Question

Consider the nonhomogeneous differential equation \(x^{\prime \prime}-3 x^{\prime}+2 x=6 e^{3 t}\). a. Find the general solution of the homogenous equation. b. Find a particular solution using the Method of Undetermined Coefficients by guessing \(x_{p}(t)=A e^{3 t}\). c. Use your answers in the previous parts to write the general solution for this problem.

Short answer

The general solution of the differential equation is \(x(t) = c_1e^{2t} + c_2e^{t} + 3e^{3t}\).

Step-by-step solution

Solve the homogeneous equation

The homogeneous equation is obtained by setting the right hand side of the given equation to zero. Hence the homogeneous equation we will solve is \(x'' - 3x' + 2x = 0\). The characteristic equation of this homogeneous differential equation is \(m^2 - 3m + 2 = 0\). Factoring this equation gives us \((m-2)(m-1) = 0\). Hence, the roots are \(m = 2, 1\). This means the general solution to the homogeneous equation is \(x_h(t) = c_1e^{2t} + c_2e^{t}\), where \(c_1\) and \(c_2\) are constants to be determined by the initial conditions.

Guess and verify a particular solution

We now find a particular solution to the non-homogeneous equation. For the non-homogeneous term in the differential equation, which takes the form \(6e^{3t}\), we can guess a particular solution of the form \(x_p(t) = Ae^{3t}\). Plug \(x_p(t)\) back into the original differential equation and solve for the constant \(A\). After calculation, we get \(A = \frac{6}{2}=3\). So a particular solution to the non-homogeneous equation is \(x_p(t) = 3e^{3t}\).

Write the general solution for the problem

The general solution of a nonhomogeneous differential equation is a combination of the homogeneous solution and the particular solution. Hence the general solution is \(x(t) = x_h(t) + x_p(t) = c_1e^{2t} + c_2e^{t} + 3e^{3t}\).

Key concepts

Homogeneous Equations

In differential equations, a homogeneous equation is one where every term is a function of the dependent variable and its derivatives. In simpler terms, it's an equation that equates to zero when solved.
For example, consider the equation from the exercise: \(x'' - 3x' + 2x = 0\). This is a linear homogeneous differential equation, where no external terms or constants are part of the equation, only the function \(x\) and its derivatives.
To solve homogeneous equations:

• Derive the characteristic equation, which is found by replacing the derivatives with powers of a variable (commonly \(m\)). For \(x'' - 3x' + 2x = 0\), the characteristic equation is \(m^2 - 3m + 2 = 0\).
• Factor the characteristic equation to discover its roots, which give the solution to the differential equation. Here, \((m-2)(m-1) = 0\) leads to roots \(m=2\) and \(m=1\).
• The general solution comes from these roots: \(x_h(t) = c_1e^{2t} + c_2e^{t}\), with \(c_1\) and \(c_2\) as constants determined by initial conditions.

Homogeneous equations form the backbone of solving more complex non-homogeneous equations, as seen in their application in obtaining the general solution.

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a clever technique used to find particular solutions of non-homogeneous linear differential equations. The idea is to "guess" a form for the particular solution based on the non-homogeneous part of the equation.
For instance, if the non-homogeneous equation includes a term like \(6e^{3t}\), you might guess a particular solution of the form \(x_p(t) = Ae^{3t}\). This is because the form of the solution should reflect the form of the non-homogeneous component.
Here are the main steps to follow:

• Guess: Propose a particular solution \(x_p(t)\) that resembles the non-homogeneous term in the differential equation. In our case, \(x_p(t) = Ae^{3t}\).
• Substitute: Insert \(x_p(t)\) into the original equation to solve for \(A\) (or other constants). This is done by equating the coefficients of like terms.
• Calculate: Perform calculations to find \(A\). For the exercise, after substitution, \(A = 3\) was found.
• Verify: Ensure \(x_p(t)\) satisfies the non-homogeneous equation.

Once \(A\) is found, you have a particular solution, which blends with the homogeneous solution for a comprehensive result.

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies the non-homogeneous part of the equation. It complements the solution of the homogeneous equation to form the full solution.
In the exercise, the non-homogeneous term \(6e^{3t}\) required us to find a specific function that, when plugged into the differential equation, makes it true. Using what we've learned from determining coefficients, we found that \(x_p(t) = 3e^{3t}\) serves as this particular solution.
The purpose of the particular solution is to account for effects not explained by the homogeneous equation alone.
Combining the homogeneous solution and the particular solution results in the general solution for the differential equation:

• The general solution combines \(x_h(t) = c_1e^{2t} + c_2e^{t}\) with \(x_p(t) = 3e^{3t}\), resulting in \(x(t) = c_1e^{2t} + c_2e^{t} + 3e^{3t}\).
• This general solution represents all possible solutions of the original differential equation, balanced by the particular influences of the non-homogeneous term.

Finding particular solutions is crucial in understanding and solving non-homogeneous linear differential equations effectively.