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

Answer

We have,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x e }^\text{x}\log\text{x}+\text{e}^\text{x}}{\text{x}\cos\text{y}}$

$\Rightarrow\text{x}\cos\text{y dy}=(\text{x e}^\text{x}\log\text{x}+\text{e}^\text{x})\ \text{dx}$

$\Rightarrow\cos\text{y dy}=\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\ \text{dx}$

Integrating both sides, we get

$\int\cos\text{y dy}=\int\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$

$\Rightarrow\sin\text{y}=\log\text{x}\int\text{e} ^\text{x}\text{dx}-\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}+\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}$

$\Rightarrow\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$

Hence, $\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ is the required solution.

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