The solution of frac{dy}{dx} = frac{x log x^2 | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ By separating the variables,we get: $(\sin y + y \cos

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

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ By separating the variables,we get: $(\sin y + y \cos y) dy = (x \log x^2 + x) dx$ Integrating both sides: $\int (\sin y + y \cos y) dy = \int (x \log x^2 + x) dx$ For the left side,using integration by parts on $\int y \cos y dy$: $\int \sin y dy + (y \sin y - \int \sin y dy) = y \sin y$ For the right side,$\int (2x \log x + x) dx = 2 \int x \log x dx + \int x dx$ Using integration by parts for $\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$ Thus,$\int (x \log x^2 + x) dx = 2(\frac{x^2}{2} \log x - \frac{x^2}{4}) + \frac{x^2}{2} + C = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C$ Therefore,the solution is $y \sin y = x^2 \log x + C$.

The solution of $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ is

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