Let the solution curve of the differential equation $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=\sqrt{{\mathrm{y}}^2+16{\mathrm{x}}^2},\mathrm{y}\left(1\right)=3$ be $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$. Then $y\left(2\right)$ is equal to:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}+\sqrt{{\mathrm{y}}^2+16{\mathrm{x}}^2}}{\mathrm{x}}$  Let y = vx
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}+\sqrt{{\mathrm{v}}^2{\mathrm{x}}^2+16{\mathrm{x}}^2}}{\mathrm{x}}=\mathrm{v}+\sqrt{{\mathrm{v}}^2+16}\\ \int \frac{\mathrm{dv}}{\sqrt{{\mathrm{v}}^2+16}}=\int \frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{\ell n}\left|\mathrm{v}+\sqrt{{\mathrm{v}}^2+16}\right|=\ln \mathrm{x}+\mathrm{\ell nc}\end{array}$
$\begin{array}{r}\frac{\mathrm{y}}{\mathrm{x}}+\frac{\sqrt{{\mathrm{y}}^2+16{\mathrm{x}}^2}}{\mathrm{x}}=\mathrm{cx}\\ \mathrm{y}+\sqrt{{\mathrm{y}}^2+16{\mathrm{x}}^2}={\mathrm{cx}}^2\end{array}$
y(1) = 3$\Rightarrow$ c = 8 and
x = 2 $\Rightarrow \mathrm{y}+\sqrt{{\mathrm{y}}^2+64}=32$
$\begin{array}{l}\therefore {\mathrm{y}}^2+64=(32-\mathrm{y}{)}^2\\ 64(1+\mathrm{y})=32\times 32\Rightarrow 1+\mathrm{y}=16\\ \therefore \mathrm{y}=15\end{array}$