10 It is given that \(x = t ^ { \frac { 1 } { 2 } }\), where \(x > 0\) and \(t > 0\), and \(y\) is a function of \(x\).
- Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
- Hence show that the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$
reduces to the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
- Find the general solution of ( \(*\) ), giving \(y\) in terms of \(x\).