Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^\text{y}\text{x}^3$

Answer

$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^\text{y}\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{y}(\text{e}^\text{x}+\text{x}^3)$
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^\text{x}+\text{x}^3)\text{dx}$
$-\text{e}^{-\text{y}}-\text{e}^\text{x}+\frac{\text{x}^4}{4}+\text{C}_1$
$\text{e}^\text{x}+\frac{\text{x}^4}{4}+\text{e}^{-\text{y}}=\text{C}$

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