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given polynomials (t)2 – 3t+2 (t – 1) (t – 2) t= 1,2 independent solutions t = 3,4,6…..; are dependent solutions

(t)2 – 5t+6 (t – 2)(t – 3) t= 2,3 independent solutions t = 4,6…..; are dependent solutions

(t)2 – 7t+12 (t – 3)(t – 4) t= 3,4 independent solutions t = 6,8,…..; are dependent solutions

(t)2 – 10t+24 (t – 4)(t – 6) t= 4,6 independent solutions t = 8,12,…..; are dependent solutions

Therefore the above solutions are linear independent solutions for R(t).

(t)2 – 5t+6 (t – 2)(t – 3) t= 2,3 independent solutions t = 4,6…..; are dependent solutions

(t)2 – 7t+12 (t – 3)(t – 4) t= 3,4 independent solutions t = 6,8,…..; are dependent solutions

(t)2 – 10t+24 (t – 4)(t – 6) t= 4,6 independent solutions t = 8,12,…..; are dependent solutions

Therefore the above solutions are linear independent solutions for R(t).