Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation
$$\text { into } \quad \begin{aligned}
x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x )
\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t
\end{aligned}$$
(7 marks)
Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$
(11 marks)
Write down the general solution of the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$