If dx + dy = (x + y)(dx - dy), then log(x + y | vedclass

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(C) Given that $dx + dy = (x + y)(dx - dy)$.Dividing by $dx$, we get:$1 + \frac{dy}{dx} = (x + y)(1 - \frac{dy}{dx})$$1 + \frac{dy}{dx} = x + y - (x +

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$x - y + c$

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(C) Given that $dx + dy = (x + y)(dx - dy)$.Dividing by $dx$, we get:$1 + \frac{dy}{dx} = (x + y)(1 - \frac{dy}{dx})$$1 + \frac{dy}{dx} = x + y - (x + y)\frac{dy}{dx}$$\frac{dy}{dx}(1 + x + y) = x + y - 1$$\frac{dy}{dx} = \frac{x + y - 1}{x + y + 1} \quad \dots(i)$Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$, so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.Substituting into Eq. $(i)$:$\frac{dt}{dx} - 1 = \frac{t - 1}{t + 1}$$\frac{dt}{dx} = \frac{t - 1}{t + 1} + 1 = \frac{t - 1 + t + 1}{t + 1} = \frac{2t}{t + 1}$Separating variables:$\frac{t + 1}{2t} dt = dx$$\frac{1}{2}(1 + \frac{1}{t}) dt = dx$Integrating both sides:$\frac{1}{2}(t + \log|t|) = x + C_1$$t + \log|t| = 2x + 2C_1$Substituting $t = x + y$:$x + y + \log(x + y) = 2x + C$$\log(x + y) = x - y + C$

If $dx + dy = (x + y)(dx - dy)$, then $\log(x + y)$ is equal to

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