Chapter 18: Problem 4
Solve the following difference equations: \((a) y_{t+2}+3 y_{t+1}-\frac{7}{4} y_{t}=9 \quad\left(y_{0}=6 ; y_{1}=3\right)\) (b) \(y_{t+2}-2 y_{t+1}+2 y_{t}=1 \quad\left(y_{0}=3 ; y_{1}=4\right)\) (c) \(y_{i+2}-y_{t+1}+\frac{1}{4} y_{t}=2 \quad\left(y_{0}=4 ; y_{1}=7\right)\)
Short Answer
Expert verified
(a) Solve using characteristic equation and particular solution; (b) use complex exponential form; (c) address repeated root in solution.
Step by step solution
01
Solve part (a) - Characteristic equation
For the equation \( y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 9 \), assume the homogeneous solution \( y_h = r^t \). Substitute into the homogeneous equation \( r^2 + 3r - \frac{7}{4} = 0 \). Use the quadratic formula to find roots: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a=1, b=3, c=-\frac{7}{4} \).
02
Calculate the roots for part (a)
Calculate the discriminant \( b^2 - 4ac = 3^2 - 4(1)(-\frac{7}{4}) = 9 + 7 = 16 \). So \( r = \frac{-3 \pm \sqrt{16}}{2} = \frac{-3 \pm 4}{2} \). The roots are \( r_1 = \frac{1}{2}, r_2 = -\frac{7}{2} \).
03
General solution for part (a)
The homogeneous solution is \( y_h = A \left(\frac{1}{2}\right)^t + B \left(-\frac{7}{2}\right)^t \). The particular solution can be assumed constant \( y_p = C \). Substitute into non-homogeneous part to solve for \( C \): \( C + 3C - \frac{7}{4}C = 9 \), solving yields \( C = -12 \).
04
Complete solution for part (a)
The complete solution is \( y_t = A \left(\frac{1}{2}\right)^t + B \left(-\frac{7}{2}\right)^t - 12 \). Use the initial conditions \( y_0 = 6 \) and \( y_1 = 3 \) to solve for \( A \) and \( B \).
05
Particular solution for part (b)
For the equation \( y_{t+2} - 2y_{t+1} + 2y_t = 1 \), the homogeneous solution is found using the characteristic equation \( r^2 - 2r + 2 = 0 \). Calculate roots to find \( r = 1 \pm i \), which implies a complex solution. The general form is \( y_h = e^t(C_1 \cos(t) + C_2 \sin(t)) \).
06
Solve for the particular solution of part (b)
Assume a constant particular solution \( y_p = C \). Substituting back gives \( C - 2C + 2C = 1 \), yielding no information about \( C \). Use differential guesses for a more accurate approximate solution if needed.
07
Solve part (c) - Characteristic equation
For the equation \( y_{t+2} - y_{t+1} + \frac{1}{4}y_t = 2 \), solve the characteristic equation \( r^2 - r + \frac{1}{4} = 0 \). Use the quadratic formula: \( r = \frac{1 \pm 0}{2} = \frac{1}{2} \), a repeated root.
08
General solution for part (c)
Since there is a repeated root \( r = \frac{1}{2} \), the solution form is \( y_h = (A + Bt) \left(\frac{1}{2}\right)^t \). Assume a particular solution in the form of a constant: \( y_p = C \), resulting in no additional form needed from differential guesses.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Solution
When solving difference equations, a critical step is finding the homogeneous solution. This solution addresses the equation when the right-hand side equals zero. It means we're considering only the inherent behavior of the equation without any external influence, such as constants or other terms outside the equation itself.
For instance, in the equation from part (a) of the original problem: \[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 9, \]we start by setting the equation to zero to find the homogeneous solution:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 0. \]We assume that the solution can be expressed in the form \(y_h = r^t\) and substitute this into the homogeneous equation. This substitution leads us to the characteristic equation, which we solve to find the roots, like \(r_1\) and \(r_2\). These roots allow us to express the homogeneous solution as a linear combination of these roots:
- If the roots are real and distinct, the solution is \(y_h = A r_1^t + B r_2^t\).- In cases of complex roots or repeated roots, the form of the solution changes but still revolves around these core concepts.
Understanding the homogeneous solution is key because it provides insights into the natural behavior of the system without external factors.
For instance, in the equation from part (a) of the original problem: \[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 9, \]we start by setting the equation to zero to find the homogeneous solution:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 0. \]We assume that the solution can be expressed in the form \(y_h = r^t\) and substitute this into the homogeneous equation. This substitution leads us to the characteristic equation, which we solve to find the roots, like \(r_1\) and \(r_2\). These roots allow us to express the homogeneous solution as a linear combination of these roots:
- If the roots are real and distinct, the solution is \(y_h = A r_1^t + B r_2^t\).- In cases of complex roots or repeated roots, the form of the solution changes but still revolves around these core concepts.
Understanding the homogeneous solution is key because it provides insights into the natural behavior of the system without external factors.
Particular Solution
The particular solution component of a difference equation accounts for the non-homogeneous part. This segment considers the specific external factors or inputs that affect the system. Unlike the homogeneous solution, which involves solving a set zero right-hand side, the particular solution deals with the right-hand side of the equation as given.
For example, in part (a), after solving for the homogeneous solution of the equation:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 9, \]we assume a particular solution of the form \(y_p = C\). This substitution simplifies the problem because we treat the unknown as a constant affecting the system.
- We substitute our proposed particular solution back into the equation.- This allows us to solve for \(C\) by equating coefficients or balancing equations.Once found, this particular solution, combined with the homogeneous solution, forms the complete solution. It's crucial because it directly adjusts the model to fit the actual behavior, including all external influences known from the problem.
For example, in part (a), after solving for the homogeneous solution of the equation:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 9, \]we assume a particular solution of the form \(y_p = C\). This substitution simplifies the problem because we treat the unknown as a constant affecting the system.
- We substitute our proposed particular solution back into the equation.- This allows us to solve for \(C\) by equating coefficients or balancing equations.Once found, this particular solution, combined with the homogeneous solution, forms the complete solution. It's crucial because it directly adjusts the model to fit the actual behavior, including all external influences known from the problem.
Characteristic Equation
The characteristic equation is a cornerstone in solving difference equations, specifically in deriving the homogeneous solution. It emerges from assuming a solution of the form \( y_h = r^t \) and substituting this form into the equation:
Say we have a difference equation, such as:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 0. \]By replacing \(y_t\) with \(r^t\), we derive:\[ r^{t+2} + 3r^{t+1} - \frac{7}{4}r^t = 0, \]which simplifies through a division by \(r^t\) (assuming \(r eq 0\)) to yield the characteristic equation:\[ r^2 + 3r - \frac{7}{4} = 0. \]
- The roots of this polynomial equation determine the forms \(r_1\) and \(r_2\).- Depending on these roots (real, complex, or repeated), different homogeneous solution forms will emerge.The characteristic equation is essential as it succinctly captures the equation's structural properties necessary to find the homogeneous behavior.
Say we have a difference equation, such as:\[ y_{t+2} + 3y_{t+1} - \frac{7}{4}y_t = 0. \]By replacing \(y_t\) with \(r^t\), we derive:\[ r^{t+2} + 3r^{t+1} - \frac{7}{4}r^t = 0, \]which simplifies through a division by \(r^t\) (assuming \(r eq 0\)) to yield the characteristic equation:\[ r^2 + 3r - \frac{7}{4} = 0. \]
- The roots of this polynomial equation determine the forms \(r_1\) and \(r_2\).- Depending on these roots (real, complex, or repeated), different homogeneous solution forms will emerge.The characteristic equation is essential as it succinctly captures the equation's structural properties necessary to find the homogeneous behavior.
Quadratic Formula
The quadratic formula is a fundamental tool in solving the characteristic equation for difference equations. The characteristic equation often takes the form of a quadratic:\[ ar^2 + br + c = 0, \]where \(a\), \(b\), and \(c\) are constants derived from the original difference equation.
The quadratic formula gives us a straightforward way to find the roots \( r \) of this equation:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Consider part (a) of our example, where:- \(a = 1\), \(b = 3\), \(c = -\frac{7}{4}\).- The discriminant \( b^2 - 4ac \) helps determine the nature of the roots.- If \( b^2 - 4ac > 0 \), the roots are real and distinct.- If \( b^2 - 4ac = 0 \), the roots are real and repeated.- If \( b^2 - 4ac < 0 \), the roots are complex.
Utilizing the quadratic formula allows us to systematically find these roots, which are critical for constructing the homogeneous solution. It's a reliable method essential for students to master when dealing with quadratic equations in difference equations.
The quadratic formula gives us a straightforward way to find the roots \( r \) of this equation:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Consider part (a) of our example, where:- \(a = 1\), \(b = 3\), \(c = -\frac{7}{4}\).- The discriminant \( b^2 - 4ac \) helps determine the nature of the roots.- If \( b^2 - 4ac > 0 \), the roots are real and distinct.- If \( b^2 - 4ac = 0 \), the roots are real and repeated.- If \( b^2 - 4ac < 0 \), the roots are complex.
Utilizing the quadratic formula allows us to systematically find these roots, which are critical for constructing the homogeneous solution. It's a reliable method essential for students to master when dealing with quadratic equations in difference equations.
