The general solution of the differential equa | vedclass

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(C) Given differential equation is $(2x-y)^2 dy - 2(2x-y)^2 dx - 2 dx = 0$. Rearranging the terms: $(2x-y)^2 dy = [2(2x-y)^2 + 2] dx$. So,$\frac{dy}{d

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$(2x-y)^3 + 6x = c$

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(C) Given differential equation is $(2x-y)^2 dy - 2(2x-y)^2 dx - 2 dx = 0$. Rearranging the terms: $(2x-y)^2 dy = [2(2x-y)^2 + 2] dx$. So,$\frac{dy}{dx} = \frac{2(2x-y)^2 + 2}{(2x-y)^2} = 2 + \frac{2}{(2x-y)^2}$. Let $v = 2x-y$. Then $\frac{dv}{dx} = 2 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 2 - \frac{dv}{dx}$. Substituting this into the equation: $2 - \frac{dv}{dx} = 2 + \frac{2}{v^2}$. $-\frac{dv}{dx} = \frac{2}{v^2} \implies -v^2 dv = 2 dx$. Integrating both sides: $-\int v^2 dv = \int 2 dx$. $-\frac{v^3}{3} = 2x + c_1$. $v^3 = -6x + c$ (where $c = -3c_1$). Substituting $v = 2x-y$ back: $(2x-y)^3 = -6x + c$,which simplifies to $(2x-y)^3 + 6x = c$.

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

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