The general solution of the differential equa | vedclass

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(B) 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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(B) 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: $\int \frac{dy}{y} = \int \frac{dx}{x}$ This yields: $\ln|y| = \ln|x| + C$ Using the property of logarithms,we can write the constant $C$ as $\ln|c|$: $\ln|y| = \ln|x| + \ln|c|$ $\ln|y| = \ln|cx|$ Taking the exponential of both sides: $y = cx$ This equation represents a family of straight lines passing through the origin.

The general solution of the differential equation $x dy - y dx = 0$ represents

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