7. (a) Given that \(x = e ^ { t }\), show that
- $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \frac { \mathrm {~d} y } { \mathrm {~d} t }$$
- $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$
(b) Use you answers to part (a) to show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = \mathrm { e } ^ { 3 t }$$
(c) Hence find the general solution of
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$