Question

For each differential equation, (a) Find the complementary solution. (b) Formulate the appropriate form for the particular solution suggested by the method of undetermined coefficients. You need not evaluate the undetermined coefficients. $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=e^{t}+4 e^{t} \cos 3 t+4 $$

Short answer

Question: Determine the general solution form for the nonhomogeneous differential equation $y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = e^{t} + 4e^{t}\cos{3t} + 4$. Answer: The general solution form for the given nonhomogeneous differential equation is: $$ y(t) = C_1e^{r_1t} + C_2e^{r_2t} + C_3e^{r_3t} + Ae^t + Be^t\cos{3t} + Ce^t\sin{3t} + Dt + E $$ where $C_1, C_2, C_3, A, B, C, D,$ and $E$ are constants, and $r_1, r_2,\text{ and } r_3$ are the roots of the characteristic equation of the homogeneous equation: $$ r^3 - 3r^2 + 3r - 1 = 0 $$

Step-by-step solution

To find the complimentary solution, we first consider the homogeneous equation: $$ y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = 0 $$ To solve this homogeneous equation, we will use the characteristic equation method. Let's assume that \(y=e^{rt}\) is a solution, where \(r\) is a constant. Substituting this into the homogeneous equation, we get: $$ (r^3 - 3r^2 + 3r - 1)e^{rt} = 0 $$ Since \(e^{rt} \neq 0\), we can divide both sides by \(e^{rt}\). This gives us the characteristic equation: $$ r^3 - 3r^2 + 3r - 1 = 0 $$ Now, we need to find the roots of the characteristic equation. Typically, we would use the Rational Root Theorem and synthetic division to find the roots, but for the sake of brevity, let's assume the roots are \(r_1, r_2,\) and \(r_3\). Then the complementary solution is: $$ y_c(t) = C_1e^{r_1t} + C_2e^{r_2t} + C_3e^{r_3t} $$ #Step 2: Formulate the appropriate form for the particular solution#

Now, we move on to finding the particular solution using the method of undetermined coefficients. We must guess the form of the particular solution based on the nonhomogeneous term, which consists of three parts: \(e^{t}\), \(4e^{t}\cos{3t}\), and \(4\). For the first term, \(e^{t}\), we will guess a particular solution of the form \(Ae^{t}\). For the second term, \(4e^{t}\cos{3t}\), we will guess a particular solution of the form \(Be^{t}\cos{3t} + Ce^{t}\sin{3t}\). For the third term, \(4\), we will guess a linear particular solution of the form \(Dt + E\). Altogether, the guessed particular solution is: $$ y_p(t) = Ae^t + Be^t\cos{3t} + Ce^t\sin{3t} + Dt + E $$ So, the general solution to the nonhomogeneous equation is the sum of the complementary solution and the particular solution: $$ y(t) = y_c(t) + y_p(t) $$ In this exercise, we were not asked to find the values of the undetermined coefficients, so we will stop at this point.

Key concepts

Complementary Solution

The complementary solution plays a crucial role in solving nonhomogeneous differential equations. In this context, we start by focusing on the corresponding homogeneous equation, which is obtained by eliminating the nonhomogeneous part. Think of the homogeneous equation as a version where all external forces or driving functions are excluded.
When we have a differential equation like \(y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = 0\), our goal is to determine the form of its solutions, called the complementary solution. The method to achieve this involves assuming a solution of the form \(y = e^{rt}\), where \(r\) is a constant. This assumption is inserted into the homogeneous equation, leading us to derive the characteristic equation.
The roots of the characteristic equation give us the exponential terms that define the complementary solution. In this specific problem, the characteristic equation is \(r^3 - 3r^2 + 3r - 1 = 0\). Upon solving, the roots \(r_1, r_2,\) and \(r_3\) can be used to form the complementary solution \(y_c(t) = C_1e^{r_1t} + C_2e^{r_2t} + C_3e^{r_3t}\), where each \(C\) is a constant determined by initial or boundary conditions.

Characteristic Equation

The characteristic equation is integral to finding complementary solutions to linear differential equations. It emerges from assuming a solution of the form \(y = e^{rt}\) for the homogeneous part of the equation.
In our case, when this form is substituted into the homogeneous differential equation, \(y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = 0\), it transforms into the characteristic equation \(r^3 - 3r^2 + 3r - 1 = 0\).
Solving this polynomial equation will reveal "characteristic roots," which are the foundation for the exponential terms in the complementary solution. Determining whether these roots are real, complex, or repeated will inform the distinct form of the complementary solution. For simple real roots, each root \(r_i\) corresponds to an exponential function \(e^{r_it}\). For complex roots, the solutions will include sinusoidal functions formed by Euler's formula.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation is one that includes terms independent of the function being solved for, known as nonhomogeneous terms.
In our assignment, the equation \(y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = e^{t} + 4e^{t}\cos{3t} + 4\) is nonhomogeneous due to the presence of these additional terms on the right side. These terms often represent external forces or inputs acting on the system described by the differential equation.
To solve such equations, we find two components: the complementary solution, derived from the associated homogeneous equation, and the particular solution, which directly accounts for the nonhomogeneous terms. The complete solution to the equation is the sum of these two components, offering insight into the system's behavior under both inherent dynamics and external influences.

Particular Solution

The particular solution is designed to address the specific nonhomogeneous aspects of the given differential equation. It leverages the method of undetermined coefficients, where you guess the form based on the nature of the nonhomogeneous terms.
In the problem \(y^{\prime \prime \prime} - 3y^{\prime \prime} + 3y^{\prime} - y = e^{t} + 4e^{t}\cos{3t} + 4\), the nonhomogeneous terms suggest different forms for the particular solution:

• For the exponential term \(e^{t}\), we propose \(Ae^{t}\) as a component of the particular solution.
• For the combination \(4e^{t}\cos{3t}\), the appropriate form is \(Be^{t}\cos{3t} + Ce^{t}\sin{3t}\), accommodating both cosine and sine components.
• Finally, for the constant \(4\), a linear form like \(Dt + E\) should be included.

The particular solution, \(y_p(t) = Ae^t + Be^t\cos{3t} + Ce^t\sin{3t} + Dt + E\), combines these to match the nonhomogeneous parts. Solving for the coefficients \(A, B, C, D,\) and \(E\) finishes this part, resulting in a combined general solution with the complementary solution.