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(A) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.Since this is a homogeneous differential equation,we substitute $y = vx$,wh

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$ay^2 = e^{x^2/y^2}$

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(A) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2}$.Now,$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.Separating the variables: $\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$,which simplifies to $(\frac{1}{v^3} + \frac{1}{v}) dv = -\frac{dx}{x}$.Integrating both sides: $\int (v^{-3} + v^{-1}) dv = -\int \frac{1}{x} dx$.This gives $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.Substituting $v = \frac{y}{x}$: $-\frac{x^2}{2y^2} + \ln(\frac{y}{x}) = -\ln|x| + C$.$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$,which simplifies to $\ln|y| - \frac{x^2}{2y^2} = C$.Rearranging gives $\ln|y| = \frac{x^2}{2y^2} + C$,so $y = e^{\frac{x^2}{2y^2} + C} = k e^{\frac{x^2}{2y^2}}$.Squaring both sides: $y^2 = k^2 e^{\frac{x^2}{y^2}}$. Letting $a = k^2$,we get $ay^2 = e^{x^2/y^2}$.

The solution of the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ is

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