MCQ
The general solution of the dofferential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$ is:
  • $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
  • B
    $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
  • C
    $\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
  • D
    $\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Answer

Correct option: A.
$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\times\text{e}^{\text{y}}$
$\Rightarrow \text{e}^{-\text{y}}\text{dy}=\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^{\text{x}}\text{dx}$
$\Rightarrow \text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{C}$

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