The general solution of the differential equa | vedclass

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Question

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

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$y \log x = x + cy$

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(C) Given differential equation: $x^2 dy - (xy - y^2) dx = 0$ $\Rightarrow x^2 dy = (xy - y^2) dx$ $\Rightarrow \frac{dy}{dx} = \frac{xy - y^2}{x^2}$ This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x(vx) - (vx)^2}{x^2} = \frac{vx^2 - v^2x^2}{x^2} = v - v^2$ $x \frac{dv}{dx} = v - v^2 - v = -v^2$ Separating the variables: $\frac{dv}{-v^2} = \frac{dx}{x}$ Integrating both sides: $\int -v^{-2} dv = \int \frac{1}{x} dx$ $\frac{1}{v} = \ln|x| + C$ Since $v = \frac{y}{x}$,we have: $\frac{x}{y} = \ln|x| + C$ $x = y \ln|x| + Cy$ $y \ln x = x - Cy$ (or $y \ln x = x + C'y$ where $C' = -C$).

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

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