Find x in terms of t given that d²x/dt² + 4dx/dt + x = 6e⁻²ᵗ, and that, when t = 0, x = 5/3 and dx/dt = 7/6. State lim_{t→∞} x.

Mark scheme

Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.

1. M1 Finds complementary function. 4m² + 4m + 1 = 0 ⇒ (2m + 1)² = 0 ⇒ m = -1/2
2. A1 C.F.: x = Ae^(-t/2) + Bte^(-t/2)
3. M1 Finds particular integral. P.I.: x = ke^(-2t) ⇒ ẋ = -2ke^(-2t) ⇒ ẍ = 4ke^(-2t)
4. A1 16k - 8k + k = 6 ⇒ k = 6/9 = 2/3 ⇒ x = (2/3)e^(-2t)
5. A1 Adds for general solution. G.S.: x = Ae^(-t/2) + Bte^(-t/2) + (2/3)e^(-2t)
6. B1 Uses initial conditions to find constants. x(0) = 5/3 = A + 2/3 ⇒ A = 1
7. M1 ẋ = -1/2 Ae^(-t/2) + B(e^(-t/2) - 1/2 te^(-t/2)) - 4/3 e^(-2t)
8. A1 ẋ(0) = 7/6 = -1/2 A + B - 4/3 ⇒ B = 3
9. A1 Gives particular solution. x = e^(-t/2) + 3te^(-t/2) + (2/3)e^(-2t)
10. B1 States limit. lim x = 0