I think because a full proof covering existence and uniqueness will either be really long or require tools from outside the scope of the text. E.g. there's a somewhat concise proof using linear algebra which I'll partially reproduce below. (I like this proof because the equation is derived from first principles rather than starting with an ansatz.)

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Let x_n be a sequence defined by the recurrence relation:

    x_{n+1} = a * x_{n-1} + b * x_n

Observe that if we define a sequence of two-element vectors of successive elements:

    [x_0]  [x_1]  [x_2]
    [x_1], [x_2], [x_3], ...

then we can form the relation in terms of matrix/vector multiplication:

    [x_1] = [[0  1]] [x_0]
    [x_2]   [[a  b]] [x_1]

Let's name the sequence of vectors as y_n and call the matrix M:

    y_1 = M * y_0

We can get the next term in the sequence with another multiplication:

    y_2 = M * y_1
        = M * (M * y_0)
        = M^2 * y_0

By induction we have:

    y_n = M^n * y_0

M has characteristic polynomial:

    r^2 - br - a = 0

with roots:

    r_1 = (b - c)/2
    r_2 = (b + c)/2
    c   = √(b^2 + 4a)

Therefore we have by diagonalization:

    y_n = S * [[r_1^n  0    ]] * S^(-1) * y_0
              [[0      r_2^n]]

where S is the matrix of eigenvectors. From here, we can finish our existence and uniqueness proofs from the existence and uniqueness of the eigenvalues of M.