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