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

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$3$

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(D) Given the differential equation: $y^2 dx + (x^2 - xy - y^2) dy = 0$.Rearranging, we get $\frac{dx}{dy} = \frac{-(x^2 - xy - y^2)}{y^2} = \frac{-x^2 + xy + y^2}{y^2} = -(\frac{x}{y})^2 + (\frac{x}{y}) + 1$.Let $x = vy$, then $\frac{dx}{dy} = v + y \frac{dv}{dy}$.Substituting into the equation: $v + y \frac{dv}{dy} = -v^2 + v + 1$.$y \frac{dv}{dy} = 1 - v^2$.Separating variables: $\frac{dv}{1 - v^2} = \frac{dy}{y}$.Integrating both sides: $\frac{1}{2} \ln |\frac{1 + v}{1 - v}| = \ln |y| + C$.Substituting $v = \frac{x}{y}$: $\frac{1}{2} \ln |\frac{1 + x/y}{1 - x/y}| = \ln |y| + C \Rightarrow \frac{1}{2} \ln |\frac{y + x}{y - x}| = \ln |y| + C$.At $(2, 1)$: $\frac{1}{2} \ln |\frac{1 + 2}{1 - 2}| = \ln |1| + C \Rightarrow \frac{1}{2} \ln |3| = C$.Thus, $\frac{1}{2} \ln |\frac{y + x}{y - x}| = \ln |y| + \frac{1}{2} \ln 3$.Multiplying by $2$: $\ln |\frac{y + x}{y - x}| = 2 \ln |y| + \ln 3 = \ln |3y^2|$.Taking exponential: $\frac{y + x}{y - x} = 3y^2$ or $\frac{y + x}{x - y} = -3y^2$.Rearranging: $y + x = 3y^2(x - y) = 3(xy^2 - y^3)$.Comparing with $x + y = k(xy^2 - y^3)$, we get $k = 3$.

If the solution for the differential equation $y^2 dx + (x^2 - xy - y^2) dy = 0$ at $(2, 1)$ is $x + y = k(xy^2 - y^3)$, then $k =$

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