The general solution of the differential equa | vedclass

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(C) Given the differential equation: $(x+y) y dx + (y-x) x dy = 0$. Rearranging the terms: $(y-x) x dy = -(x+y) y dx$,which gives $\frac{dy}{dx} = \fr

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$x+y \log (cxy) = 0$

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(C) Given the differential equation: $(x+y) y dx + (y-x) x dy = 0$. Rearranging the terms: $(y-x) x dy = -(x+y) y dx$,which gives $\frac{dy}{dx} = \frac{(x+y) y}{(x-y) x}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{(x+vx) vx}{(x-vx) x} = \frac{v(1+v)}{1-v} = \frac{v+v^2}{1-v}$. $x \frac{dv}{dx} = \frac{v+v^2}{1-v} - v = \frac{v+v^2-v+v^2}{1-v} = \frac{2v^2}{1-v}$. Separating the variables: $\int \frac{1-v}{v^2} dv = \int \frac{2}{x} dx$. Integrating both sides: $\int (v^{-2} - v^{-1}) dv = 2 \int \frac{1}{x} dx$. $-v^{-1} - \ln|v| = 2 \ln|x| + C$. Substituting $v = \frac{y}{x}$: $-\frac{x}{y} - \ln(\frac{y}{x}) = 2 \ln|x| + C$. $-\frac{x}{y} - \ln|y| + \ln|x| = 2 \ln|x| + C$. $-\frac{x}{y} = \ln|y| + \ln|x| + C = \ln|xy| + C$. Multiplying by $-y$: $x = -y \ln|xy| - yC$. $x + y(\ln|xy| + C) = 0$. $x + y \ln|cxy| = 0$,where $C = \ln|c|$.

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

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