NPTEL Question

Question

Find the particular solution of the given Recurrence relation or Difference equation.

a_r - 5a_r-1 + 6a_r-2 = 2^r + r

Answer 1

Characteristic equation: r² - 5r + 6 = 0,

Solving the equation gives r = 2, 3.

f(r) = 2^r+ r,

Particular solution, a^(p) = A₂(r)(2^r) + A₁r + A₀

Putting this trial solution into the recurrence equation gives

A₂r2^r + A₁r + A₀ - 5[A₂(r-1)2^r-1 + A₁(r-1) + A₀] + 6[A₂(r-2)2^r-2 + A₁(r-2) + A₀] = 2^r + r

or, (2^r-2)(A₂r2² - 5A₂r.2 + 5A₂.2 + 6A₂.r - 12A₂) + r(A₁ - 5A₁ + 6A₁) + (A₀ +5A₁ - 5A₀ - 12A₁ +6A₀) = 2^r-2(2²) + r(1) + (0)

Matching the coefficients give,

A₂r2² - 5A₂r.2 + 5A₂.2 + 6A₂.r - 12A₂ = 4   => A₂ = -2

A₁ - 5A₁ + 6A₁ = 1  => A₁ = 1/2

A₀ +5A₁ - 5A₀ - 12A₁ +6A₀ = 0   => A₀ = 7/4

So the particular solution is

a^(p) = -2(r)(2^r) + (1/2)r + 7/4