Answer :
Certainly! Let's solve the given difference equation step by step.
The difference equation is given by:
[tex]\[ 2 y_t - y_{t-1} = 2^t \][/tex]
with the initial condition:
[tex]\[ y_0 = \frac{8}{3} \][/tex]
### Step 1: Understanding the Difference Equation
Firstly, let's rewrite the equation in a standard form:
[tex]\[ 2 y_t = y_{t-1} + 2^t \][/tex]
[tex]\[ y_t = \frac{1}{2} y_{t-1} + \frac{1}{2} 2^t \][/tex]
So our difference equation becomes:
[tex]\[ y_t = \frac{1}{2} y_{t-1} + 2^{t-1} \][/tex]
### Step 2: Solving the Homogeneous Part
The homogeneous part of the difference equation is:
[tex]\[ 2 y_t = y_{t-1} \][/tex]
or,
[tex]\[ y_t - \frac{1}{2} y_{t-1} = 0 \][/tex]
This suggests a solution of the form:
[tex]\[ y_t = A \left(\frac{1}{2}\right)^t \][/tex]
for some constant [tex]\(A\)[/tex].
### Step 3: Solving the Non-Homogeneous Part
We need to find a particular solution to the non-homogeneous equation:
[tex]\[ y_t = \frac{1}{2} y_{t-1} + 2^{t-1} \][/tex]
Assume a particular solution of the form:
[tex]\[ y_t^p = B 2^t \][/tex]
Substitute [tex]\(y_t^p\)[/tex] into the non-homogeneous equation:
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
Divide by [tex]\(2^{t-1}\)[/tex] to simplify:
[tex]\[ B \cdot 2 = \frac{1}{2} B + 1 \][/tex]
[tex]\[ 2B = \frac{1}{2}B + 1 \][/tex]
[tex]\[ 4B = B + 2 \][/tex]
[tex]\[ 3B = 2 \][/tex]
[tex]\[ B = \frac{2}{3} \][/tex]
So, the particular solution is:
[tex]\[ y_t^p = \frac{2}{3} 2^t \][/tex]
### Step 4: General Solution
The general solution to the difference equation is the sum of the homogeneous solution and the particular solution:
[tex]\[ y_t = A \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t \][/tex]
### Step 5: Applying the Initial Condition
We use the initial condition [tex]\( y_0 = \frac{8}{3} \)[/tex]:
[tex]\[ y_0 = A \cdot \left( \frac{1}{2} \right)^0 + \frac{2}{3} \cdot 2^0 = \frac{8}{3} \][/tex]
[tex]\[ A + \frac{2}{3} = \frac{8}{3} \][/tex]
[tex]\[ A = \frac{8}{3} - \frac{2}{3} = 2 \][/tex]
### Final Solution
Substituting the constant [tex]\(A\)[/tex] back into the general solution:
[tex]\[ y_t = 2 \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t \][/tex]
[tex]\[ y_t = 2 \cdot 2^{-t} + \frac{2}{3} \cdot 2^t \][/tex]
[tex]\[ y_t = 2^{1-t} + \frac{2}{3} \cdot 2^t \][/tex]
So, the final solution is:
[tex]\[ y_t = 2 \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t = 2^{1-t} + \frac{2}{3} 2^t \][/tex]
Therefore, the solution to the given difference equation with the initial condition is:
[tex]\[ y_t = 2^{1-t} + \frac{2}{3} 2^t \][/tex]
The difference equation is given by:
[tex]\[ 2 y_t - y_{t-1} = 2^t \][/tex]
with the initial condition:
[tex]\[ y_0 = \frac{8}{3} \][/tex]
### Step 1: Understanding the Difference Equation
Firstly, let's rewrite the equation in a standard form:
[tex]\[ 2 y_t = y_{t-1} + 2^t \][/tex]
[tex]\[ y_t = \frac{1}{2} y_{t-1} + \frac{1}{2} 2^t \][/tex]
So our difference equation becomes:
[tex]\[ y_t = \frac{1}{2} y_{t-1} + 2^{t-1} \][/tex]
### Step 2: Solving the Homogeneous Part
The homogeneous part of the difference equation is:
[tex]\[ 2 y_t = y_{t-1} \][/tex]
or,
[tex]\[ y_t - \frac{1}{2} y_{t-1} = 0 \][/tex]
This suggests a solution of the form:
[tex]\[ y_t = A \left(\frac{1}{2}\right)^t \][/tex]
for some constant [tex]\(A\)[/tex].
### Step 3: Solving the Non-Homogeneous Part
We need to find a particular solution to the non-homogeneous equation:
[tex]\[ y_t = \frac{1}{2} y_{t-1} + 2^{t-1} \][/tex]
Assume a particular solution of the form:
[tex]\[ y_t^p = B 2^t \][/tex]
Substitute [tex]\(y_t^p\)[/tex] into the non-homogeneous equation:
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
[tex]\[ B 2^t = \frac{1}{2} B 2^{t-1} + 2^{t-1} \][/tex]
Divide by [tex]\(2^{t-1}\)[/tex] to simplify:
[tex]\[ B \cdot 2 = \frac{1}{2} B + 1 \][/tex]
[tex]\[ 2B = \frac{1}{2}B + 1 \][/tex]
[tex]\[ 4B = B + 2 \][/tex]
[tex]\[ 3B = 2 \][/tex]
[tex]\[ B = \frac{2}{3} \][/tex]
So, the particular solution is:
[tex]\[ y_t^p = \frac{2}{3} 2^t \][/tex]
### Step 4: General Solution
The general solution to the difference equation is the sum of the homogeneous solution and the particular solution:
[tex]\[ y_t = A \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t \][/tex]
### Step 5: Applying the Initial Condition
We use the initial condition [tex]\( y_0 = \frac{8}{3} \)[/tex]:
[tex]\[ y_0 = A \cdot \left( \frac{1}{2} \right)^0 + \frac{2}{3} \cdot 2^0 = \frac{8}{3} \][/tex]
[tex]\[ A + \frac{2}{3} = \frac{8}{3} \][/tex]
[tex]\[ A = \frac{8}{3} - \frac{2}{3} = 2 \][/tex]
### Final Solution
Substituting the constant [tex]\(A\)[/tex] back into the general solution:
[tex]\[ y_t = 2 \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t \][/tex]
[tex]\[ y_t = 2 \cdot 2^{-t} + \frac{2}{3} \cdot 2^t \][/tex]
[tex]\[ y_t = 2^{1-t} + \frac{2}{3} \cdot 2^t \][/tex]
So, the final solution is:
[tex]\[ y_t = 2 \left( \frac{1}{2} \right)^t + \frac{2}{3} 2^t = 2^{1-t} + \frac{2}{3} 2^t \][/tex]
Therefore, the solution to the given difference equation with the initial condition is:
[tex]\[ y_t = 2^{1-t} + \frac{2}{3} 2^t \][/tex]
