- (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y + 4 x ^ { 2 } y ^ { 3 } \ln x = 0 \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { 2 z } { x } = 8 x \ln x \quad x > 0$$
(b) By solving differential equation (II), determine the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\)