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

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

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

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(A) Given differential equation is $y^2 dx + (x^2 - xy + y^2) dy = 0$.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: $v + x \frac{dv}{dx} = \frac{-(vx)^2}{x^2 - x(vx) + (vx)^2} = \frac{-v^2}{1 - v + v^2}$.Then $x \frac{dv}{dx} = \frac{-v^2}{1 - v + v^2} - v = \frac{-v^2 - v + v^2 - v^3}{1 - v + v^2} = \frac{-(v^3 + v)}{v^2 - v + 1}$.Separating variables: $\frac{v^2 - v + 1}{v^3 + v} dv = -\frac{1}{x} dx$.Using partial fractions: $\frac{v^2 - v + 1}{v(v^2 + 1)} = \frac{A}{v} + \frac{Bv + C}{v^2 + 1}$. Solving gives $A=1, B=0, C=-1$.So,$(\frac{1}{v} - \frac{1}{v^2 + 1}) dv = -\frac{1}{x} dx$.Integrating both sides: $\log|v| - 	an^{-1}(v) = -\log|x| + C$.$\log|vx| = 	an^{-1}(v) + C$.Since $y = vx$,we have $\log|y| = 	an^{-1}(\frac{y}{x}) + C$,or $	an^{-1}(\frac{y}{x}) = \log y + C$.

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

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