Question

Consider the differential equation \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\). In each exercise, the nonhomogeneous term, \(g(t)\), and the form of the particular solution prescribed by the method of undetermined coefficients are given. Determine the constants \(\alpha\) and \(\beta\). $$ g(t)=t+e^{3 t}, \quad y_{P}(t)=A_{1} t^{2}+A_{0} t+B_{0} t e^{3 t} $$

Short answer

Question: Given the second-order linear differential equation y''(t) + αy'(t) + βy(t) = g(t), where g(t) = t + e^(3t), and a particular solution yP(t) = A1t^2 + A0t + B0te^(3t), find the relationship between the constants α and β. Answer: The relationship between the constants α and β is given by α = -⅓β.

Step-by-step solution

Differentiate the given particular solution

First, we need to find the first and second derivatives of the given particular solution, yP(t)=A1t^2+A0t+B0te^(3t). The first derivative is: $$ y_P'(t) = 2A_1t + A_0 + B_0(3te^{3t} + e^{3t}) $$ The second derivative is: $$ y_P''(t) = 2A_1 + B_0(6e^{3t} + 9t^2e^{3t} + 9te^{3t}) $$

Substitute yP(t), yP'(t), and yP''(t) into the differential equation

Now, substitute yP(t), its first and second derivatives yP'(t), and yP''(t) into the given differential equation: $$ y''(t) + \alpha y'(t) +\beta y(t) = g(t) $$ So we have: $$ (2A_1 + B_0(6e^{3t} + 9t^2e^{3t} + 9te^{3t})) + \alpha(2A_1t + A_0 + B_0(3te^{3t} + e^{3t})) +\beta(A_1t^2+A_0t+B_0te^{3t}) = t + e^{3t} $$

Match the coefficients of g(t) to the coefficients of the left side of the equation

Expand the equation above and equate the coefficients of the respective terms in the equation to match the coefficients of g(t): For the t^2 term, $$ \beta A_1 = 0 $$ For the t term, $$ 2A_1 + \alpha A_0 +\beta A_0 = 1 $$ For the te^(3t) term, $$ \alpha (3B_0) + \beta B_0 = 0 $$ For the e^(3t) term, $$ 6B_0 = 1 $$ Solve the last equation, we get B0. $$ B_0 = \frac{1}{6} $$ Now, substitute B0 into the third equation to solve for α and β: $$ \alpha \bigg(\frac{1}{2}\bigg) + \beta \bigg(\frac{1}{6}\bigg) = 0 $$ Since we cannot uniquely solve for α and β (there are infinitely many solutions for α and β), we cannot determine their exact values. However, we have found the relation: $$ \alpha = -\frac{1}{3}\beta $$

Key concepts

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to nonhomogeneous linear differential equations. This method especially works well when the nonhomogeneous term, such as a polynomial, an exponential function, or a sine or cosine function, is involved.

When employing this method, we assume a form for the particular solution based on the nonhomogeneous term, ensuring that the terms in the particular solution are not solutions to the corresponding homogeneous equation. In the case of the given exercise, where the nonhomogeneous term is a combination of a polynomial and an exponential function, the assumed particular solution is a mix of corresponding polynomial and exponential terms multiplied by undetermined coefficients. These coefficients (A_1, A_0, and B_0) will eventually be found by applying the differential equation and matching coefficients.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation is characterized by an equation that equals a non-zero function, which is termed the nonhomogeneous term. Unlike homogeneous equations that have solutions which can be superposed, nonhomogeneous equations require a particular solution to represent the nonhomogeneous term effect. This equation takes the form: \( y'' + \: \alpha y' + \beta y = g(t) \) where \(g(t) \) is the nonhomogeneous part. In this context, finding a solution comprises two parts: the general solution to the corresponding homogeneous equation and the particular solution that we find using methods like undetermined coefficients to address the nonhomogeneity.

It's essential to comprehend that the particular solution does not contain arbitrary constants; it's designed to fit the nonhomogeneity directly and, once discovered, can be added to the general homogeneous solution to produce the most general solution to the nonhomogeneous equation.

Particular Solution

The particular solution of a differential equation is a specific solution that satisfies not only the differential equation but also fits the nonhomogeneous term of the equation. This solution, denoted as \(y_P(t)\), does not contain any constants which would represent the family of solutions to the homogeneous part of the equation. Instead, it is perfectly tailored to fulfill the non-zero \(g(t)\).

In the exercise, the particular solution is given by \(y_P(t)=A_{1}t^2+A_0t+B_0te^{3t}\). This formula is structured to address the specific nonhomogeneous term provided, \(g(t)=t+e^{3t}\), with the aim being to determine the coefficients that will make \(y_P(t)\) a valid component of the overall solution to the differential equation.

Matching Coefficients

The technique of matching coefficients involves equating the coefficients from both sides of an equation to find values for the undetermined coefficients in the particular solution of a differential equation. By setting up equivalences for each degree of the term, we create a system of equations that, once solved, provides the needed coefficients to form a particular solution that satisfies the equation.

In this exercise, after substituting in the particular solution and its derivatives into the differential equation, we match the coefficients of like terms from both sides. For example, we match the coefficients of \(t^2, t, te^{3t}\), and \(e^{3t}\) separately, creating a system of equations. These can then be solved sequentially, beginning typically with the simplest equation. Matching coefficients is a critical step in applying the method of undetermined coefficients, and the successful application of this technique in such exercises depends on careful expansion and organization of the terms in the differential equation.