To solve the recurrence relation $a_n + 5a_{n-1} + 6a_{n-2} = 3n^2$, we can break this down into two parts:

1. Solve the homogeneous recurrence relation $a_n + 5a_{n-1} + 6a_{n-2} = 0$.
2. Find a particular solution to the non-homogeneous equation $a_n + 5a_{n-1} + 6a_{n-2} = 3n^2$.

Step 1: Solve the homogeneous part

The homogeneous equation is: $a_n + 5a_{n-1} + 6a_{n-2} = 0.$ Assume the solution is of the form $a_n = r^n$. Substituting this into the homogeneous equation gives the characteristic equation: $r^n + 5r^{n-1} + 6r^{n-2} = 0.$ Dividing through by $r^{n-2}$ (assuming $r \neq 0$): $r^2 + 5r + 6 = 0.$ Factoring the quadratic equation: $(r + 2)(r + 3) = 0.$ Thus, the roots are $r = -2$ and $r = -3$.

So, the general solution to the homogeneous recurrence relation is: $a_n^{(h)} = C_1(-2)^n + C_2(-3)^n,$ where $C_1$ and $C_2$ are constants to be determined later from initial conditions.

Step 2: Find a particular solution

Now we find a particular solution to the non-homogeneous recurrence relation: $a_n + 5a_{n-1} + 6a_{n-2} = 3n^2.$ We use the method of undetermined coefficients. Since the non-homogeneous term is a polynomial of degree 2 ($3n^2$), we assume the particular solution is of the form: $a_n^{(p)} = An^2 + Bn + C.$ Substitute this into the original recurrence relation:

$(An^2 + Bn + C) + 5(A(n-1)^2 + B(n-1) + C) + 6(A(n-2)^2 + B(n-2) + C) = 3n^2.$

Expand each term and equate coefficients of $n^2$, $n$, and constant terms on both sides to solve for $A$, $B$, and $C$. After calculating, you will get specific values for $A$, $B$, and $C$.

Step 3: General solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $a_n = a_n^{(h)} + a_n^{(p)} = C_1(-2)^n + C_2(-3)^n + (An^2 + Bn + C).$

You can determine $C_1$, $C_2$, and any constants from the particular solution by using initial conditions provided for $a_n$ (if given).

Would you like further details on expanding the particular solution or any initial conditions you have?

Related Questions:

1. How do you find the particular solution using undetermined coefficients?
2. What are the initial conditions for this recurrence relation?
3. Can we use another method, like generating functions, for solving recurrence relations?
4. How does the characteristic equation help in solving homogeneous recurrence relations?
5. How do the roots of the characteristic equation affect the form of the homogeneous solution?

Tip:

When solving recurrence relations, always factor the characteristic equation carefully, as incorrect roots can lead to errors in the general solution.