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Magazine Chapter 6: Problem 19
Solve the difference equations $$ y(t+1)=3 y(t)+1, y_{0}=0 $$
Short Answer
Expert verified
The explicit solution is \( y(t) = \frac{1}{2}\cdot 3^t - \frac{1}{2} \).
Step by step solution
01
Identify Initial Condition and Form of Solution
We start by identifying the initial condition given as \( y_0 = 0 \) and the recursive relationship \( y(t+1) = 3y(t) + 1 \). Our goal is to find the explicit form of \( y(t) \) as a function of \( t \).
02
Find Homogeneous Solution
First, solve the homogeneous part of the equation: \( y_h(t+1) = 3y_h(t) \). This is a geometric sequence which can be solved as \( y_h(t) = C\cdot 3^t \), where \( C \) is a constant.
03
Find Particular Solution
Next, find a particular solution to the equation \( y(t+1) = 3y(t) + 1 \). Assume a constant solution \( y_p = a \). Substitute into the equation: \( a = 3a + 1 \). Solving gives \( a = -\frac{1}{2} \).
04
General Solution
Combine the homogeneous and particular solutions: \( y(t) = y_h(t) + y_p \). Therefore, the general solution is \( y(t) = C\cdot 3^t - \frac{1}{2} \).
05
Apply Initial Condition
Use the initial condition \( y(0) = 0 \) to solve for \( C \). Substitute \( t = 0 \) into the general solution: \( 0 = C\cdot 3^0 - \frac{1}{2} \). Solve for \( C \): \( C = \frac{1}{2} \).
06
Explicit Solution
Substitute \( C = \frac{1}{2} \) back into the general solution, yielding the explicit formula \( y(t) = \frac{1}{2}\cdot 3^t - \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recursive Sequences
Recursive sequences are a type of mathematical expression that define a sequence of numbers based on previous terms. In our difference equation, the recursive sequence is illustrated by the relationship \( y(t+1) = 3y(t) + 1 \). Here, we can see that each term is determined by applying a specific formula to the preceding term, which allows us to calculate the series in an orderly fashion.
Key points about recursive sequences include:
Key points about recursive sequences include:
- They rely on initial values, from which all subsequent values are derived.
- They often require fewer computations compared to explicitly computing terms directly from a formula.
- They are widely used in various applications, like computing compound interest or modeling population growth.
Homogeneous Solutions
In solving difference equations, a homogeneous solution refers to solving the equation as if there were no additional terms or constants. In our example, the homogeneous part is \( y_h(t+1) = 3y_h(t) \). This results in a geometric sequence due to the constant ratio between consecutive terms.
It's important to note that:
It's important to note that:
- The solution takes the form \( y_h(t) = C \cdot 3^t \), where \( C \) is a constant determined by initial conditions.
- The homogeneous solution represents the underlying behavior or structure of the sequence without external influences.
Particular Solutions
The particular solution is a specific sequence that satisfies the non-homogeneous part of the difference equation. In our example, this involves solving \( y(t+1) = 3y(t) + 1 \) with the assumption that the solution is constant (i.e., attempting \( y_p = a \)).
This process gives us:
This process gives us:
- The equation \( a = 3a + 1 \), simplifying to \( a = -\frac{1}{2} \).
- The significance of the particular solution is that it accounts for specific terms or external conditions impacting the sequence.
Initial Conditions
Initial conditions provide the starting point necessary to solve a difference equation fully. In our case, \( y(0) = 0 \) knows where the sequence should begin. This condition was used to solve for the constant \( C \) in the general solution.
Consider these points:
Consider these points:
- Initial conditions anchor the solution in reality, making sure it meets the specific criteria given in a problem.
- They help to find necessary constants, unifying the homogeneous and particular solutions into one full model.
