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(C) Given differential equation is $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.Let $v = x-y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\fr

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

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(C) Given differential equation is $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.Let $v = x-y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.Substituting this into the equation: $1 - \frac{dv}{dx} = \frac{v+3}{2v+5}$.Rearranging gives $\frac{dv}{dx} = 1 - \frac{v+3}{2v+5} = \frac{2v+5-v-3}{2v+5} = \frac{v+2}{2v+5}$.Separating variables: $\int \frac{2v+5}{v+2} dv = \int dx$.Rewriting the integrand: $\int \left( \frac{2(v+2) + 1}{v+2} \right) dv = \int dx$.This simplifies to $\int (2 + \frac{1}{v+2}) dv = \int dx$.Integrating both sides: $2v + \log|v+2| = x + c$.Substituting $v = x-y$ back: $2(x-y) + \log|x-y+2| = x + c$.

The solution of the differential equation $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$ is

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