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 $\int y \cos y dy$: $\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y$.Thus, $\int (\sin y + y \cos y) dy = -\cos y + y \sin y + \cos y = y \sin y$.For the right side, $\int (x \log x^2 + x) dx = \int (2x \log x + x) dx$.Using integration by parts for $\int 2x \log x dx$: Let $u = \log x$, $dv = 2x dx$, then $du = \frac{1}{x} dx$, $v = x^2$.$\int 2x \log x dx = x^2 \log x - \int x^2 \cdot \frac{1}{x} dx = x^2 \log x - \int x dx = x^2 \log x - \frac{x^2}{2}$.Adding the integral of $x$: $\int (2x \log x + x) dx = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C$.Equating both sides: $y \sin y = x^2 \log x + C$.

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

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