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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$x y \frac{d y}{d x}=2 y^2-x^2$ $\frac{d y}{d x}=\frac{2 y}{x}-\frac{x}{y}$ $\frac{d y}{d x}-\frac{2 y}{x}=\frac{-x}{y} \Rightarrow y \frac{d y}{d x} \frac{-2 y^2}{x}=-x$ $\ldots$ (i) Put, $\quad v=y^2$ $\ldots$ (ii) $\Rightarrow\left(\frac{d v}{d x}=2 y \frac{d y}{d x}\right) \Rightarrow\left(\frac{1}{2} \frac{d v}{d x}=y \frac{d y}{d x}\right)$ From Eq. (i) $\frac{1}{2} \frac{d v}{d x}-\frac{2 v}{x}=-x$ $\frac{d v}{d x}-\frac{4 v}{x}=-2 x$ $\mathrm{IF}=e^{\int p d x}=e^{\int-\frac{4}{x} d x}=e^{-4 \log x}$ $=e^{\log x^{-4}}=x^{-4}$ Complete solution is $x^{-4} \cdot v=\int(-2 x) \cdot x^{-4} d x+c$ $\frac{v}{x^4}=-2 \int \frac{d x}{x^3}+c \Rightarrow \frac{v}{x^4}=\frac{1}{x^2}+c$ $v=x^2+c x^4$ $y^2=c x^4+x^2 \quad[$ from Eq. (ii) $]$ which is the required family of curves.

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…

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