The solution of the equation (2y - 1) , dx - | vedclass

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Question

(C) Given the differential equation: $(2y - 1) \, dx = (2x + 3) \, dy$ Separating the variables,we get: $\frac{dx}{2x + 3} = \frac{dy}{2y - 1}$ Integr

Vedclass · Accepted Answer

$\frac{2x + 3}{2y - 1} = c$

Vedclass · Answer

(C) Given the differential equation: $(2y - 1) \, dx = (2x + 3) \, dy$ Separating the variables,we get: $\frac{dx}{2x + 3} = \frac{dy}{2y - 1}$ Integrating both sides: $\int \frac{dx}{2x + 3} = \int \frac{dy}{2y - 1}$ $\frac{1}{2} \ln|2x + 3| = \frac{1}{2} \ln|2y - 1| + C'$ Multiply by $2$: $\ln|2x + 3| = \ln|2y - 1| + 2C'$ $\ln|2x + 3| - \ln|2y - 1| = \ln c$ (where $\ln c = 2C'$) $\ln \left| \frac{2x + 3}{2y - 1} \right| = \ln c$ Therefore,$\frac{2x + 3}{2y - 1} = c$.

The solution of the equation $(2y - 1) \, dx - (2x + 3) \, dy = 0$ is

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