8. (a) Show that the substitution \(v = y ^ { - 2 }\) transforms the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 6 x y = 3 x \mathrm { e } ^ { x ^ { 2 } } y ^ { 3 } \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} x } - 12 v x = - 6 x \mathrm { e } ^ { x ^ { 2 } } \quad x > 0$$
(b) Hence find the general solution of the differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
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