A2 October 2020 Q2
2. Solve the recurrence system
\[u_1 = 1 \qquad u_2 = 4\]\[9u_{n+2} - 12u_{n+1} + 4u_n = 3n\](9)
| Scheme | Marks | AO |
|---|---|---|
| Auxiliary equation is \(9r^2 - 12r + 4 = 0\), so \(r = \ldots\) | M1 | 1.1b |
| \((3r - 2)^2 = 0 \Rightarrow r = \dfrac{2}{3}\) is repeated root. | A1 | 1.1b |
| Complementary function is \(x_n = (A + Bn)\left(\frac{2}{3}\right)^n\) or \(A\left(\frac{2}{3}\right)^n + Bn\left(\frac{2}{3}\right)^n\) | M1 | 2.2a |
| Try particular solution \(y_n = an + b \Rightarrow 9(a(n+2) + b) - 12(a(n+1) + b) + 4(an + b) = 3n\) | M1 | 2.1 |
| \(\Rightarrow an + 6a + b = 3n \Rightarrow a = \ldots, b = \ldots\) | dM1 | 1.1b |
| \(a = 3, b = -18\) | A1 | 1.1b |
| General solution is \(u_n = x_n + y_n = (A + Bn)\left(\frac{2}{3}\right)^n + 3n - 18\) | B1ft | 2.2a |
| \(\left.\begin{aligned} u_1 = 1 &\Rightarrow 1 = \left(\frac{2}{3}\right)(A + B) - 15 \\ u_2 = 4 &\Rightarrow 4 = \left(\frac{4}{9}\right)(A + 2B) - 12 \end{aligned}\right\} A = \ldots, B = \ldots\) | M1 | 2.1 |
| \(u_n = 12(n + 1)\left(\frac{2}{3}\right)^n + 3n - 18\) oe | A1 | 1.1b |
| (9) | ||
| (9 marks) |
Notes
M1: Forms and solves the auxiliary equation.
A1: Correct (repeated) root found.
M1: Forms the correct complementary function for their (real) root(s) to the equation, \((A + Bn)r^n\) if repeated root, or allow \(Ar_1^{\,n} + Br_2^{\,n}\) if distinct real roots are found.
M1: Attempts to use a particular solution of the correct form (ie \(an + b\) or a higher order polynomial in \(n\) containing this) in the recurrence relation.
dM1: Expands and solves for \(a\) and \(b\)
A1: Correct values for \(a\) and \(b\)
B1ft: Forms the general solution as the sum of their complementary function and a particular solution of correct form with their \(a\) and \(b\)
M1: Applies the initial values and solves for the constants
A1: Correct answer.