Solve The Given Differential Equation By Undetermined Coefficients

Solving Differential Equations Using the Method of Undetermined Coefficients

The method of undetermined coefficients is a powerful technique for solving non-homogeneous linear ordinary differential equations (ODEs). It's particularly useful when the non-homogeneous term (the forcing function) is a relatively simple function, such as a polynomial, exponential, sine, cosine, or a combination thereof. This method leverages the structure of the forcing function to guess a particular solution, which is then combined with the complementary solution (obtained from the homogeneous equation) to find the general solution. This article will provide a comprehensive guide to understanding and applying this method, covering various scenarios and complexities.

Understanding the Basics

Before diving into the method, let's establish some foundational concepts:

1. Linear Ordinary Differential Equations (ODEs): These are equations involving a function and its derivatives, where the function and its derivatives appear linearly (no squares, products, or other non-linear terms). A general form is:

a_n(x)y⁽ⁿ⁾ + a_n-1(x)y^(n-1) + ... + a₁(x)y' + a₀(x)y = f(x)

where:

• y is the dependent variable (the function we're trying to find)
• x is the independent variable
• a_i(x) are coefficient functions
• f(x) is the forcing function or non-homogeneous term
• y⁽ⁿ⁾ denotes the nth derivative of y with respect to x

2. Homogeneous vs. Non-homogeneous ODEs: If f(x) = 0, the ODE is homogeneous. Otherwise, it's non-homogeneous. The solution to a non-homogeneous ODE consists of two parts:

• Complementary Solution (y_c): The solution to the associated homogeneous equation (set f(x) = 0). This often involves finding the roots of the characteristic equation.
• Particular Solution (y_p): A solution that satisfies the non-homogeneous equation. This is where the method of undetermined coefficients comes into play.

3. The General Solution: The general solution to the non-homogeneous ODE is the sum of the complementary and particular solutions: y = y_c + y_p.

The Method of Undetermined Coefficients: A Step-by-Step Guide

The core idea behind this method is to make an educated guess about the form of the particular solution (y_p) based on the form of the forcing function f(x). The guess will include undetermined coefficients that we'll solve for by substituting the guess into the original ODE.

1. Find the Complementary Solution (y_c):

First, solve the associated homogeneous equation. This typically involves finding the characteristic equation, determining its roots, and constructing the complementary solution based on the nature of the roots (real distinct, real repeated, complex conjugate).

Example: For the homogeneous equation y'' + 2y' + y = 0, the characteristic equation is r² + 2r + 1 = 0, which factors to (r+1)² = 0. This has a repeated root r = -1. The complementary solution is y_c = c₁e^-x + c₂xe^-x, where c₁ and c₂ are arbitrary constants.

2. Determine the Form of the Particular Solution (y_p):

This is the crucial step. The form of y_p depends on the form of f(x):

• If f(x) is a polynomial of degree n: y_p should be a polynomial of degree n with undetermined coefficients.
• If f(x) = e^ax: y_p = Ae^ax, where A is an undetermined coefficient.
• If f(x) = sin(bx) or cos(bx): y_p = Acos(bx) + Bsin(bx), where A and B are undetermined coefficients.
• If f(x) is a product or sum of these functions: y_p will be a combination of the corresponding forms. For instance, if f(x) = xe^2xcos(3x), the form of y_p will be a more complex combination including polynomials, exponentials, sines, and cosines.

Important Note: If any term in your assumed y_p is already present in y_c, you must modify your guess. Multiply the repeated term(s) by x (or x² if the repeated term appears twice in y_c, and so on).

3. Substitute y_p into the ODE and Solve for the Undetermined Coefficients:

Substitute your assumed y_p (including its derivatives) into the original non-homogeneous ODE. Equate coefficients of like terms on both sides of the equation. This will give you a system of algebraic equations that you can solve for the undetermined coefficients.

4. Combine y_c and y_p to Obtain the General Solution:

Finally, combine the complementary solution (y_c) and the particular solution (y_p) to get the general solution: y = y_c + y_p.

Advanced Scenarios and Examples

Let's explore some more complex scenarios and illustrate the method with detailed examples.

Example 1: Polynomial Forcing Function

Solve: y'' - y' - 2y = 4x²

1. Complementary Solution: The characteristic equation is r² - r - 2 = 0, which factors to (r-2)(r+1) = 0. The roots are r = 2 and r = -1. Thus, y_c = c₁e^2x + c₂e^-x.

2. Particular Solution: Since f(x) = 4x², we assume y_p = Ax² + Bx + C.

3. Substitution and Solving for Coefficients:

Substituting y_p into the ODE:

2A - (2Ax + B) - 2(Ax² + Bx + C) = 4x²

Equating coefficients:

• x²: -2A = 4 => A = -2
• x: -2A - 2B = 0 => B = 2
• constant: 2A - B - 2C = 0 => C = -3

Therefore, y_p = -2x² + 2x - 3.

4. General Solution: y = y_c + y_p = c₁e^2x + c₂e^-x - 2x² + 2x - 3

Example 2: Exponential Forcing Function

Solve: y'' + 4y' + 4y = 6e^-2x

1. Complementary Solution: The characteristic equation is r² + 4r + 4 = 0, which factors to (r+2)² = 0. The repeated root is r = -2. Therefore, y_c = c₁e^-2x + c₂xe^-2x.

2. Particular Solution: Since f(x) = 6e^-2x, we initially might guess y_p = Ae^-2x. However, e^-2x and xe^-2x are already present in y_c. We must modify our guess to y_p = Ax²e^-2x.

3. Substitution and Solving for Coefficients: (This step involves lengthy differentiation and substitution; the details are omitted for brevity. The result is A = 3).

4. General Solution: y = c₁e^-2x + c₂xe^-2x + 3x²e^-2x

Example 3: Trigonometric Forcing Function

Solve: y'' + y = 2sin(x)

1. Complementary Solution: The characteristic equation is r² + 1 = 0, with roots r = ±i. Thus, y_c = c₁cos(x) + c₂sin(x).

2. Particular Solution: Since f(x) = 2sin(x), we initially assume y_p = Acos(x) + Bsin(x). However, both cos(x) and sin(x) are present in y_c. We modify our guess to y_p = Axcos(x) + Bxsin(x).

3. Substitution and Solving for Coefficients: (Again, the detailed substitution is omitted. The result is A = -1 and B = 0).

4. General Solution: y = c₁cos(x) + c₂sin(x) - xcos(x)

Conclusion

The method of undetermined coefficients provides an efficient approach to solving non-homogeneous linear ODEs with relatively simple forcing functions. Understanding the process of finding the complementary solution, correctly assuming the form of the particular solution (handling repeated terms carefully), and solving for the undetermined coefficients are key to mastering this technique. Remember that the complexity of the calculations increases with the complexity of the forcing function, but the underlying principle remains the same. Practicing with various examples is crucial to building proficiency in this important method for solving differential equations.