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 \(\beta\). $$ \begin{aligned} &g(t)=-e^{t}+\sin 2 t+e^{t} \sin 2 t \\ &y_{P}(t)=A_{0} e^{t}+B_{0} t \cos 2 t+C_{0} t \sin 2 t+D_{0} e^{t} \cos 2 t+E_{0} e^{t} \sin 2 t \end{aligned} $$

Short answer

Short answer: After finding the first and second derivatives of the particular solution and plugging them back into the given nonhomogeneous differential equation, we can compare coefficients of each term to create a system of equations. The values of α and β can be found by solving this system, resulting in: $$ \alpha = \frac{1}{A_0} - \frac{4}{E_0} - 1 \\ \beta = -\frac{1}{A_0} - \alpha $$

Step-by-step solution

Differentiate the particular solution with respect to t

First, we need to find the first and second derivatives of the given particular solution, \(y_P(t)\). $$ y_P'(t) = A_0 e^t - 2B_0 t \sin 2t + 2B_0 \cos 2t + 2C_0 t \cos 2t - 2C_0 \sin 2t + (2D_0 - 4E_0) e^t \sin 2t + (4D_0 - 2E_0) e^t \cos 2t $$ $$ y_P''(t) = A_0 e^t - 4B_0 t \cos 2t - 4B_0 \sin 2t - 4C_0 t \sin 2t - 4C_0 \cos 2t - 4(2D_0 + 2E_0) e^t \sin 2t - 4(2D_0 - 2E_0) e^t \cos 2t $$

Plug the derivatives and particular solution back into the differential equation

Now, we need to substitute the particular solution and its derivatives back into the differential equation: \(y'' + \alpha y' + \beta y = g(t)\). $$ (A_0 e^t - 4B_0 t \cos 2t - 4B_0 \sin 2t - 4C_0 t \sin 2t - 4C_0 \cos 2t - 4(2D_0 + 2E_0) e^t \sin 2t - 4(2D_0 - 2E_0) e^t \cos 2t) + \alpha (A_0 e^t - 2B_0 t \sin 2t + 2B_0 \cos 2t + 2C_0 t \cos 2t - 2C_0 \sin 2t + (2D_0 - 4E_0) e^t \sin 2t + (4D_0 - 2E_0) e^t \cos 2t) + \beta (A_{0} e^{t}+B_{0} t \cos 2 t+C_{0} t \sin 2 t+D_{0} e^{t} \cos 2 t+E_{0} e^{t} \sin 2 t) = -e^{t}+\sin 2 t+e^{t} \sin 2 t $$

Compare the coefficients of each term

Next, we need to compare the coefficients of each term in the particular solution, its derivatives, and the given function g(t). From the coefficients of \(e^t\), we get: $$ \beta A_0 + \alpha A_0 + A_0 = -1 $$ From the coefficients of \(\sin 2t\), we get: $$ \beta (-4C_0) + \alpha (2B_0) - 4B_0 = 1 $$ From the coefficients of \(e^t \sin 2t\), we get: $$ \beta E_0 + \alpha (-4E_0) + 4E_0 - 8E_0 = 1 $$

Solve the system of equations for α and β

Finally, we need to solve this system of linear equations to find the values of α and β. $$ \begin{cases} \beta A_0 + \alpha A_0 + A_0 = -1 \\ \beta (-4C_0) + \alpha (2B_0) - 4B_0 = 1 \\ \beta E_0 + \alpha (-4E_0) + 4E_0 - 8E_0 = 1 \end{cases} $$ After solving the system of equations, we find the values of α and β are: $$ \alpha = \frac{1}{A_0} - \frac{4}{E_0} - 1 \\ \beta = -\frac{1}{A_0} - \alpha $$ These are the values of α and β for the given differential equation.

Key concepts

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to nonhomogeneous linear differential equations. It's powerful when dealing with nonhomogeneous terms that are exponential, sinusoidal, or polynomial in nature.

Here's how it works: you guess a form for the particular solution—this guess is informed by the type of function you're dealing with. For example, if the nonhomogeneous term, denoted as \(g(t)\), includes an exponential term, your guess should include similar exponential components. The trick is to include undetermined coefficients in your guess—these are placeholders that you solve for later.

During this process, you would differentiate your guess and substitute it into the original differential equation. By comparing coefficients from both sides of the equation, you create a system of linear equations that can be solved to find the actual values of the undetermined coefficients. These values make your guess into an actual particular solution, specifically tailored to the nonhomogeneous part of the differential equation.

Particular Solution

In the context of differential equations, a particular solution is a specific solution to a nonhomogeneous equation. It's not the general solution of the related homogeneous equation, but rather one that satisfies the entirety of the nonhomogeneous equation including the non-zero term \(g(t)\).

To identify a particular solution, you look at the form of the nonhomogeneous term and then, especially when using the method of undetermined coefficients, you guess a solution that mirrors the structure of that term but with undetermined coefficients. Once you have found the appropriate coefficients that make your equation valid, you've obtained your particular solution. Remember, this solution may not represent the general solution to the equation, as it does not incorporate the complementary solution which is derived from the associated homogeneous equation.

Nonhomogeneous Equation

A nonhomogeneous differential equation is characterized by the presence of a non-zero term, which often represents an external force or input in physical systems. Unlike homogeneous equations where solutions can be simply added together due to the superposition principle, nonhomogeneous equations require special methods, like the method of undetermined coefficients, to find their particular solutions.

For example, an equation of the form \(y'' + \beta y = g(t)\), where \(g(t)\) is non-zero and can be any function of \(t\), is nonhomogeneous. The solution to this equation consists of two parts: the complementary solution, which solves the corresponding homogeneous equation (where \(g(t)\) would be zero), and the particular solution, which is specific to the nonhomogeneous component of the equation. It's essential to understand the distinction between the homogeneous and nonhomogeneous parts when solving these types of equations.

Linear Differential Equations

Linear differential equations are fundamental in mathematics and its applications since many natural phenomena can be described by linear relationships. These equations involve an unknown function and its derivatives, and they satisfy the property of superposition – the sum of any two solutions is also a solution.

They can be either homogeneous, where the equation equals zero, or nonhomogeneous, which includes a non-zero term. In the given exercise, we are dealing with a nonhomogeneous linear differential equation with constant coefficients. The process of solving such an equation starts with finding the complementary solution of its homogeneous counterpart and then a particular solution corresponding to the nonhomogeneous part. Combining these two provides the general solution to the initial nonhomogeneous linear differential equation.