Math Problem Statement

t n -2t n - 1 =(n+5)3^ n n >= 1 solve using Inhomogenous Recuurence.

Solution

To solve the recurrence relation tn2tn1=(n+5)3nt_n - 2t_{n-1} = (n+5)3^n for n1n \geq 1 using the method of solving inhomogeneous recurrence relations, we will follow these steps:

Step 1: Solve the Homogeneous Part

First, solve the corresponding homogeneous recurrence relation: tn2tn1=0t_n - 2t_{n-1} = 0

The characteristic equation is: r2=0r - 2 = 0 So, r=2r = 2.

The general solution to the homogeneous equation is: tn(h)=C2nt_n^{(h)} = C \cdot 2^n where CC is a constant to be determined by initial conditions.

Step 2: Find a Particular Solution

Next, we need to find a particular solution tn(p)t_n^{(p)} to the inhomogeneous equation: tn(p)2tn1(p)=(n+5)3nt_n^{(p)} - 2t_{n-1}^{(p)} = (n+5)3^n

Since the non-homogeneous term is of the form (n+5)3n(n+5)3^n, we try a particular solution of the form: tn(p)=(An+B)3nt_n^{(p)} = (An + B) \cdot 3^n where AA and BB are constants to be determined.

Substitute tn(p)=(An+B)3nt_n^{(p)} = (An + B)3^n into the original recurrence relation: (An+B)3n2[A(n1)+B]3n1=(n+5)3n(An + B)3^n - 2 \cdot \left[ A(n-1) + B \right]3^{n-1} = (n+5)3^n

Simplifying, we get: (An+B)3n2A(n1)3n2B3n1=(n+5)3n(An + B)3^n - 2A(n-1) \cdot 3^n - 2B \cdot 3^{n-1} = (n+5)3^n

3n(An+B2A(n1)3)2B33n=(n+5)3n3^n \left( An + B - \frac{2A(n-1)}{3} \right) - \frac{2B}{3} \cdot 3^n = (n+5)3^n

Simplifying further: 3n(An+B2An3+2A32B3)=(n+5)3n3^n \left( An + B - \frac{2An}{3} + \frac{2A}{3} - \frac{2B}{3} \right) = (n+5)3^n

Now, equate coefficients of like terms:

  1. Coefficient of n3nn \cdot 3^n: A2A3=1    A3=1    A=3A - \frac{2A}{3} = 1 \implies \frac{A}{3} = 1 \implies A = 3
  2. Constant term: B+2A32B3=5    B+22B3=5B + \frac{2A}{3} - \frac{2B}{3} = 5 \implies B + 2 - \frac{2B}{3} = 5 Simplifying: 3B2B3=3    B=9\frac{3B - 2B}{3} = 3 \implies B = 9

Thus, the particular solution is: tn(p)=(3n+9)3nt_n^{(p)} = (3n + 9)3^n

Step 3: General Solution

The general solution to the original recurrence relation is the sum of the homogeneous and particular solutions: tn=tn(h)+tn(p)=C2n+(3n+9)3nt_n = t_n^{(h)} + t_n^{(p)} = C \cdot 2^n + (3n + 9)3^n

Step 4: Apply Initial Conditions (If Given)

If an initial condition like t0t_0 is given, you can substitute n=0n = 0 and solve for the constant CC.

Summary

The general solution to the recurrence relation tn2tn1=(n+5)3nt_n - 2t_{n-1} = (n+5)3^n is: tn=C2n+(3n+9)3nt_n = C \cdot 2^n + (3n + 9)3^n where CC is determined by initial conditions.


Would you like more details or have any questions? Here are some related questions to consider:

  1. How would you solve a recurrence relation with a different inhomogeneous term, like 3n3^n instead of (n+5)3n(n+5)3^n?
  2. What if the characteristic equation had complex roots? How would the solution differ?
  3. How can you determine the initial conditions for this problem?
  4. What if the recurrence relation had a different coefficient, like 3tn13t_{n-1} instead of 2tn12t_{n-1}?
  5. How do you verify that the solution satisfies the original recurrence relation?

Tip: Always check your particular solution by substituting it back into the original recurrence to verify that it satisfies the equation.

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Math Problem Analysis

Mathematical Concepts

Inhomogeneous Recurrence Relations
Homogeneous Recurrence Relations
Algebraic Equations

Formulas

Characteristic equation: r - 2 = 0
General solution for homogeneous part: t_n^{(h)} = C * 2^n
Particular solution form: t_n^{(p)} = (An + B) * 3^n

Theorems

Method of Undetermined Coefficients

Suitable Grade Level

Undergraduate Mathematics