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The given equation is $ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + (\frac{y}{x})^2 $. Since the right-hand side is a homogeneous function of degree 0, this is a homogeneous differential equation.
Let $ y = vx $. Then, differentiating with respect to $ x $, we get $ \frac{dy}{dx} = v + x\frac{dv}{dx} $. Substituting these into the equation:
$$ v + x\frac{dv}{dx} = 1 + v + v^2 $$Subtracting $ v $ from both sides, we get $ x\frac{dv}{dx} = 1 + v^2 $. Separating the variables:
$$ \frac{dv}{1 + v^2} = \frac{dx}{x} $$Integrating both sides:
$$ \int \frac{dv}{1 + v^2} = \int \frac{dx}{x} $$ $$ \tan^{-1}(v) = \ln|x| + C $$Substitute $ v = \frac{y}{x} $ back into the equation:
$$ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C $$Final Answer: \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C
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