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

beljuA 2021-09-06 Answered

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

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Expert Answer

yunitsiL
Answered 2021-09-07 Author has 12690 answers

Solve the separable equation \(-{e}^{x}+\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{\left({e}^{x}+{1}\right)}{y}{\left({x}\right)}={0}:\)
Solve for \(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}:\)
\(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}=\frac{{e}^{x}}{{{\left({e}^{x}+{1}\right)}{y}{\left({x}\right)}}}\)
INTERMEDIATE STEPS:
Solve for \(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}:\)
\({y}{\left({x}\right)}\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{\left({e}^{x}+{1}\right)}-{e}^{x}={0}\)
Hint: Isolate terms with \(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}\) to the left hand side.
Add \(\displaystyle{e}^{{{x}}}\) to both sides:
\({y}{\left({x}\right)}\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{\left({e}^{x}+{1}\right)}={e}^{x}\)
Solve for \(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}.\)
Divide both sides by \({\left({e}^{x}+{1}\right)}{y}{\left({x}\right)}:\)
\(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}=\frac{{e}^{x}}{{{y}{\left({x}\right)}{\left({e}^{x}+{1}\right)}}}\)
Multiply both sides by y(x):
\(\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{y}{\left({x}\right)}=\frac{{e}^{x}}{{{e}^{x}+{1}}}\)
Integrate both sides with respect to x:
\(\int\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}{y}{\left({x}\right)}{\left.{d}{x}\right.}=\int\frac{{e}^{x}}{{{e}^{x}+{1}}}{\left.{d}{x}\right.}\)
Evaluate the integrals:
\({y}\frac{{\left({x}\right)}^{2}}{{2}}= \log{{\left({e}^{x}+{1}\right)}}+{c}_{{1}}\), where \(\displaystyle{c}_{{{1}}}\) is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
integral \(\frac{{e}^{x}}{{{e}^{x}+{1}}}{\left.{d}{x}\right.}\)
For the integrand \(\frac{{e}^{x}}{{{e}^{x}+{1}}}\), substitute \({u}={e}^{x}+{1}{\quad\text{and}\quad}{d}{u}={e}^{x}{\left.{d}{x}\right.}:\)
\(=\int\frac{1}{{u}}{d}{u}\)
The integral of \(\frac{1}{{u}}\) is \(\log(u)\):
\(= \log(u) + \text{constant}\)
Substitute back for \({u}={e}^{x}+{1}\):
\(= \log{{\left({e}^{x}+{1}\right)}}+\text{constant}\)
Solve for y(x):
Answer:
\({y}{\left({x}\right)}=-\sqrt{{{2}}}\sqrt{{ \log{{\left({e}^{x}+{1}\right)}}+{c}_{{1}}}}{\quad\text{or}\quad}{y}{\left({x}\right)}=\sqrt{{{2}}}\sqrt{{ \log{{\left({e}^{x}+{1}\right)}}+{c}_{{1}}}}\)

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