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(D) Given differential equation is $x y \frac{d y}{d x}=2 y^2-x^2$. Dividing by $x$,we get $\frac{d y}{d x}=\frac{2 y}{x}-\frac{x}{y}$. Rearranging th

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

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(D) Given differential equation is $x y \frac{d y}{d x}=2 y^2-x^2$. Dividing by $x$,we get $\frac{d y}{d x}=\frac{2 y}{x}-\frac{x}{y}$. Rearranging the terms: $y \frac{d y}{d x}-\frac{2 y^2}{x}=-x$ $\ldots$ $(i)$. Let $v=y^2$,then differentiating with respect to $x$,we get $\frac{d v}{d x}=2 y \frac{d y}{d x}$,which implies $y \frac{d y}{d x}=\frac{1}{2} \frac{d v}{d x}$. Substituting this into $(i)$: $\frac{1}{2} \frac{d v}{d x}-\frac{2 v}{x}=-x$. Multiplying by $2$: $\frac{d v}{d x}-\frac{4 v}{x}=-2 x$. This is a linear differential equation of the form $\frac{d v}{d x}+P v=Q$,where $P=-\frac{4}{x}$ and $Q=-2 x$. Integrating factor $IF=e^{\int P d x}=e^{\int-\frac{4}{x} d x}=e^{-4 \log x}=x^{-4}$. The solution is $v \cdot IF = \int Q \cdot IF d x + c$. $v \cdot x^{-4} = \int (-2 x) \cdot x^{-4} d x + c = -2 \int x^{-3} d x + c$. $v x^{-4} = -2 \left( \frac{x^{-2}}{-2} \right) + c = x^{-2} + c$. $v = x^2 + c x^4$. 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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