Find y(n), if y(n) satisfies the recurrence equation 6 y(n+2) - 5 y(n+1) + y(n) = 1, y(1) = 1, y(2) = 1?

Being a second order linear recurrence relation with constant coefficients, the homogeneous part has characteristic equation

6r^2 - 5r + 1 = 0

==> (3r - 1)(2r - 1) = 0

==> r = 1/3, 1/2.

So, y_h(n) = A(1/3)^n + B(2/3)^n.

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Particular solution.

Suppose that y_p(n) = C for some constant C (since the right side of the recurrence is constant). Substituting this into it yields

6C - C + C = 1 ==> C = 1/6.

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Hence, a general solution is y(n) = A(1/3)^n + B(1/2)^n + 1/6.

To find A and B, use the given initial conditions.

y(1) = 1 ==> A/3 + B/2 + 1/6 = 1 ==> 2A + 3B = 5

y(2) = 1 ==> A/9 + B/4 + 1/6 = 1 ==> 4A + 9B = 30.

Solving for A and B yields A = -15/2 and B = 20/3.

So, y(n) = (-15/2)(1/3)^n + (20/3)(1/2)^n + 1/6.

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I hope this helps!