11 It is given that \(x \neq 0\) and $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 x y = 8 x ^ { 2 } + 16$$ Show that if \(z = x y\) then $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 z = 8 x ^ { 2 } + 16$$ Find \(y\) in terms of \(x\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2\) when \(x = \frac { 1 } { 2 } \pi\).