Solve the following differential equation x dy=(xe^x +e^x)dx - Vidyadip

Question

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

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Answer

We have $\text{x}\cos\text{y dy}=(\text{xe}^\text{x}\log\text{x}+\text{e}^\text{x})\text{dx}$ $\Rightarrow\cos\text{y dy}\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$Integrating both sides, we get
$\int\cos\text{y dy}\int\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$ $\sin\text{y}=\log\text{x}\int\text{e}^\text{ x dx} -\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}+\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}$ $\Rightarrow\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ Hence, $\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ is the required solution.

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