The solution of the differential equation x , | vedclass

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(B) Given differential equation is $x \, dy - y \, dx = 0$. Rearranging the terms,we get $x \, dy = y \, dx$. Separating the variables,we have $\frac{

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Straight line passing through origin

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(B) Given differential equation is $x \, dy - y \, dx = 0$. Rearranging the terms,we get $x \, dy = y \, dx$. Separating the variables,we have $\frac{dy}{y} = \frac{dx}{x}$. Integrating both sides,$\int \frac{dy}{y} = \int \frac{dx}{x}$. This gives $\ln |y| = \ln |x| + \ln |c|$,where $\ln |c|$ is the constant of integration. Using logarithmic properties,$\ln |y| = \ln |cx|$. Taking the exponential of both sides,$y = cx$. This is the equation of a straight line passing through the origin.

The solution of the differential equation $x \, dy - y \, dx = 0$ represents:

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