A-LevelFurther MathematicsDifferential equations9231/12 · May/June 2014 · Q46 Marks

Obtain the general solution of the differential equation d²x/dt² – 6dx/dt + 25x = 195 sin 2t.

Obtain the general solution of the differential equation d²x/dt² – 6dx/dt + 25x = 195 sin 2t.[6]

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

[M1] m² – 6m + 25 = 0 ⇒ m = 3 + 4i
[A1] CF: x = e^(3t) (A cos 4t + B sin 4t)
[M1] PI: x = p sin 2t + q cos 2t ⇒ dx/dt = 2p cos 2t – 2q sin 2t ⇒ d²x/dt² = −4p sin 2t – 4q cos 2t
[M1] ⇒ -4p sin 2t – 4q cos 2t – 6(2p cos 2t – 2q sin 2t) + 25(p sin 2t + q cos 2t) = 195 sin 2t (or implies equations)
[A1] ⇒ (21p + 12q)sin 2t + (21q - 12p)cos 2t = 195 sin 2t. Compare coefficients: 21p + 12q = 195, -12p + 21q = 0 ⇒ p=7, q=4
[A1] GS: x = e^(3t) (A cos 4t + B sin 4t) + 7 sin 2t + 4 cos 2t

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Topic: Differential equations, A-Level Further Mathematics (syllabus 9231). Structured question, 6 marks.
Source: A-Level Further Mathematics, May/June 2014 Paper 1 Variant 2, question 4.

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