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(D) Given the differential equation: $x \log x \, dy = (x \log x - y) \, dx$. Dividing both sides by $dx$ and $x \log x$,we get: $\frac{dy}{dx} = \fra

Vedclass · Accepted Answer

$(y-x) \log x + x = c$

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(D) Given the differential equation: $x \log x \, dy = (x \log x - y) \, dx$. Dividing both sides by $dx$ and $x \log x$,we get: $\frac{dy}{dx} = \frac{x \log x - y}{x \log x} = 1 - \frac{y}{x \log x}$. Rearranging the terms: $\frac{dy}{dx} + \frac{1}{x \log x} y = 1$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x \log x}$ and $Q(x) = 1$. The integrating factor $(IF)$ is given by $e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log x} \, dx}$. Let $u = \log x$,then $du = \frac{1}{x} \, dx$. Thus,$\int \frac{1}{x \log x} \, dx = \int \frac{1}{u} \, du = \log |u| = \log |\log x|$. Therefore,$IF = e^{\log |\log x|} = \log x$. The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + c$. $y \log x = \int 1 \cdot \log x \, dx + c$. Using integration by parts: $\int \log x \, dx = x \log x - x$. So,$y \log x = x \log x - x + c$. Rearranging gives $(y-x) \log x + x = c$.

The general solution of the differential equation $x \log x \, dy = (x \log x - y) \, dx$ is

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