Solve the following differential equations: dy dx =e^ x+y +e^ -x+… - Vidyadip

Question

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

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Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{y}}(\text{e}^{\text{x}}+\text{e}^{-\text{x}})$
$\Rightarrow\text{e}^{-\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
$\Rightarrow-\text{e}^{-\text{y}}=\text{e}^{\text{x}}-\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$
Hence, $\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$ is the required solution.

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