The general solution of the differential equa | vedclass

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(D) Given the differential equation: $(x - (x+y) \log(x+y)) dx + x dy = 0$. Rearranging the terms: $x dy = ((x+y) \log(x+y) - x) dx$. $\frac{dy}{dx} =

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$\log (x+y) = cx$

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(D) Given the differential equation: $(x - (x+y) \log(x+y)) dx + x dy = 0$. Rearranging the terms: $x dy = ((x+y) \log(x+y) - x) dx$. $\frac{dy}{dx} = \frac{(x+y) \log(x+y) - x}{x}$. Let $v = x+y$,then $dv = dx + dy$,so $dy = dv - dx$. Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v \log v - x}{x} = \frac{v \log v}{x} - 1$. $\frac{dv}{dx} = \frac{v \log v}{x}$. Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{v \log v} = \int \frac{dx}{x}$. Let $u = \log v$,then $du = \frac{1}{v} dv$. $\int \frac{du}{u} = \int \frac{dx}{x} \implies \log|u| = \log|x| + \log|c|$. $u = cx \implies \log v = cx$. Substituting $v = x+y$ back: $\log(x+y) = cx$.

The general solution of the differential equation $(x-(x+y) \log (x+y)) dx + x dy = 0$ is

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