The general solution of y^2 dx + (x^2 - xy + | vedclass

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(A) Given the differential equation: $y^2 dx + (x^2 - xy + y^2) dy = 0$. Rearranging,we get: $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$. This is a

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$	an^{-1}\left(\frac{y}{x}ight) = \log y + C$

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(A) Given the differential equation: $y^2 dx + (x^2 - xy + y^2) dy = 0$. Rearranging,we get: $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{-(vx)^2}{x^2 - x(vx) + (vx)^2} = \frac{-v^2}{1 - v + v^2}$. $x \frac{dv}{dx} = \frac{-v^2}{v^2 - v + 1} - v = \frac{-v^2 - v(v^2 - v + 1)}{v^2 - v + 1} = \frac{-v^2 - v^3 + v^2 - v}{v^2 - v + 1} = \frac{-(v^3 + v)}{v^2 - v + 1}$. Separating variables: $\frac{v^2 - v + 1}{v^3 + v} dv = -\frac{1}{x} dx$. Using partial fractions: $\frac{v^2 - v + 1}{v(v^2 + 1)} = \frac{A}{v} + \frac{Bv + D}{v^2 + 1}$. Solving gives $A = 1, B = 0, D = -1$. So,$\left(\frac{1}{v} - \frac{1}{v^2 + 1}ight) dv = -\frac{1}{x} dx$. Integrating both sides: $\log|v| - 	an^{-1}(v) = -\log|x| + C$. $\log|v| + \log|x| = 	an^{-1}(v) + C$. $\log|vx| = 	an^{-1}(v) + C$. Since $y = vx$,we have $\log|y| = 	an^{-1}\left(\frac{y}{x}ight) + C$,or $	an^{-1}\left(\frac{y}{x}ight) = \log y + C$.

The general solution of $y^2 dx + (x^2 - xy + y^2) dy = 0$ is:

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