The solution of the differential equation x d | vedclass

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(C) Given the differential equation: $x dy - y dx = 0$ Rearranging the terms,we get: $x dy = y dx$ Dividing both sides by $xy$ (assuming $x, y 
eq 0$

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$A$ straight line passing through the origin.

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(C) Given the differential equation: $x dy - y dx = 0$ Rearranging the terms,we get: $x dy = y dx$ Dividing both sides by $xy$ (assuming $x, y 
eq 0$): $\frac{dy}{y} = \frac{dx}{x}$ Integrating both sides: $\int \frac{1}{y} dy = \int \frac{1}{x} dx$ This gives: $\ln|y| = \ln|x| + C$ Taking the exponential of both sides: $y = cx$,where $c$ is an arbitrary constant. 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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