The solution of the differential equation e^{ | vedclass

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(A) Given the differential equation: $e^{2y}(1 + \ln x)dx + \csc y(2 + \cot y)dy = 0$.Rearranging the terms,we get: $\csc y(2 + \cot y)dy = -e^{2y}(1

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$x \ln x + e^{-\pi} = \frac{e^{-2y}}{\sin y}$

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(A) Given the differential equation: $e^{2y}(1 + \ln x)dx + \csc y(2 + \cot y)dy = 0$.Rearranging the terms,we get: $\csc y(2 + \cot y)dy = -e^{2y}(1 + \ln x)dx$.Dividing by $e^{2y}$,we have: $e^{-2y}(\csc y(2 + \cot y))dy = -(1 + \ln x)dx$.Integrating both sides: $\int e^{-2y}(2\csc y + \csc y \cot y)dy = -\int (1 + \ln x)dx$.Let $f(y) = e^{-2y} \csc y$. Then $f'(y) = e^{-2y}(-\csc y \cot y) + (-2)e^{-2y} \csc y = -e^{-2y}(\csc y \cot y + 2 \csc y)$.Thus,the integral of the left side is $-e^{-2y} \csc y$.The integral of the right side is $-\int (1 + \ln x)dx = -(x \ln x) + C$.So,$-e^{-2y} \csc y = -x \ln x + C$,which simplifies to $x \ln x + C = e^{-2y} \csc y = \frac{e^{-2y}}{\sin y}$.Using the condition $y(1) = \frac{\pi}{2}$,we substitute $x=1$ and $y=\frac{\pi}{2}$:$1 \cdot \ln(1) + C = \frac{e^{-2(\pi/2)}}{\sin(\pi/2)} \Rightarrow 0 + C = \frac{e^{-\pi}}{1} \Rightarrow C = e^{-\pi}$.Therefore,the solution is $x \ln x + e^{-\pi} = \frac{e^{-2y}}{\sin y}$.

The solution of the differential equation $e^{2y} (1 + \ln x)dx + \csc y (2 + \cot y)dy = 0$ satisfying $y(1) = \frac{\pi}{2}$ is

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