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The general solution of the differential equation $\frac{d y}{d x}=e^{y+x}+e^{y-x}$ is
where $c$ is an arbitrary constant
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where $c$ is an arbitrary constant
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Verified Answer
The correct answer is:
$\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
Hints: $e^{-y} d y=\left(e^x+e^{-x}\right) d x$ Integrate
$-\mathrm{e}^{-\mathrm{y}}=e^x-e^{-x}+c, \quad \mathrm{e}^{-\mathrm{y}}=e^{-x}-e^{+x}+c$
$-\mathrm{e}^{-\mathrm{y}}=e^x-e^{-x}+c, \quad \mathrm{e}^{-\mathrm{y}}=e^{-x}-e^{+x}+c$
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