The general solution of the differential equa | vedclass

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Question

(C) Given differential equation is $(x + y)dx + xdy = 0$. Rearranging the terms,we get $xdy = -(x + y)dx$,which implies $\frac{dy}{dx} = -\frac{x + y}

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$x^2 + 2xy = c$

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(C) Given differential equation is $(x + y)dx + xdy = 0$. Rearranging the terms,we get $xdy = -(x + y)dx$,which implies $\frac{dy}{dx} = -\frac{x + y}{x} = -(1 + \frac{y}{x})$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $v + x\frac{dv}{dx} = -(1 + v) = -1 - v$. $x\frac{dv}{dx} = -1 - 2v$. Separating the variables: $\frac{dv}{1 + 2v} = -\frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{1 + 2v} = -\int \frac{dx}{x}$. $\frac{1}{2} \ln|1 + 2v| = -\ln|x| + C_1$. $\ln|1 + 2v| = -2\ln|x| + 2C_1 = \ln|x^{-2}| + \ln|C|$. $1 + 2v = \frac{C}{x^2}$. Substituting $v = \frac{y}{x}$: $1 + 2(\frac{y}{x}) = \frac{C}{x^2}$. $\frac{x + 2y}{x} = \frac{C}{x^2} \implies x(x + 2y) = C \implies x^2 + 2xy = C$.

The general solution of the differential equation $(x + y)dx + xdy = 0$ is

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