Problem 45 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}-2 y^{\prime}-3 y=4 e^{x}-9$$

Step-by-Step Solution

Answer

The general solution is \( y = C_1e^{3x} + C_2e^{-x} - e^x + 3 \).

Step 1: Identify the homogeneous equation

The given differential equation is \( y'' - 2y' - 3y = 4e^x - 9 \). First, consider the homogeneous part \( y'' - 2y' - 3y = 0 \). This is the associated homogeneous equation.

Step 2: Solve the homogeneous equation

To solve \( y'' - 2y' - 3y = 0 \), find the characteristic equation: \( r^2 - 2r - 3 = 0 \). Factor this equation to get \( (r-3)(r+1) = 0 \), so \( r = 3 \) and \( r = -1 \). Thus, the complementary solution is \( y_c = C_1e^{3x} + C_2e^{-x} \).

Step 3: Formulate the particular solution

For the non-homogeneous part \( 4e^x - 9 \), assume a particular solution of the form \( y_p = Ae^x + B \). This form is chosen because it covers both the exponential and constant function on the right-hand side.

Step 4: Determine coefficients for particular solution

Substitute \( y_p = Ae^x + B \) into the original equation: \( y_p'' - 2y_p' - 3y_p = 4e^x - 9 \). Calculate derivatives: \( y_p' = Ae^x \) and \( y_p'' = Ae^x \). Substitute: \( Ae^x - 2Ae^x - 3(Ae^x + B) = 4e^x - 9 \). Simplify to obtain \( -4Ae^x - 3B = 4e^x - 9 \).

Step 5: Solve the system of equations

Equate the coefficients from both sides: \(-4A = 4\) and \(-3B = -9\). Solving these gives \( A = -1 \) and \( B = 3 \). Thus, \( y_p = -e^x + 3 \).

Step 6: Write the general solution

Combine the complementary and particular solutions to form the general solution: \( y = y_c + y_p = C_1e^{3x} + C_2e^{-x} - e^x + 3 \).

Key Concepts

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions of linear differential equations with constant coefficients. This method assumes a specific form for the particular solution based on the function on the right-hand side of the differential equation.
For example, if the non-homogeneous term is of the form \(4e^x - 9\), we assume the particular solution to be of the form \(Ae^x + B\). This is because the non-homogeneous term consists of an exponential function \(e^x\) and a constant \(-9\).
To apply this method effectively:

• Determine the form of the non-homogeneous term.
• Choose a particular solution that mirrors the form of this term.
• Calculate derivatives, substitute into the equation, and simplify.
• Compare coefficients to solve for the unknown constants.

Undetermined coefficients is a powerful technique, especially when the differential equation fits neatly into its assumptions.

Homogeneous Equations

A homogeneous differential equation refers to an equation where the right-hand side is zero, such as \( y'' - 2y' - 3y = 0 \). The solution to this part of the differential equation provides the 'complementary solution,' which is crucial for forming the final general solution.
The steps to handle a homogeneous equation include finding the characteristic equation. For the example given, the characteristic equation becomes \( r^2 - 2r - 3 = 0 \), which can be factored into \((r-3)(r+1) = 0\). Solving this gives the roots \( r = 3 \) and \( r = -1 \).
The complementary solution uses these roots to express the solution in terms of exponentials: \( y_c = C_1e^{3x} + C_2e^{-x} \). Each root corresponds to a term in this solution, where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions or boundary values.

Particular Solutions

Finding a particular solution for a differential equation involves guessing a function that, when substituted into the equation, satisfies the non-homogeneous part of the equation. In our example, the non-homogeneous term is \(4e^x - 9\).
The trick is to choose a form for the particular solution that reflects the structure of the non-homogeneous term. Here, an assumed solution form was \( y_p = Ae^x + B \). This selection is because it covers the exponential function \(e^x\) and the constant part \(-9\).
After assuming this form, the next steps are:

• Differentiate the assumed solution as required.
• Substitute the derivatives back into the original differential equation.
• Simplify and compare coefficients to find the values of \(A\) and \(B\).

In the example, this process resulted in \( A = -1 \) and \( B = 3 \), leading to a particular solution of \( y_p = -e^x + 3 \).

General Solution

The general solution of a linear differential equation is formed by combining the complementary solution, which solves the homogeneous part, and the particular solution, which addresses the non-homogeneous part.
In this exercise, the complementary solution \( y_c = C_1e^{3x} + C_2e^{-x} \) was derived from solving the homogeneous equation. The particular solution \( y_p = -e^x + 3 \) was identified through the method of undetermined coefficients.
Therefore, the general solution is a combination: \( y = y_c + y_p = C_1e^{3x} + C_2e^{-x} - e^x + 3 \).
This solution encompasses all possible solutions based on various initial or boundary conditions. The constants \( C_1 \) and \( C_2 \) allow for adaptability to specific situations, offering a comprehensive solution to the differential equation.