Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
Find the general solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10\).
Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 8\) when \(x = 1\).