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 & m+x \cdot \frac{dm}{dx}=\frac{m\left(5-3 m^2\right)}{2\left(1-m^2\right)} \\ & x \cdot \frac{dm}{dx}=\frac{\left(5-3 m^2\right) m-2 m\left(1-m^2\right)}{2\left(1-m^2\right)} \\ \Rightarrow & \frac{dx}{x}=\frac{2\left(m^2-1\right)}{m\left(m^2-3\right)} dm \\ \Rightarrow & \frac{dx}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{m^2-3}\right) dm \\ \Rightarrow & \int \frac{dx}{x}=\int \frac{\left(\frac{2}{3}\right)}{m} dm+\int \frac{2}{3}\left(\frac{2 m}{m^2-3}\right) dm \\ \Rightarrow & \ln |x|=\frac{2}{3} \ln |m|+\frac{2}{3} \ln \left|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 $(x=1, y=1):$ we get $c=-\frac{2}{3} \ln (2)$ $ \Rightarrow \ln |x|=\frac{2}{3} \ln \left|\frac{y}{x}\right|+\frac{2}{3} \ln \left|\left(\frac{y}{x}\right)^2-3\right|-\frac{2}{3} \ln (2 $ $ \Rightarrow\left(\frac{y}{x}\right)\left[\left(\frac{y}{x}\right)^2-3\right]=2 \cdot\left(x^{3 / 2}\right) $

Put $x=2$ to get $y(2)$ $ \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} $

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