The particular solution of the differential e | vedclass

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(A) Given the differential equation: $(x+y) dy + (x-y) dx = 0$. Rearranging,we get $\frac{dy}{dx} = -\frac{x-y}{x+y} = \frac{y-x}{y+x}$. This is a hom

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

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(A) Given the differential equation: $(x+y) dy + (x-y) dx = 0$. Rearranging,we get $\frac{dy}{dx} = -\frac{x-y}{x+y} = \frac{y-x}{y+x}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx-x}{vx+x} = \frac{v-1}{v+1}$. $x \frac{dv}{dx} = \frac{v-1}{v+1} - v = \frac{v-1-v^2-v}{v+1} = -\frac{1+v^2}{1+v}$. Separating variables: $\int \frac{1+v}{1+v^2} dv = -\int \frac{dx}{x}$. $\int \frac{1}{1+v^2} dv + \frac{1}{2} \int \frac{2v}{1+v^2} dv = -\log|x| + C$. $	an^{-1}(v) + \frac{1}{2} \log(1+v^2) = -\log|x| + C$. Substituting $v = \frac{y}{x}$: $	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$. At $x=1, y=1$: $	an^{-1}(1) + \frac{1}{2} \log(1^2+1^2) = C \Rightarrow \frac{\pi}{4} + \frac{1}{2} \log(2) = C$. Substituting $C$ back: $	an^{-1}(\frac{y}{x}) + \frac{1}{2} \log(x^2+y^2) = \frac{\pi}{4} + \frac{1}{2} \log(2)$. $\frac{1}{2} \log(\frac{x^2+y^2}{2}) = \frac{\pi}{4} - 	an^{-1}(\frac{y}{x})$. Multiplying by $2$: $\log(\frac{x^2+y^2}{2}) = \frac{\pi}{2} - 2 	an^{-1}(\frac{y}{x})$.

The particular solution of the differential equation $(x+y) dy + (x-y) dx = 0$ at $x=1, y=1$ is

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