Question

Given the difference equation x(k+2) + 5x(k+1) + 4x(k)= e(k)

where
e(k)= 1,0,
​k=0 otherwise ​x(0)= 0, x(1)=−1​
solve for x(k) as a function of k.

Answer 1

To solve the given difference equation
`\(x(k+2) + 5x(k+1) + 4x(k) = e(k)\)` with the provided initial conditions
`\(x(0) = 0\) and \(x(1) = -1\)`, and
`\(e(k) = 1\) for \(k = 0\) and \(e(k) = 0\)` otherwise, we can follow these steps:

Step-by-step explanation:

1. Form the Characteristic Equation:

Write down the characteristic equation associated with the homogeneous part of the difference equation:

`\[ r^2 + 5r + 4 = 0 \]`

2. Solve for the Roots (r):

Factorize the characteristic equation or use the quadratic formula to find the roots.

`\[ (r+1)(r+4) = 0 \]`

`\[ r_1 = -1, \quad r_2 = -4 \]`

3. Write the Homogeneous Solution:

The homogeneous solution is given by:

`\[ x_h(k) = A(-1)^k + B(-4)^k \]`

where
`\(A\)` and
`\(B\)` are constants determined by the initial conditions.

4. **Apply Initial Conditions:**

- Use the provided initial conditions
`\(x(0) = 0\) and \(x(1) = -1\)` to find the values of
`\(A\)` and
`\(B\)`.

5. Write the Particular Solution:

Write down the particular solution associated with the non-homogeneous part of the difference equation. Since
`\(e(k) = 1\) for \(k = 0\) and \(e(k) = 0\)` otherwise, the particular solution is:
`\[ x_p(k) = 1 \]`

6. Write the General Solution:

Combine the homogeneous and particular solutions to get the general solution:

`\[ x(k) = x_h(k) + x_p(k) \]`

7. Final Result:

Plug in the values of
`\(A\),`
`\(B\),` and
`\(x_p(k)\)` to get the complete solution as a function of
`\(k\)`.

Therefore, the solution to the given difference equation is
`\(x(k) = A(-1)^k + B(-4)^k + 1\)`, where
`\(A\)` and
`\(B\)` are determined by the initial conditions.