If y' = frac{x - y}{x + y},then its solution | vedclass

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(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{d

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$y^2 + 2xy - x^2 = c$

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(A) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x - vx}{x + vx} = \frac{1 - v}{1 + v}$.Rearranging terms: $x \frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$.Separating variables: $\frac{1 + v}{1 - 2v - v^2} dv = \frac{dx}{x}$.Multiply by $-1/2$: $-\frac{1}{2} \frac{-2 - 2v}{1 - 2v - v^2} dv = \frac{dx}{x}$.Integrating both sides: $-\frac{1}{2} \ln|1 - 2v - v^2| = \ln|x| + C_1$.Multiply by $-2$: $\ln|1 - 2v - v^2| = -2 \ln|x| + C_2 = \ln|x^{-2}| + C_2$.Thus,$1 - 2v - v^2 = \frac{C}{x^2}$.Substituting $v = \frac{y}{x}$: $1 - 2(\frac{y}{x}) - (\frac{y}{x})^2 = \frac{C}{x^2}$.Multiplying by $x^2$: $x^2 - 2xy - y^2 = C$,which can be rewritten as $y^2 + 2xy - x^2 = c$.

If $y' = \frac{x - y}{x + y}$,then its solution is

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