33. Find the unique solution to each recurrence relation with the given initi. (a) $$ a_{n}=10 a_{n-1}-32 a_{n-2}+32 a_{n-3} $$ with $$ a_{0}=5, a_{1}=18, a_{2}=76 $$

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1. **Find the characteristic polynomial.**  
   Assume a solution of the form $$ a_n = r^n $$. Then:
   $$
   r^n = 10r^{n-1} - 32r^{n-2} + 32r^{n-3}
   $$
   Divide by $$ r^{n-3} $$ (with $$ r \neq 0 $$):
   $$
   r^3 = 10r^2 - 32r + 32
   $$
   The characteristic equation is:
   $$
   r^3 - 10r^2 + 32r - 32 = 0
   $$

2. **Factor the characteristic polynomial.**  
   Test $$ r = 2 $$:
   $$
   2^3 - 10\cdot2^2 + 32\cdot2 - 32 = 8 - 40 + 64 - 32 = 0
   $$
   Since $$ r = 2 $$ is a root, factor $$ r-2 $$ from the cubic. Dividing gives:
   $$
   r^3 - 10r^2 + 32r - 32 = (r-2)(r^2 - 8r + 16)
   $$
   Notice that:
   $$
   r^2 - 8r + 16 = (r-4)^2
   $$
   So the full factorization is:
   $$
   (r-2)(r-4)^2=0
   $$

3. **Write the general solution.**  
   The roots are $$ r=2 $$ (simple) and $$ r=4 $$ (double). Therefore, the general solution is:
   $$
   a_n = A\cdot2^n + (B+Cn) \cdot 4^n
   $$
   where $$ A $$, $$ B $$, and $$ C $$ are constants.

4. **Determine the constants using the initial conditions.**  
   The given initial conditions are:
   $$
   a_0 = 5,\quad a_1 = 18,\quad a_2 = 76.
   $$
   
   - For $$ n=0 $$:
     $$
     a_0 = A\cdot 2^0 + (B+C\cdot0) \cdot 4^0 = A + B = 5.
     $$
     
   - For $$ n=1 $$:
     $$
     a_1 = A\cdot 2^1 + (B+C\cdot1) \cdot 4^1 = 2A + 4(B+C) = 18.
     $$
     Divide by 2:
     $$
     A + 2(B+C) = 9.
     $$
     
   - For $$ n=2 $$:
     $$
     a_2 = A\cdot 2^2 + (B+2C) \cdot 4^2 = 4A + 16(B+2C) = 76.
     $$
     Divide by 4:
     $$
     A + 4(B+2C) = 19.
     $$

5. **Solve the system of equations.**  
   We have:
   $$
   \begin{cases}
   A + B = 5 & \text{(i)} \
   A + 2B + 2C = 9 & \text{(ii)} \
   A + 4B + 8C = 19 & \text{(iii)}
   \end{cases}
   $$
   Subtract equation (i) from (ii):
   $$
   (A+2B+2C) - (A+B) = 9 - 5 \quad \Rightarrow \quad B+2C = 4. \quad \text{(iv)}
   $$
   Subtract equation (ii) from (iii):
   $$
   (A+4B+8C) - (A+2B+2C) = 19 - 9 \quad \Rightarrow \quad 2B+6C = 10.
   $$
   Divide by 2:
   $$
   B+3C = 5. \quad \text{(v)}
   $$
   Subtract (iv) from (v):
   $$
   (B+3C) - (B+2C) = 5 - 4 \quad \Rightarrow \quad C = 1.
   $$
   Substitute $$ C=1 $$ into (iv):
   $$
   B + 2(1) = 4 \quad \Rightarrow \quad B = 2.
   $$
   Finally, substitute $$ B=2 $$ into (i):
   $$
   A + 2 = 5 \quad \Rightarrow \quad A = 3.
   $$

6. **Write the unique solution.**  
   The unique solution is:
   $$
   a_n = 3\cdot2^n + (2+n)\cdot4^n.
   $$
The unique solution to the recurrence relation is: $$ a_n = 3 \cdot 2^n + (2 + n) \cdot 4^n $$

33. Find the unique solution to each recurrence relation with the given initi. (a) \( a_{n}=10 a_{n-1}-32 a_{n-2}+32 a_{n-3} \) with \( a_{0}=5, a_{1}=18, a_{2}=76 \)

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