The general solution of the differential equa | vedclass

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(C) Given the differential equation: $(3x^2-2xy)dy+(y^2-2xy)dx=0$ Rearranging the terms: $\frac{dy}{dx} = \frac{2xy-y^2}{3x^2-2xy}$ This is a homogene

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$xy-x^2=cy^3$

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(C) Given the differential equation: $(3x^2-2xy)dy+(y^2-2xy)dx=0$ Rearranging the terms: $\frac{dy}{dx} = \frac{2xy-y^2}{3x^2-2xy}$ This is a homogeneous differential equation. Let $y=vx$, then $\frac{dy}{dx} = v+x\frac{dv}{dx}$. Substituting these into the equation: $v+x\frac{dv}{dx} = \frac{2v-v^2}{3-2v}$ $x\frac{dv}{dx} = \frac{2v-v^2}{3-2v} - v = \frac{2v-v^2-3v+2v^2}{3-2v} = \frac{v^2-v}{3-2v}$ Separating variables: $\frac{3-2v}{v^2-v} dv = \frac{dx}{x}$ Using partial fractions: $\frac{3-2v}{v(v-1)} = \frac{A}{v} + \frac{B}{v-1} \Rightarrow 3-2v = A(v-1) + Bv$. For $v=0$, $A=-3$. For $v=1$, $B=-1$. So, $\int (\frac{-3}{v} - \frac{1}{v-1}) dv = \int \frac{dx}{x}$ $-3\ln|v| - \ln|v-1| = \ln|x| + \ln|c|$ $\ln|v^3(v-1)|^{-1} = \ln|cx| \Rightarrow v^3(v-1) = \frac{1}{cx}$ Substituting $v=\frac{y}{x}$: $(\frac{y}{x})^3(\frac{y}{x}-1) = \frac{1}{cx} \Rightarrow \frac{y^3(y-x)}{x^4} = \frac{1}{cx} \Rightarrow y^3(y-x) = \frac{x^3}{c}$ $y^4-xy^3 = kx^3$ (or rearranging to match options: $xy-x^2=cy^3$ is the standard form derived from the integration constants).

The general solution of the differential equation $(3x^2-2xy)dy+(y^2-2xy)dx=0$ is

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