1. §7.8 # 12. We look for solutions of the form hn
= qn. We find q by finding roots of the
characteristic equation x 2 − 4. The roots are x = 2 and x = −2 so we
are trying to
find c1 and c2 such that hn = c1(2)n
+c2(−2)n satisfies the initial conditions. Plugging
in n = 0 and n = 1 gives us the two equations c1 + c2 = 0
and 2c1 − 2c2 = 1. We can
just plug c1 = −c2 into the second equation to get c2
= −1/4 and so c1 = 1/4 which
gives us
2. §7.8 # 14. Again, looking for solutions of the form hn
= qn is equivalent to finding roots
of the characteristic equation x3 − x2 − 9x + 9. To find
the roots we use factoring by
grouping : x3−x2−9x+9 = x2(x−1)−9(x−1) = (x−1)(x2−9)
= (x−1)(x−3)(x+3) so
the roots are x = 1, x = 3, x = −3. The general solution will be c1(1)n+c2(3)n+c3(−3)n
and so to satisfy the initial conditions we are trying to solve the following
equations:
c1+c2+c3 = 0, c1+3c2−3c3=
1, and c1+9c2+9c3 = 2. This is equivalent to
putting
the first matrix in row eschelon form (described below):
To go from the first to second matrix we take (−1) times
first row and add it to each
of second and third row. To go from second to third matrix we take (−4) times
second
row and add it to third row and also divide second row by 2.
From the third row of the last matrix we know 
The second row then
tells us 
and the first row tells us that c1
= −c2 − c3 =
So the solution is
3. §7.8 # 17. We are looking for roots of the polynomial x 4−5x3+6x2+4x−8.
We check
small numbers. x = −1 and x = 2 both are roots. You can take either x+1 or x−2
(or
even their product ) and perform long division by the characteristic polynomial
which
will give you (x + 1)(x − 2)(x2 − 4x − 4) = (x + 1)(x − 2)3.
So we are in the case of
repeated roots as in Theorem 7.2.2. q1 is −1 and s1 = 1
but q2 = 2 and s2 = 3 so the
general solution is
Considering the initial conditions, we are trying to solve
the fol lowing equations c1 +
c2 = 0, −c1 +2c2 +2c3 +2c4
= 1, c1 +4c2 +8c3 +16c4 = 1 and −c1
+8c2 +24c3 +72c4.
The first equation lets us replace c1 with −c2 in the
other three equations and so we
are doing row reduction on the following matrix:
Second row of second matrix is (−1) times first row plus
second row of first matrix,
third row of second matrix is (−3) times first row plus third row of first
matrix.
Third row of matrix is (−3) times second row plus third row of second matrix. So
c4 = −1/24, and 6c3 = 14/24 so c3 = 7/72 and 3c2
= −2c3 − 2c4 + 1 = 8/27 which
means c1 = −c2 = −8/27. The solution to this recurrence
relation with the specified
initial conditions is then
5. §7.8 # 29 (b) This is equal to the number of solutions
to e1+e2+e3+e4 = n where each
ei occurs a multiple of three times. This is (1 +x3 +x6
+x9 +· · · )4 (the fourth power
counts each of the 4 ei’s). But since 
we get
(d). Again, the number of n- combinations of S with these conditions is
equivalent to
the number of non- negative integral solutions with the conditions. So the
generating
function for e1 occuring 1, 3, or 11 times is (x+x3+x11)
while the generating function
for e2 to occur 2, 4, 5 times is (x2 + x4 + x5).
Since x3 and x4 have no restrictions we
have (1+x+x2 +· · · ) as their generating functions. Using the
multiplicitve principle
this gives:
6. §7.8 # 34. Various answers but you need at most 2 of
the first object, an even number
and at most 6 of the second object, an even number of the third object, and and
least
1 of the last object.
7. §7.8 # 36. We change variables so f1 = 2e1
and f2 = 5e2, f3 = e3 and f4
= 7e4. Then
we are looking for the generating function for solutions to f1 + f2
+ f3 + f4 = n in
non-negative integers where f1 is even, f2 is a multiple
of
5, and f4 is a multiple of 7.
This will be
