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Question: Answered & Verified by Expert
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
MathematicsDifferential EquationsWBJEEWBJEE 2009
Options:
  • A $\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c}$
  • B $\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
  • C $e^{-y}=e^x+e^{-x}+c$
  • D $e^y=e^x+e^{-x}+c$
Solution:
2113 Upvotes 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
$-\mathrm{e}^{-\mathrm{y}}=e^x-e^{-x}+c, \quad \mathrm{e}^{-\mathrm{y}}=e^{-x}-e^{+x}+c$

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