rafaltatara919

(a) To compute the values of a1, a2, a3, a4, and a5, we use the formula derived for by: by = an+1 - an = n - 2 Starting with a = 0, we have: a1 = a + b1 = 0 + (1 - 2) = -1 a2 = a + b2 = 0 + (2 - 2) = 0 a3 = a + b3 = 0 + (3 - 2) = 1 a4 = a + b4 = 0 + (4 - 2) = 2 a5 = a + b5 = 0 + (5 - 2) = 3

Therefore, the values of {an} are: 0, -1, 0, 1, 2, 3.

(b) The points with coordinates (n, an) for n = 0, 1, 2, 3, 4, and 5 are: (0, 0), (1, -1), (2, 0), (3, 1), (4, 2), and (5, 3). Plotting these points in the Cartesian plane gives the following graph:

        |
   4    |      o
        |
   3    |             o
        |
   2    |                      o
        |
   1    |            o
        |
   0    | o
        |_________________________
            0   1   2   3   4   5

(c) To find a formula for an, we can use the formula derived for by: by = an+1 - an = n - 2 Rearranging this equation, we get: an+1 = an + (n - 2) Substituting a = 0 and expanding the recurrence relation, we get: a1 = -1 a2 = a1 + (2 - 2) = -1 a3 = a2 + (3 - 2) = 0 a4 = a3 + (4 - 2) = 1 a5 = a4 + (5 - 2) = 3

Therefore, we can see that an = n(n-1)/2 for n >= 2, and a1 = -1.

(d) To check if the formula derived in part (c) gives the correct values of a, we plug in n = 1, 2, 3, 4, and 5: a1 = -1, a2 = 0, a3 = 1, a4 = 2, a5 = 3. These values match the values found in part (a), so we can be confident that the formula is correct for all values of n.

If the Wronskian of two functions y₁ and y₂ is identically zero,show by direct integration that y₁=cy₂ , that is, that y₁ and y₂ are linearly dependent. Assume the functions have continuous derivatives and that at least one of the functions does not vanish in the interval under consideration.