The solution of the differential equation y(1 | vedclass

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(B) Given differential equation is $y(1+\log x) \frac{dx}{dy} = x \log x$. Rearranging the terms to separate the variables,we get: $\frac{1+\log x}{x

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$x \log x = yc$

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(B) Given differential equation is $y(1+\log x) \frac{dx}{dy} = x \log x$. Rearranging the terms to separate the variables,we get: $\frac{1+\log x}{x \log x} dx = \frac{1}{y} dy$. Integrating both sides: $\int \frac{1+\log x}{x \log x} dx = \int \frac{1}{y} dy$. Split the integral on the left side: $\int \frac{1}{x \log x} dx + \int \frac{1}{x} dx = \int \frac{1}{y} dy$. Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes: $\int \frac{1}{u} du + \int \frac{1}{x} dx = \int \frac{1}{y} dy$. $\log |u| + \log |x| = \log |y| + \log |c|$. Substituting $u = \log x$ back: $\log |\log x| + \log |x| = \log |y| + \log |c|$. 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 solution of the differential equation $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ is

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