Answer: (1)

Formula: Homogeneous equations

$ \begin{aligned} & \left(3 y^2-5 x^2\right) y \cdot d x+2 x\left(x^2-y^2\right) d y=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^2-3 y^2\right)}{2 x\left(x^2-y^2\right)} \end{aligned} $

Put $y=m x$ $ \begin{aligned} \Rightarrow & \mathrm{m}+\mathrm{x} \cdot \frac{\mathrm{dm}}{\mathrm{dx}}=\frac{\mathrm{m}\left(5-3 \mathrm{~m}^2\right)}{2\left(1-\mathrm{m}^2\right)} \\ & \mathrm{x} \cdot \frac{\mathrm{dm}}{\mathrm{dx}}=\frac{\left(5-3 \mathrm{~m}^2\right) \mathrm{m}-2 \mathrm{~m}\left(1-\mathrm{m}^2\right)}{2\left(1-\mathrm{m}^2\right)} \\ \Rightarrow & \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^2-1\right)}{\mathrm{m}\left(\mathrm{m}^2-3\right)} \mathrm{dm} \\ \Rightarrow & \frac{\mathrm{dx}}{\mathrm{x}}=\left(\frac{2}{\mathrm{~m}}-\frac{\frac{4}{3}}{\mathrm{~m}}+\frac{\frac{4 \mathrm{~m}}{3}}{\mathrm{~m}^2-3}\right) \mathrm{dm} \\ \Rightarrow & \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\left(\frac{2}{3}\right)}{\mathrm{m}} \mathrm{dm}+\int \frac{2}{3}\left(\frac{2 \mathrm{~m}}{\mathrm{~m}^2-3}\right) \mathrm{dm} \\ \Rightarrow & \ln |\mathrm{x}|=\frac{2}{3} \ln |\mathrm{m}|+\frac{2}{3} \ln \left|\mathrm{m}^2-3\right|+C \end{aligned} $

Or, $\ln |x|=\frac{2}{3} \ln \left|\frac{y}{x}\right|+\frac{2}{3} \ln \left|\left(\frac{y}{x}\right)^2-3\right|+C$ Put $(\mathrm{x}=1, \mathrm{y}=1):$ we get $\mathrm{c}=-\frac{2}{3} \ln (2)$ $ \Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^2-3\right|-\frac{2}{3} \ln (2 $ $ \Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^2-3\right]=2 \cdot\left(\mathrm{x}^{3 / 2}\right) $

Put $\mathrm{x}=2$ to get $\mathrm{y}(2)$ $ \begin{aligned} & \Rightarrow y\left(y^2-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2} \\ & \Rightarrow y^3-12 y=32 \sqrt{2} \end{aligned} $

$\therefore |y^3-12 y|=32 \sqrt{2}$