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The constant C is determined from the initial condition:

( 1− )
y()0 = 0 = C( 1− ) +
0
4
giving for the constant C the value 1/4.

In-Class Exercises
Pb. 2.9 Use the following program to model Example 2.3:

N=19;
y(1)=0;
for k=1:N
y(k+1)=k-y(k);
end
y

Verify the closed-form answer.
Pb. 2.10 Find, for k ≥ 2, the general solution of the second-order difference
equation:

y(k) – 3y(k – 1) – 4y(k – 2) = 4 + 2 × 4 k–1
k
with the initial conditions y(0) = 1 and y(1) = 9. (Hint: When the functional
form of the homogeneous and particular solutions are the same, use the same
functional form for the solutions as in the case of multiple roots for the char-
acteristic polynomial.)

Answer: yk() =− 1 (−1 ) + 26 ( ) 4 k  6 k4 k  
k

 
 5
25
 25

Homework Problems
Pb. 2.11 Given the general geometric series y(k), where:

y(k) = 1 + a + a + … + a k
2
show that y(k) obeys the first-order equation:

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