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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log(x^2) + x}{\sin y + y \cos y}$.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,note that $\log(x^2) = 2 \log x$:$\int (2x \log x + x) dx = 2 \int x \log x dx + \int x dx$.Using integration by parts for $\int x \log x dx$:$2 [\frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx] + \frac{x^2}{2} = 2 [\frac{x^2}{2} \log x - \frac{x^2}{4}] + \frac{x^2}{2} = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} = x^2 \log x$.Thus,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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