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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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}$$1 - \frac{dz}{dx} = \frac{z+1}{z}$$1 - \frac{dz}{dx} = 1 + \frac{1}{z}$$-\frac{dz}{dx} = \frac{1}{z}$$z dz = -dx$Integrating both sides:$\int z dz = \int -dx$$\frac{z^2}{2} = -x + C_1$$z^2 = -2x + 2C_1$Substituting $z = x-2y$ back:$(x-2y)^2 = -2x + C$$(x-2y)^2 + 2x = C$Thus,the correct option is $A$.

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

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