$\text{ If }{x}^3\frac{dy}{dx}=y^3+y^2\sqrt{y^2-x^2}\text{ and }y=1\text{ when }x=1\text{ then }y=2\text{ when }x\text{ is }$

$\frac{dy}{dx}=\frac{y^3}{x^3}+\frac{y^2}{x^3}x\sqrt{\frac{y^2}{x^2}-1}$

$\begin{array}{l}\text{ Put }\mathrm{y}=\mathrm{Vx}\to \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \to \mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}={\mathrm{v}}^3+{\mathrm{v}}^2\sqrt{{\mathrm{v}}^2-1}\\ x\frac{dv}{dx}=v^3-v+{\mathrm{v}}^2\sqrt{{\mathrm{v}}^2-1}\\ \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}\left({\mathrm{v}}^2-1\right)+{\mathrm{v}}^2\sqrt{{\mathrm{v}}^2-1}\\ x\frac{dv}{dx}=\sqrt{{\mathrm{v}}^2-1}\left(v\sqrt{{\mathrm{v}}^2-1}+v^2\right)\\ \frac{dv}{\sqrt{{\mathrm{v}}^2-1}\left(v\sqrt{{\mathrm{v}}^2-1}+v^2\right)}=\frac{dx}{x}\\ \to \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\mathrm{dv}}{\mathrm{v}\sqrt{{\mathrm{v}}^2-1} \left(\mathrm{v}+\sqrt{{\mathrm{v}}^2-1}\right)}\times \frac{\left(\mathrm{v}-\sqrt{{\mathrm{v}}^2-1}\right)}{\left(\mathrm{v}-\sqrt{{\mathrm{v}}^2-1}\right)}\\ \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\left(\mathrm{v}-\sqrt{{\mathrm{v}}^2-1}\right)}{\mathrm{v}\sqrt{{\mathrm{v}}^2-1}}dv\\ \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{1}{\sqrt{{\mathrm{v}}^2-1}}dv-\int \frac{1}{\mathrm{v}}\mathrm{dv}\\ \log x+\log c=\log (v+\sqrt{{\mathrm{v}}^2-1})-\log v\\ \Rightarrow cx=\frac{v+\sqrt{v^2-1}}{v}=\frac{y+\sqrt{y^2-x^2}}{y}\\ \text{ Put }x=y=1\Rightarrow c=1\\ \Rightarrow put \mathrm{y}=2 \Rightarrow x=\frac{2+\sqrt{4-x^2}}{2}\\ 2x=2+\sqrt{4-x^2} \Rightarrow {\left(2x-2\right)}^2=4-x^2\\ \Rightarrow (5x-8)x=0\to x=\frac{8}{5}\end{array}$