The general solution of the differential equa | vedclass

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(C) Given differential equation is $\frac{dy}{dx} = \frac{2x(y-2) + 1(y-2)}{2y(x-2) + 1(x-2)} = \frac{(2x+1)(y-2)}{(2y+1)(x-2)}$.Separating the variab

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$2(y-x) + 5 \log \left| \frac{y-2}{x+1} \right| = c$

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(C) Given differential equation is $\frac{dy}{dx} = \frac{2x(y-2) + 1(y-2)}{2y(x-2) + 1(x-2)} = \frac{(2x+1)(y-2)}{(2y+1)(x-2)}$.Separating the variables,we get $\frac{2y+1}{y-2} dy = \frac{2x+1}{x-2} dx$.Rewrite the fractions: $\frac{2(y-2)+5}{y-2} dy = \frac{2(x-2)+5}{x-2} dx$.This simplifies to $(2 + \frac{5}{y-2}) dy = (2 + \frac{5}{x-2}) dx$.Integrating both sides: $\int (2 + \frac{5}{y-2}) dy = \int (2 + \frac{5}{x-2}) dx$.$2y + 5 \log |y-2| = 2x + 5 \log |x-2| + C$.Rearranging terms: $2(y-x) + 5 \log |y-2| - 5 \log |x-2| = C$.$2(y-x) + 5 \log \left| \frac{y-2}{x-2} \right| = C$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{2xy - 4x + y - 2}{2xy + x - 4y - 2}$ is

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