Question

Question: find a general solution of homogeneous linear difference equation of the 2nd order with constant coefficients. How to solve it? Thank you. (pictures below)

task:

this should be solution:

Thank you!

Answer 1

`a_n = 4^n(An+b)`

Explanation:

We seek a solution to the Linear difference equation:

`a_(n+2) -8_(n+1) + 16a_n = 0` ..... [A]

This is a Second Order Homogeneous linear difference equations, and is solved in a similar way to a second order linear differential equations with constant coefficients, by forming the Auxiliary Equation, which is the polynomial equation with the coefficients of the difference terms:

So, the associated Auxiliary equation is:

`m^2-8m+16 = 0`

`:. (m-4)^2 = 0`

And so we have the solutions:

`m = +-4` (repeated real root)

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

• Real distinct roots `m=alpha,beta, ...` will yield independent solutions of the form `a_alpha=Aalpha^n`, `a_beta=Bbeta^n`, ...
• Real repeated roots `m=alpha`, will yield a solution of the form `a_n=(An+B)alpha^n` where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) `m=R(costheta+isintheta)` will yield solutions of the form `a_n=R^n(Acost theta n+Bsin theta n)`

Thus the General Solution of the homogeneous equation [A] is:

`a_n = 4^n(An+b) \ \ \ \` QED