The general solution of the differential equa | vedclass

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

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$x^2-y^2=Cx$

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(C) Given differential equation: $x^2+y^2-2xy \frac{dy}{dx}=0$ Rearranging the terms: $2xy \frac{dy}{dx} = x^2+y^2$ $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$ This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$ $x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$ Separating the variables: $\frac{2v}{1-v^2} dv = \frac{dx}{x}$ Integrating both sides: $\int \frac{2v}{1-v^2} dv = \int \frac{dx}{x}$ Let $1-v^2 = t$,then $-2v dv = dt$,so $2v dv = -dt$. $-\int \frac{dt}{t} = \int \frac{dx}{x} \Rightarrow -\ln|t| = \ln|x| + \ln|C_1|$ $-\ln|1-v^2| = \ln|x| + \ln|C_1| = \ln|C_1 x|$ $\ln|1-v^2|^{-1} = \ln|C_1 x| \Rightarrow \frac{1}{1-v^2} = C_1 x$ Substituting $v = y/x$: $\frac{1}{1-(y^2/x^2)} = C_1 x \Rightarrow \frac{x^2}{x^2-y^2} = C_1 x$ $\frac{x}{x^2-y^2} = C_1 \Rightarrow x = C_1(x^2-y^2)$ Rearranging gives $x^2-y^2 = Cx$ (where $C = 1/C_1$ is an arbitrary constant).

The general solution of the differential equation $x^2+y^2-2xy \frac{dy}{dx}=0$ is (where $C$ is a constant of integration.)

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