Let $y=y\left(x\right)$ be the solution of the differential equation  $\left(3{\mathrm{y}}^2-5{\mathrm{x}}^2\right)\mathrm{ydx}+2\mathrm{x}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)\mathrm{dy}=0$ such that $\mathrm{y}\left(1\right)=1$. Then $\left|\mathrm{y}(2{)}^3-12\mathrm{y}(2)\right|$is equal to :

$\left(3{\mathrm{y}}^2-5{\mathrm{x}}^2\right)\mathrm{ydx}+2\mathrm{x}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)\mathrm{dy}=0,\mathrm{y}\left(1\right)=1$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}\left(3{\mathrm{y}}^2-5{\mathrm{x}}^2\right)}{2\mathrm{x}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}$
put y = Vx  & simplify

$\Rightarrow V+x\cdot \frac{dV}{dx}=\frac{V\left(5-3V^2\right)}{2\left(1-V^2\right)}$

$x\cdot \frac{dV}{dx}=\frac{\left(5-3V^2\right){\displaystyle V}-2V\left(1-V^2\right)}{2\left(1-V^2\right)}$
$\frac{2\left(1-{\mathrm{V}}^2\right)}{\mathrm{V}\left(3-{\mathrm{V}}^2\right)}\mathrm{dV}=\frac{1}{\mathrm{x}}\mathrm{dx}$
Integrate both sides
$\log \mathrm{x}+\log \mathrm{c}=\int \frac{2\mathrm{V}\left(1-{\mathrm{V}}^2\right)}{{\mathrm{V}}^2\left(3-{\mathrm{V}}^2\right)}\mathrm{dV}$
Put ${\mathrm{V}}^2=\mathrm{t}$
$\begin{array}{l}\Rightarrow \int \frac{1-\mathrm{t}}{\mathrm{t}(3-\mathrm{t})}\mathrm{dt}=\frac{1}{3}\int \left(\frac{1}{\mathrm{t}}+\frac{2}{\mathrm{t}-3}\right)\mathrm{dt}\\ \Rightarrow {\mathrm{C}}^3{\mathrm{x}}^3=\mathrm{t}(\mathrm{t}-3{)}^2\text{ when }\mathrm{x}=1,\mathrm{y}=1\Rightarrow \mathrm{t}={\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2=1\\ {\mathrm{C}}^3\cdot 1=4\Rightarrow {\mathrm{C}}^3=4,4{\mathrm{x}}^3=\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}{\left(\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}-3\right)}^2\Rightarrow 2\mathrm{x}\sqrt{\mathrm{x}}=\frac{\mathrm{y}}{\mathrm{x}}\left(\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}-3\right)\end{array}$
Put $\mathrm{x}=2\Rightarrow 4\sqrt{2}=\frac{\mathrm{y}}{2}\left(\frac{{\mathrm{y}}^2-12}{4}\right)\Rightarrow 32\sqrt{2}={\mathrm{y}}^3-12\mathrm{y}\text{ at }\mathrm{x}=2$