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(A) The given differential equation is: $y \log y \, dx - x \, dy = 0$. Rearranging the terms,we get: $y \log y \, dx = x \, dy$. Separating the varia

Vedclass · Accepted Answer

$y = e^{Cx}$

Vedclass · Answer

(A) The given differential equation is: $y \log y \, dx - x \, dy = 0$. Rearranging the terms,we get: $y \log y \, dx = x \, dy$. Separating the variables,we have: $\frac{dy}{y \log y} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dy}{y \log y} = \int \frac{dx}{x}$. Let $t = \log y$,then $dt = \frac{1}{y} \, dy$. Substituting this into the integral: $\int \frac{dt}{t} = \int \frac{dx}{x}$. Integrating,we get: $\log |t| = \log |x| + \log |C|$. $\log |\log y| = \log |Cx|$. Taking the exponential of both sides: $\log y = Cx$. Thus,the general solution is: $y = e^{Cx}$.

Find the general solution of the differential equation: $y \log y \, dx - x \, dy = 0$.

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