The general solution of the differential equa | vedclass

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(B) Given differential equation: $(xy + y^2) dx - (x^2 - 2xy) dy = 0$ Rearranging the terms: $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - 2xy}$ Divide numer

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$cxy^2 e^{\frac{x}{y}} = 1$

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(B) Given differential equation: $(xy + y^2) dx - (x^2 - 2xy) dy = 0$ Rearranging the terms: $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - 2xy}$ Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{\frac{y}{x} + (\frac{y}{x})^2}{1 - 2(\frac{y}{x})}$ Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting into the equation: $v + x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v}$ $x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v} - v = \frac{v + v^2 - v + 2v^2}{1 - 2v} = \frac{3v^2}{1 - 2v}$ Separating variables: $(\frac{1 - 2v}{3v^2}) dv = \frac{dx}{x} \Rightarrow (\frac{1}{3v^2} - \frac{2}{3v}) dv = \frac{dx}{x}$ Integrating both sides: $\int (\frac{1}{3}v^{-2} - \frac{2}{3v}) dv = \int \frac{1}{x} dx$ $-\frac{1}{3v} - \frac{2}{3} \ln|v| = \ln|x| + C_1$ Substitute $v = \frac{y}{x}$: $-\frac{x}{3y} - \frac{2}{3} \ln(\frac{y}{x}) = \ln|x| + C_1$ Multiply by $3$: $-\frac{x}{y} - 2(\ln y - \ln x) = 3\ln x + 3C_1$ $-\frac{x}{y} = 3\ln x + 2\ln y - 2\ln x + C_2 = \ln x + 2\ln y + C_2 = \ln(xy^2) + C_2$ $-\frac{x}{y} = \ln(Cxy^2) \Rightarrow e^{-\frac{x}{y}} = Cxy^2$ Therefore,$Cxy^2 e^{\frac{x}{y}} = 1$.

The general solution of the differential equation $(xy + y^2) dx - (x^2 - 2xy) dy = 0$ is

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