Question

Let a second order difference equation be
\(y_{n+2} + 4y_n = 4y_{n+1}, \, n=2,3,4,......, \,\, y_0=1, y_1=4\)
Then the general solution is

• A. \((1+n^2)2^n\)
• B. \((1+n)2^n\)
• C. \((1+\frac1{n})2^n\)
• D. \((n^2+n+1)2^n\)

Answer 1

The Correct Option is B

To find the general solution of the given second-order difference equation:

\(y_{n+2} + 4y_n = 4y_{n+1}\)

with initial conditions \(y_0 = 1\) and \(y_1 = 4\), we follow these steps:

1. Find the characteristic equation:

Assume a solution of the form \(y_n = r^n\). Substituting this into the difference equation gives:

\(r^{n+2} + 4r^n = 4r^{n+1}\)

Divide the entire equation by \(r^n\) (assuming \(r \neq 0\)):

\(r^2 + 4 = 4r\)

This simplifies to the characteristic equation:

\(r^2 - 4r + 4 = 0\)

1. Solve the characteristic equation:

The characteristic equation \(r^2 - 4r + 4 = 0\) can be factored as:

\((r-2)^2 = 0\)

This gives a repeated root, \(r = 2\).

1. Write the general solution for the difference equation:

Since we have a repeated root, the general solution for the difference equation is of the form:

\(y_n = (A + Bn) \cdot 2^n\)

where \( A \) and \( B \) are constants that we need to determine using the initial conditions.

1. Use the initial conditions to find constants:

Using \(y_0 = 1\), we get:

\(A + B \cdot 0 = 1 \Rightarrow A = 1\)

Using \(y_1 = 4\), we substitute into the general solution:

\((A + B \cdot 1) \cdot 2^1 = 4\)

\(2(A + B) = 4\)

Substituting \(A = 1\), we get:

\(2(1 + B) = 4 \Rightarrow 1 + B = 2 \Rightarrow B = 1\)

1. Write the complete general solution:

Substituting \(A = 1\) and \(B = 1\) back into the general form, we get:

\(y_n = (1 + n) \cdot 2^n\)

This matches the given correct answer option.

Thus, the general solution to the difference equation is:

\((1+n)2^n\)