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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 7

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Answer: \sin y=e^{x} \log x+c

Hint:Separate the terms of x and y and then integrate them.

Given: x \cos y d y=\left(x e^{x} \log x+e^{x}\right) d x

Solution: x \cos y d y=\left(x e^{x} \log x+e^{x}\right) d x

        \cos y d y=\left[\frac{\left(x e^{x} \log x+e^{x}\right)}{x}\right] d x

        Integrating both sides,           

        \begin{aligned} &\int \cos y d y=\int \frac{\left(x e^{x} \log x+e^{x}\right)}{x} d x \\\\ &\sin y=\int\left(e^{x} \log x+\frac{e^{x}}{x}\right) d x \\\\ &\sin y=\int e^{x} \log x d x+\int \frac{e^{x}}{x} d x \end{aligned}                [Integrating by parts]

        \begin{aligned} &\sin y=\log x \int e^{x} d x-\int \frac{1}{x}\left(\int e^{x} d x\right) d x+\int \frac{e^{x}}{x} d x \\\\ &\sin y=\log x e^{x}-\int \frac{e^{x}}{x} d x+\int \frac{e^{x}}{x} d x+c \\\\ &\sin y=e^{x} \log x+c \end{aligned}

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