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Question
x cos y dy = (xex log x + ex) dx
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Solution
We have,
\[x \cos y dy = \left( x e^x \log x + e^x \right) dx\]
\[ \Rightarrow \cos y dy = \left( e^x \log x + \frac{1}{x} e^x \right)dx\]
Integrating both sides, we get
\[\int \cos y dy = \int\left( e^x \log x + \frac{1}{x} e^x \right)dx\]
\[ \Rightarrow \sin y = \log x \int e^x dx - \int\frac{1}{x} e^x dx + \int\frac{1}{x} e^x dx\]
\[ \Rightarrow \sin y = e^x \log x + C\]
\[\text{ Hence, }\sin y = e^x \log x +\text{ C is the required solution }.\]
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