The solution of the differential equation (x^ | vedclass

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(B) The given differential equation is $(x^2 - xy)dy = (xy + y^2)dx$.This can be written as $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - xy}$.This is a homo

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$xy = ce^{-x/y}$

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(B) The given differential equation is $(x^2 - xy)dy = (xy + y^2)dx$.This can be written as $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - xy}$.This is a homogeneous differential equation.Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x(vx) + (vx)^2}{x^2 - x(vx)} = \frac{v + v^2}{1 - v}$.$x\frac{dv}{dx} = \frac{v + v^2}{1 - v} - v = \frac{v + v^2 - v + v^2}{1 - v} = \frac{2v^2}{1 - v}$.Separating the variables: $\frac{1 - v}{v^2} dv = \frac{2}{x} dx$.Integrating both sides: $\int (v^{-2} - v^{-1}) dv = 2 \int \frac{1}{x} dx$.$-v^{-1} - \ln|v| = 2\ln|x| + C$.Substituting $v = y/x$: $-\frac{x}{y} - \ln|y/x| = 2\ln|x| + C$.$-\frac{x}{y} - \ln|y| + \ln|x| = 2\ln|x| + C$.$-\frac{x}{y} = \ln|y| + \ln|x| + C = \ln|xy| + C$.Thus,$xy = e^{-x/y - C} = Ce^{-x/y}$.

The solution of the differential equation $(x^2 - xy)dy = (xy + y^2)dx$ is

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