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(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = v$. Then $1 + \frac{dy}{dx} = \frac{dv}{dx}$,so $\frac{dy}{dx} =

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$y - x + \frac{1}{3} = \log |x+y|$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = v$. Then $1 + \frac{dy}{dx} = \frac{dv}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$.$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$.Separating variables: $\int \frac{v-1}{2v} dv = \int dx$.$\frac{1}{2} \int (1 - \frac{1}{v}) dv = x + C$.$\frac{1}{2} (v - \log |v|) = x + C$.Substitute $v = x+y$: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x + C$.Given $x = \frac{2}{3}$ and $y = \frac{1}{3}$,then $x+y = 1$.$\frac{1}{2} - \frac{1}{2} \log |1| = \frac{2}{3} + C$.$\frac{1}{2} - 0 = \frac{2}{3} + C \Rightarrow C = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$.Substituting $C$ back: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x - \frac{1}{6}$.Multiply by $2$: $(x+y) - \log |x+y| = 2x - \frac{1}{3}$.Rearranging: $y - x + \frac{1}{3} = \log |x+y|$.

The particular solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ when $x = \frac{2}{3}$ and $y = \frac{1}{3}$ is

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