- (a) Show that the transformation \(x = \mathrm { e } ^ { t }\) transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { 2 t }$$
(b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(t\).
(c) Hence find the general solution of the differential equation (I).