The general solution of the differential equa | vedclass

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(B) Given differential equation is $(2 y-1) dx - (2 x+3) dy = 0$. Rearranging the terms,we get $(2 y-1) dx = (2 x+3) dy$. Separating the variables,we

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

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(B) Given differential equation is $(2 y-1) dx - (2 x+3) dy = 0$. Rearranging the terms,we get $(2 y-1) dx = (2 x+3) dy$. Separating the variables,we have $\frac{dx}{2 x+3} = \frac{dy}{2 y-1}$. Integrating both sides,we get $\int \frac{dx}{2 x+3} = \int \frac{dy}{2 y-1}$. This results in $\frac{1}{2} \ln|2 x+3| = \frac{1}{2} \ln|2 y-1| + C_1$. Multiplying by $2$,we get $\ln|2 x+3| = \ln|2 y-1| + 2C_1$. Let $2C_1 = \ln|c|$,then $\ln|2 x+3| - \ln|2 y-1| = \ln|c|$. Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$,we get $\ln|\frac{2 x+3}{2 y-1}| = \ln|c|$. Taking the exponential of both sides,we obtain $\frac{2 x+3}{2 y-1} = c$.

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

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