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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-2y+1}{2(x-2y)}$.Let $z = x-2y$. Then, differentiating with respect to $x$, we get $\frac

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$(x-2y)^2 + 2x = c$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-2y+1}{2(x-2y)}$.Let $z = x-2y$. Then, differentiating with respect to $x$, we get $\frac{dz}{dx} = 1 - 2\frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dz}{dx})$.Substituting these into the original equation:$\frac{1}{2}(1 - \frac{dz}{dx}) = \frac{z+1}{2z}$.Multiplying both sides by $2$, we get $1 - \frac{dz}{dx} = \frac{z+1}{z} = 1 + \frac{1}{z}$.Subtracting $1$ from both sides, we get $-\frac{dz}{dx} = \frac{1}{z}$, which simplifies to $z dz = -dx$.Integrating both sides, we get $\int z dz = \int -dx$, which results in $\frac{z^2}{2} = -x + C_1$.Multiplying by $2$, we get $z^2 = -2x + 2C_1$, or $z^2 + 2x = C$ (where $C = 2C_1$).Substituting $z = x-2y$ back, we get $(x-2y)^2 + 2x = C$.

The solution of the differential equation $\frac{dy}{dx} = \frac{x-2y+1}{2x-4y}$ is

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