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

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

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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):
$${y}{\left({x}\right)}=-\sqrt{{{2}}}\sqrt{{ \log{{\left({e}^{x}+{1}\right)}}+{c}_{{1}}}}{\quad\text{or}\quad}{y}{\left({x}\right)}=\sqrt{{{2}}}\sqrt{{ \log{{\left({e}^{x}+{1}\right)}}+{c}_{{1}}}}$$