Question

The solution of difference equation u_n+1 - 2u_ncos θ + u_n-1= 0 will be __________, where c₁ and c₂ are constants.
1. u_n = (c₁ + c₂n) sin n θ cos n θ 
2. u_n = (c_q + c₂n) cos n θ + sin n θ 
3. u_n = (-1)ⁿ [c₁ cos(n - 1)θ + c₂(n - 1)θ]
4. u_n = c₁ cos n θ + c₂ sin n θ 

Answer 1

Correct Answer - Option 4 : u_n = c₁ cos n θ + c₂ sin n θ 

Concept:

A general linear difference equation of order ‘r’ with constant coefficients is given by:

(a₀E^r + a₁E^r-1 + a₂E^r-2 +…+ a_r-1E + a_r) u_n = F(n);

Where a₀, a₁, a₂,…,a_n are constants and F(n) is a function of ‘n’ alone or constant.

In other ways, it is written as

f(E) u_n = F(n);

Complete solution of the above equation is given by u = CF + PI;

For finding a complementary function, we have to solve the auxiliary equation f(E);

(i) If n roots are real and distinct say m₁, m₂, m₃, … , m_n;

CF = c₁ m₁ⁿ + c₂ m₂ⁿ + … + c_n m_nⁿ

(ii) If two or more roots are equal i.e. m₁ = m₂ = m₃ = … = m_k;

CF = (c₁ + c₂n + c₃n² + … + c_k n^k-1) m₁ⁿ + … + c_n m_nⁿ

(iii) If a pair of imaginary roots i.e. m₁ = α + i β, m₂ = α – i β;  

CF = rⁿ (c₁ cos nθ + c₂ sin nθ) + c₃ m₃ⁿ + … + c_n m_nⁿ; 

Where \(r = \sqrt {{\alpha^2}+{\beta^2}} \; and \; θ = tan^{-1} {\frac {\beta}{\alpha}} \)

Calculation:

Given Difference equation is u_n+1 – 2u_ncos θ + u_n-1 = 0;

Comparing with f(E) u_n-1 = F(n);

⇒ (E² – 2Ecos θ + 1) u_n-1 = 0;

Since F(n) = 0, the complimentary function will be general solution;

The auxiliary equation is E² – 2Ecos θ + 1 = 0

The roots will be

\(E = {2cos θ ± \sqrt{4cos^2θ-4} \over 2}\)

⇒ E = cos θ ± i sin θ;

Comparing with α ± β ⇒ α = cos θ, β = sin θ;  

Now the CF will be

CF = rⁿ (c₁ cos nθ + c₂ sin nθ)

Where 

\(r = \sqrt {{\alpha^2}+{\beta^2}} \; = \sqrt {{cos^2 \theta} + {sin^2 \theta}} = 1\)

\(θ = tan^{-1} {\frac {\beta}{\alpha}} = tan^{-1} ({\frac {sin \theta}{cos \theta}}) = \theta\)

⇒ CF = c₁ cos nθ + c₂ sin nθ 

∴ general solution will be c1 cos nθ + c2 sin nθ