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 \(\beta\). $$ g(t)=3 e^{2 t}-e^{-2 t}+t, \quad y_{P}(t)=A_{0} t e^{2 t}+B_{0} t e^{-2 t}+C_{1} t+C_{0} $$

Short answer

Given the nonhomogeneous linear differential equation with the form \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\) and the particular solution \(y_P(t)=A_{0} t e^{2 t}+B_{0} t e^{-2 t}+C_{1} t+C_{0}\), the constants \(\alpha\) and \(\beta\) can be found as follows: 1. Calculate the first and second derivatives of the particular solution. 2. Substitute the particular solution and its derivatives into the differential equation. 3. Group the terms and compare coefficients. 4. Solve the system of equations for α and β. Following these steps, it is determined that \(\alpha = -1\) and \(\beta = -2\).

Step-by-step solution

Find the derivatives of the particular solution

Firstly, we will find the first and second derivatives of the given particular solution, \(y_P(t)=A_{0} t e^{2 t}+B_{0} t e^{-2 t}+C_{1} t+C_{0}\). To find the first derivative, \(y_P'(t)\): $$ y_P'(t) = (A_{0} e^{2t} + 2A_{0}te^{2t}) + (-2B_{0} e^{-2t} + B_{0}te^{-2t}) + C_1 $$ And for the second derivative, \(y_P''(t)\): $$ y_P''(t) = (4A_{0}e^{2t} + 4A_{0}te^{2t}) + (4B_{0}e^{-2t} - 4B_{0}te^{-2t}) $$

Substitute the particular solution and its derivatives into the differential equation

Now, we will substitute \(y_P(t)\), \(y_P'(t)\), and \(y_P''(t)\) into the given differential equation, \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\). $$ (4A_{0}e^{2t} + 4A_{0}te^{2t}) + (4B_{0}e^{-2t} - 4B_{0}te^{-2t}) + \alpha((A_{0} e^{2t}+2A_{0}te^{2t})+(-2B_{0} e^{-2t}+B_{0}te^{-2t})+C_1) + \beta(A_{0}te^{2t} + B_{0}te^{-2t} + C_{1}t + C_{0}) = 3e^{2t} - e^{-2t} + t $$

Group the terms and compare coefficients

Now, let's group all terms with similar factors: $$ (4A_{0} + \alpha A_{0}) e^{2t} + (-4B_{0} - 2\alpha B_{0})e^{-2t} + (4A_{0} + 2\alpha A_{0} + \beta A_{0})te^{2t} + (-4B_{0} + \alpha B_{0} + \beta B_{0})te^{-2t} + (\alpha C_1 + \beta C_1) t + \beta C_0 = 3e^{2t} - e^{-2t} + t $$ Now, we will compare the coefficients of each term: $$ 4A_{0} + \alpha A_{0} = 3 \\ -4B_{0} - 2\alpha B_{0} = -1 \\ 4A_{0} + 2\alpha A_{0} + \beta A_{0} = 0 \\ -4B_{0} + \alpha B_{0} + \beta B_{0} = 1 \\ \alpha C_1 + \beta C_1 = 1 \\ \beta C_0 = 0 $$

Solve the system of equations for α and β

Let's solve the system of equations using the first two equations, since they don't contain the constants \(C_0\) and \(C_1\). From the first equation: $$ 4A_{0} + \alpha A_{0} = 3 \implies \alpha = \frac{3-4A_0}{A_0} $$ From the second equation: $$ -4B_{0} - 2\alpha B_{0} = -1 \implies \alpha = \frac{1+4B_0}{2B_0} $$ As both equations equal to \(\alpha\), we can equate them to each other: $$ \frac{3-4A_0}{A_0} =\frac{1+4B_0}{2B_0} $$ For simplicity purposes, we can choose \(A_0 = 1\) and \(B_0 = 1\). Therefore: $$ \frac{3-4}{1} = \frac{1+4}{2} \implies \alpha = -1 $$ Substituting this back into our equation for \(\beta A_0\): $$ 4 + 2(-1) + \beta = 0 \implies \beta = -2 $$ So, the constants we determined are \(\alpha = -1\) and \(\beta = -2\).

Key concepts

Nonhomogeneous Differential Equation

When we talk about a nonhomogeneous differential equation, we're discussing an equation that includes a term dependent on an independent variable, which cannot be derived from the homogeneous part of the equation. Take, for instance, the equation \( y^{\prime \prime} + \alpha y^{\prime} + \beta y = g(t) \). Here, \( g(t) \) represents the nonhomogeneous part, which includes variables or functions of \( t \) – such as \( 3e^{2t} - e^{-2t} + t \) – that do not result in zero when all occurrences of \( y \) and its derivatives are set to zero.

The nonhomogeneous term compels us to seek a particular solution that complements the complementary solution found in the homogeneous equation \( y^{\prime \prime} + \alpha y^{\prime} + \beta y = 0 \). This particular solution directly addresses the specific nonhomogeneous part, \( g(t) \), enabling us to solve the overall equation.

Method of Undetermined Coefficients

The method of undetermined coefficients is a powerful tool for solving nonhomogeneous differential equations where the nonhomogeneous term is a linear combination of specific functions (e.g., exponentials, polynomials, sines, and cosines). The core idea of the method is to guess a form for the particular solution \( y_P(t) \) that resembles the nonhomogeneous term \( g(t) \) and to use its derivatives to determine the constants that will satisfy the differential equation.

This method becomes extremely practical when the nonhomogeneous term has a clear pattern. For instance, given \( g(t) = 3e^{2t} - e^{-2t} + t \), we hypothesize a particular solution that contains terms similar to those in \( g(t) \) but with undetermined coefficients, such as \( y_P(t)=A_{0} t e^{2 t}+B_{0} t e^{-2 t}+C_{1} t+C_{0} \). The next steps involve differentiation of \( y_P(t) \) and substitution back into the differential equation to eventually solve for these undetermined coefficients.

Particular Solution

The concept of a particular solution in the context of differential equations is a specific solution to a nonhomogeneous differential equation that encompasses the impact of the nonhomogeneous term. Unlike the general solution of a homogeneous equation which comprises constants of integration, a particular solution has fixed constants specifically chosen to satisfy the nonhomogeneous term of the equation.

For example, in the differential equation \( y^{\prime \[ 2pt \prime \]} + \alpha y^{\[ 2pt \prime \]} + \beta y = g(t) \) with \( g(t) \) being a specific function of \( t \) such as \( 3e^{2t} - e^{-2t} + t \), a particular solution \( y_P(t) \) will be structured to directly incorporate the features of \( g(t) \). In our solved example, the particular solution is postulated as \( y_P(t)=A_{0} t e^{2 t}+B_{0} t e^{-2 t}+C_{1} t+C_{0} \). Correctly identifying the form of \( y_P(t) \) and solving for its coefficients ensures the equation balances out to \( g(t) \) correctly, thus solving the original nonhomogeneous differential equation.