The general solution of the differential equa | vedclass

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(B) Given the differential equation: $(x^2+xy)y'=y^2$.Dividing by $(x^2+xy)$,we get: $\frac{dy}{dx} = \frac{y^2}{x^2+xy} = \frac{y^2}{x(x+y)}$.This is

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$cy=e^{-\frac{y}{x}}$

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(B) Given the differential equation: $(x^2+xy)y'=y^2$.Dividing by $(x^2+xy)$,we get: $\frac{dy}{dx} = \frac{y^2}{x^2+xy} = \frac{y^2}{x(x+y)}$.This is a homogeneous differential equation. Let $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{(vx)^2}{x^2+x(vx)} = \frac{v^2x^2}{x^2(1+v)} = \frac{v^2}{1+v}$.Rearranging the terms: $x\frac{dv}{dx} = \frac{v^2}{1+v} - v = \frac{v^2 - v - v^2}{1+v} = \frac{-v}{1+v}$.Separating the variables: $\frac{1+v}{v} dv = -\frac{1}{x} dx$.Integrating both sides: $\int (\frac{1}{v} + 1) dv = -\int \frac{1}{x} dx$.$\ln|v| + v = -\ln|x| + C$.Substituting $v = \frac{y}{x}$: $\ln|\frac{y}{x}| + \frac{y}{x} = -\ln|x| + C$.$\ln|y| - \ln|x| + \frac{y}{x} = -\ln|x| + C$.$\ln|y| + \frac{y}{x} = C$.Taking the exponential of both sides: $e^{\ln|y| + \frac{y}{x}} = e^C$.$y \cdot e^{\frac{y}{x}} = K$ (where $K = e^C$).Rearranging gives $e^{-\frac{y}{x}} = cy$ (where $c = 1/K$).

The general solution of the differential equation $(x^2+xy)y'=y^2$ is

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