# Solve the following differential equations: \frac{dy}{dx}=\frac{x-e^{-x}}{y+e^{y}}

Solve the following differential equations:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$

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Step 1
Given
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$
To Find- The value of the above differential equations.
Step 2
Explanation- Rewrite the given expression,
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$
Simplifying the above expression and integrating both sides, we get,
$$\displaystyle\int{\left({y}+{e}^{{{y}}}\right)}{\left.{d}{y}\right.}=\int{\left({x}-{e}^{{-{x}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{\frac{{{y}^{{{2}}}}}{{{2}}}}={e}^{{{y}}}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{e}^{{-{x}}}+{C}$$
$$\displaystyle{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}}}{{{2}}}}+{e}^{{{y}}}-{e}^{{-{x}}}={C}$$
Answer- Hence, the solution of the differential equation $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$ is
$$\displaystyle{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}}}{{{2}}}}+{e}^{{{y}}}-{e}^{{-{x}}}={C}$$.
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