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

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$\log (x^{2}+y^{2})+2 	an ^{-1}(\frac{y}{x})=\frac{\pi}{2}+\log 2$

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(A) Given differential equation: $(x+y) dy + (x-y) dx = 0$Rearranging the terms: $(x+y) dy = -(x-y) dx$$\Rightarrow \frac{dy}{dx} = \frac{-(x-y)}{x+y} = \frac{y-x}{x+y}$  ..........$(1)$This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting in $(1)$: $v + x \frac{dv}{dx} = \frac{vx-x}{x+vx} = \frac{v-1}{1+v}$$x \frac{dv}{dx} = \frac{v-1}{1+v} - v = \frac{v-1-v-v^{2}}{1+v} = \frac{-(1+v^{2})}{1+v}$Separating variables: $\frac{1+v}{1+v^{2}} dv = -\frac{dx}{x}$Integrating both sides: $\int \frac{1}{1+v^{2}} dv + \int \frac{v}{1+v^{2}} dv = -\int \frac{dx}{x}$$	an^{-1} v + \frac{1}{2} \log(1+v^{2}) = -\log x + C$Substitute $v = \frac{y}{x}$: $	an^{-1}(\frac{y}{x}) + \frac{1}{2} \log(1+\frac{y^{2}}{x^{2}}) = -\log x + C$$	an^{-1}(\frac{y}{x}) + \frac{1}{2} \log(\frac{x^{2}+y^{2}}{x^{2}}) = -\log x + C$$	an^{-1}(\frac{y}{x}) + \frac{1}{2} \log(x^{2}+y^{2}) - \log x = -\log x + C$$	an^{-1}(\frac{y}{x}) + \frac{1}{2} \log(x^{2}+y^{2}) = C$Multiply by $2$: $2 	an^{-1}(\frac{y}{x}) + \log(x^{2}+y^{2}) = 2C = K$Given $y=1$ at $x=1$: $2 	an^{-1}(1) + \log(1^{2}+1^{2}) = K$$2(\frac{\pi}{4}) + \log 2 = K \Rightarrow K = \frac{\pi}{2} + \log 2$Thus,the particular solution is: $\log(x^{2}+y^{2}) + 2 	an^{-1}(\frac{y}{x}) = \frac{\pi}{2} + \log 2$.

Find the particular solution satisfying the given condition: $(x+y) dy + (x-y) dx = 0$; $y=1$ when $x=1$.

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