)}{\left.{d}{y}\right.}+{x}{\left({y}^{2}+{1}\right)}{\left.{d}{x}\right.}={0}

Tazmin Horton 2021-09-09 Answered

\({y}{\left({x}^{2}-{1}\right)}{\left.{d}{y}\right.}+{x}{\left({y}^{2}+{1}\right)}{\left.{d}{x}\right.}={0}\)

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Expert Answer

Ayesha Gomez
Answered 2021-09-10 Author has 10698 answers

Solve the separable equation x \({\left({y}{\left({x}\right)}^{2}+{1}\right)}+\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{\left({x}^{2}-{1}\right)}{y}{\left({x}\right)}={0}:\)
Solve for \(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}:\)
\(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}=-\frac{{{x}{\left({y}{\left({x}\right)}^{2}+{1}\right)}}}{{{\left({x}^{2}-{1}\right)}{y}{\left({x}\right)}}}\)
Divide both sides by \(-\frac{{{y}{\left({x}\right)}^{2}+{1}}}{{y}}{\left({x}\right)}:\)
\(-\frac{{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{y}{\left({x}\right)}}}{{{y}{\left({x}\right)}^{2}+{1}}}=\frac{x}{{{x}^{2}-{1}}}\)
Integrate both sides with respect to x:
\(\int-\frac{{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{y}{\left({x}\right)}}}{{{y}{\left({x}\right)}^{2}+{1}}}{\left.{d}{x}\right.}=\int\frac{x}{{{x}^{2}-{1}}}{\left.{d}{x}\right.}\)
Evaluate the integrals:
\(-\frac{1}{{2}} \log{{\left({y}{\left({x}\right)}^{2}+{1}\right)}}=\frac{1}{{2}} \log{{\left({x}^{2}-{1}\right)}}+{c}_{{1}}\), where \(\displaystyle{c}_{{{1}}}\) is an arbitrary constant.
Solve for y(x):
\({y}{\left({x}\right)}=-\frac{\sqrt{{{e}^{{-{2}{c}_{{1}}}}-{x}^{2}+{1}}}}{\sqrt{{{x}^{2}-{1}}}}{\quad\text{or}\quad}{y}{\left({x}\right)}=\frac{\sqrt{{{e}^{{-{2}{c}_{{1}}}}-{x}^{2}+{1}}}}{\sqrt{{{x}^{2}-{1}}}}\)
INTERMEDIATE STEPS:
Solve for \(y(x)\):
\(-\frac{1}{{2}} \log{{\left({y}{\left({x}\right)}^{2}+{1}\right)}}=\frac{1}{{2}} \log{{\left({x}^{2}-{1}\right)}}+{c}_{{1}}\)
Hint: Multiply both sides by a constant to simplify the equation.
Multiply both sides by -2:
\(\log{{\left({y}{\left({x}\right)}^{2}+{1}\right)}}=- \log{{\left({x}^{2}-{1}\right)}}-{2}{c}_{{1}}\)
Hint: Eliminate the logarithm from the left hand side.
Cancel logarithms by taking exp of both sides:
\({y}{\left({x}\right)}^{2}+{1}=\frac{{e}^{{-{2}{c}_{{1}}}}}{{{x}^{2}-{1}}}\)
Hint: Isolate terms with y(x) to the left hand side.
Subtract 1 from both sides:
\({y}{\left({x}\right)}^{2}=\frac{{e}^{{-{2}{c}_{{1}}}}}{{{x}^{2}-{1}}}-{1}\)
Hint: Eliminate the exponent on the left hand side.
Take the square root of both sides:
\({y}{\left({x}\right)}=\sqrt{{\frac{{e}^{{-{2}{c}_{{1}}}}}{{{x}^{2}-{1}}}-{1}}}{\quad\text{or}\quad}{y}{\left({x}\right)}=-\sqrt{{\frac{{e}^{{-{2}{c}_{{1}}}}}{{{x}^{2}-{1}}}-{1}}}\)
Simplify the arbitrary constants:
Answer:
\({y}{\left({x}\right)}=-\frac{\sqrt{{-{x}^{2}+{c}_{{1}}}}}{\sqrt{{{x}^{2}-{1}}}}{\quad\text{or}\quad}{y}{\left({x}\right)}=\frac{\sqrt{{-{x}^{2}+{c}_{{1}}}}}{\sqrt{{{x}^{2}-{1}}}}\)

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