The solution of the differential equation (x | vedclass

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(C) Given equation is $(x + y - 1)dx + (2x + 2y - 3)dy = 0$,which can be written as $\frac{dy}{dx} = - \frac{x + y - 1}{2(x + y) - 3}$.Let $x + y = t$

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$2y + x + \log (x + y - 2) = c$

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(C) Given equation is $(x + y - 1)dx + (2x + 2y - 3)dy = 0$,which can be written as $\frac{dy}{dx} = - \frac{x + y - 1}{2(x + y) - 3}$.Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.Substituting these into the equation: $\frac{dt}{dx} - 1 = - \frac{t - 1}{2t - 3}$.$\frac{dt}{dx} = 1 - \frac{t - 1}{2t - 3} = \frac{2t - 3 - t + 1}{2t - 3} = \frac{t - 2}{2t - 3}$.Separating the variables: $\frac{2t - 3}{t - 2} dt = dx$.$\frac{2(t - 2) + 1}{t - 2} dt = dx \implies (2 + \frac{1}{t - 2}) dt = dx$.Integrating both sides: $\int (2 + \frac{1}{t - 2}) dt = \int dx$.$2t + \log |t - 2| = x + c$.Substituting $t = x + y$: $2(x + y) + \log |x + y - 2| = x + c$.$2x + 2y + \log |x + y - 2| = x + c \implies x + 2y + \log |x + y - 2| = c$.

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

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