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(A) Given the differential equation: $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$.Rearranging the terms,we get: $x \log(x^2 e) dx = (\

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$y \sin y = x^2 \log x$

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(A) Given the differential equation: $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$.Rearranging the terms,we get: $x \log(x^2 e) dx = (\sin y + y \cos y) dy$.Integrating both sides: $\int x \log(x^2 e) dx = \int (\sin y + y \cos y) dy$.For the left side,let $t = x^2 e$,then $dt = 2x e dx$,so $x dx = \frac{dt}{2e}$.$\int \log(t) \frac{dt}{2e} = \frac{1}{2e} (t \log t - t) = \frac{x^2 e}{2e} (\log(x^2 e) - 1) = \frac{x^2}{2} (\log x^2 + \log e - 1) = \frac{x^2}{2} (2 \log x + 1 - 1) = x^2 \log x$.For the right side,using integration by parts: $\int (\sin y + y \cos y) dy = y \sin y - \int \sin y dy + \int \sin y dy = y \sin y + C$.Thus,$x^2 \log x = y \sin y + C$.Given $y(1) = 0$,substitute $x = 1$ and $y = 0$: $1^2 \log(1) = 0 \cdot \sin(0) + C \implies 0 = 0 + C \implies C = 0$.Therefore,the particular solution is $x^2 \log x = y \sin y$.

Find the particular solution of the differential equation $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$,given that $y(1) = 0$.

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