The general solution of the differential equa | vedclass

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(B) Given the differential equation: $y(1+\log x)\left(\frac{dx}{dy}\right) - x \log x = 0$ Rearranging the terms,we get: $y(1+\log x) dx = x \log x d

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$x \log x=yc$

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(B) Given the differential equation: $y(1+\log x)\left(\frac{dx}{dy}\right) - x \log x = 0$ Rearranging the terms,we get: $y(1+\log x) dx = x \log x dy$ Separating the variables: $\frac{(1+\log x)}{x \log x} dx = \frac{dy}{y}$ Integrating both sides: $\int \frac{1+\log x}{x \log x} dx = \int \frac{1}{y} dy$ Split the integral on the left: $\int \frac{1}{x \log x} dx + \int \frac{\log x}{x \log x} dx = \int \frac{1}{y} dy$ $\int \frac{1}{x \log x} dx + \int \frac{1}{x} dx = \int \frac{1}{y} dy$ Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes: $\int \frac{1}{u} du + \int \frac{1}{x} dx = \int \frac{1}{y} dy$ Integrating gives: $\log|u| + \log|x| = \log|y| + \log|c|$ $\log|\log x| + \log|x| = \log|y| + \log|c|$ Using the property $\log a + \log b = \log(ab)$: $\log|x \log x| = \log|yc|$ Taking the exponential of both sides: $x \log x = yc$

The general solution of the differential equation $y(1+\log x)\left(\frac{dx}{dy}\right) - x \log x = 0$ is

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