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(D) Given the differential equation: $x y \frac{d y}{d x}=2 y^2-x^2$. Divide by $x$: $y \frac{d y}{d x} = \frac{2 y^2}{x} - x$. Rearranging: $y \frac{

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$y^2=x^2+c x^4$

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(D) Given the differential equation: $x y \frac{d y}{d x}=2 y^2-x^2$. Divide by $x$: $y \frac{d y}{d x} = \frac{2 y^2}{x} - x$. Rearranging: $y \frac{d y}{d x} - \frac{2 y^2}{x} = -x$ $\ldots$ $(i)$. Let $v = y^2$. Then $\frac{d v}{d x} = 2 y \frac{d y}{d x}$, which implies $y \frac{d y}{d x} = \frac{1}{2} \frac{d v}{d x}$. Substituting into $(i)$: $\frac{1}{2} \frac{d v}{d x} - \frac{2 v}{x} = -x$. Multiply by $2$: $\frac{d v}{d x} - \frac{4 v}{x} = -2 x$. This is a linear differential equation of the form $\frac{d v}{d x} + P(x) v = Q(x)$, where $P(x) = -\frac{4}{x}$ and $Q(x) = -2 x$. Integrating factor $IF = e^{\int P(x) d x} = e^{\int -\frac{4}{x} d x} = e^{-4 \ln |x|} = x^{-4}$. The solution is $v \cdot IF = \int Q(x) \cdot IF d x + c$. $v \cdot x^{-4} = \int (-2 x) \cdot x^{-4} d x + c = \int -2 x^{-3} d x + c$. $\frac{v}{x^4} = -2 \left( \frac{x^{-2}}{-2} \right) + c = \frac{1}{x^2} + c$. $v = x^2 + c x^4$. Since $v = y^2$, the family of curves is $y^2 = x^2 + c x^4$.

$A$ family of curves has the differential equation $x y \frac{d y}{d x}=2 y^2-x^2$. Then, the family of curves is

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