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The given problem asks to find the general solution to a non-homogeneous recurrence relation of the form:

\[
a_{n+2} - 5a_{n+1} + 6a_n = n^2 - 7n + 10
\]

This is a second-order linear recurrence relation with constant coefficients and a non-homogeneous term on the right-hand side.

### Steps to Solve:

1. **Solve the Homogeneous Equation**:
   First, solve the associated homogeneous recurrence relation:
   \[
   a_{n+2} - 5a_{n+1} + 6a_n = 0
   \]
   The characteristic equation for this is:
   \[
   r^2 - 5r + 6 = 0
   \]
   Factoring:
   \[
   (r - 2)(r - 3) = 0
   \]
   So, the roots are \(r_1 = 2\) and \(r_2 = 3\).

   Therefore, the general solution to the homogeneous equation is:
   \[
   a_n^{(h)} = C_1 \cdot 2^n + C_2 \cdot 3^n
   \]
   where \(C_1\) and \(C_2\) are constants to be determined.

2. **Solve the Non-Homogeneous Equation**:
   For the non-homogeneous part, the right-hand side is a quadratic polynomial: \(n^2 - 7n + 10\). We try a particular solution of the form:
   \[
   a_n^{(p)} = An^2 + Bn + C
   \]
   Substitute this into the non-homogeneous equation:
   \[
   a_{n+2}^{(p)} - 5a_{n+1}^{(p)} + 6a_n^{(p)} = n^2 - 7n + 10
   \]
   First, compute \(a_{n+1}^{(p)}\) and \(a_{n+2}^{(p)}\):
   \[
   a_{n+1}^{(p)} = A(n+1)^2 + B(n+1) + C = A(n^2 + 2n + 1) + B(n + 1) + C
   \]
   \[
   = An^2 + (2A + B)n + (A + B + C)
   \]
   \[
   a_{n+2}^{(p)} = A(n+2)^2 + B(n+2) + C = A(n^2 + 4n + 4) + B(n + 2) + C
   \]
   \[
   = An^2 + (4A + B)n + (4A + 2B + C)
   \]
   Substitute these into the recurrence relation:
   \[
   [An^2 + (4A + B)n + (4A + 2B + C)] - 5[An^2 + (2A + B)n + (A + B + C)] + 6[An^2 + Bn + C] = n^2 - 7n + 10
   \]
   Simplifying this gives:
   \[
   An^2 + (4A + B)n + (4A + 2B + C) - 5(An^2 + (2A + B)n + (A + B + C)) + 6(An^2 + Bn + C)
   \]
   Collect like terms:
   - \(n^2\): \(A - 5A + 6A = 2A\)
   - \(n\): \((4A + B) - 5(2A + B) + 6B = -6A + B\)
   - Constant: \((4A + 2B + C) - 5(A + B + C) + 6C = -A - 3B + C\)

   Now, set the coefficients of \(n^2\), \(n\), and the constant term equal to those on the right-hand side of the equation \(n^2 - 7n + 10\):

   \[
   2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}
   \]
   \[
   -6A + B = -7 \quad \Rightarrow \quad -6\left(\frac{1}{2}\right) + B = -7 \quad \Rightarrow \quad -3 + B = -7 \quad \Rightarrow \quad B = -4
   \]
   \[
   -A - 3B + C = 10 \quad \Rightarrow \quad -\frac{1}{2} - 3(-4) + C = 10 \quad \Rightarrow \quad -\frac{1}{2} + 12 + C = 10 \quad \Rightarrow \quad C = -\frac{3}{2}
   \]

   So, the particular solution is:
   \[
   a_n^{(p)} = \frac{1}{2}n^2 - 4n - \frac{3}{2}
   \]

3. **General Solution**:
   The general solution to the non-homogeneous recurrence relation is the sum of the homogeneous and particular solutions:
   \[
   a_n = C_1 \cdot 2^n + C_2 \cdot 3^n + \frac{1}{2}n^2 - 4n - \frac{3}{2}
   \]

Would you like further clarification or details on any step?

### Related Questions:
1. How do you determine the form of the particular solution for different types of non-homogeneous terms?
2. What if the characteristic equation had repeated roots? How would that change the solution?
3. Can you explain why we use different forms of particular solutions for different non-homogeneous terms?
4. What happens if the non-homogeneous term is an exponential or trigonometric function?
5. How can you verify the correctness of the final general solution?

#### Tip:
When solving recurrence relations, always check the characteristic equation's roots first to decide on the form of the homogeneous solution before tackling the non-homogeneous part.

Найдите общее решение неоднородного рекуррентного уравнения: a_{n+2} - 5a_{n+1} + 6a_n = n^2 - 7n + 10

Learn how to solve a non-homogeneous recurrence relation a_{n+2} - 5a_{n+1} + 6a_n = n^2 - 7n + 10 using characteristic equations and the method of undetermined coefficients. This detailed step-by-step solution covers key concepts in recurrence relations and polynomial functions.

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