Challenging +1.2 This is a structured multi-part question that guides students through a substitution method. Part (a) is a standard FP3 second-order DE with constant coefficients. Parts (b) and (c) involve algebraic manipulation to verify given results rather than discovering them independently. Part (d) simply requires back-substitution. While the topic is Further Maths, the question provides significant scaffolding and most steps are routine verification rather than problem-solving.
Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$
giving your answer in the form \(y = \mathrm { f } ( t )\).
Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$$
(5 marks)
Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$
(2 marks)
Hence write down the general solution of the differential equation
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$
(l mark)