Question
The general solution of the dofferential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$ is:
  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
  2. $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
  3. $\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
  4. $\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Answer

  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Solution:

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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