The radial displacement in a rotating disc is governed by the differential equation $$\,\,{{{d^2}u} \over {d{x^2}}} + {1 \over x}{{du} \over {dx}} - {u \over {{x^2}}} = 8x.\,\,\,$$ where $$u$$ is the displacement and $$x$$ is the radius. If $$u=0$$ at $$x=0$$ and $$u=2$$ at $$x=1,$$ calculate the displacement at $$\,x = {1 \over {2.}}$$

Solve for $$y$$ if $${{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + y = 0$$ with $$y(0)=1$$ and $${y^1}\left( 0 \right) = 2$$