Problem 14 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients.In Problems \(1-26\) solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}-4 y=\left(x^{2}-3\right) \sin 2 x$$

Step-by-Step Solution

Answer

The solution is the sum of the homogeneous solution and a particular solution found by undetermined coefficients.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y'' - 4y = (x^2 - 3)\sin(2x) \). This is a non-homogeneous linear differential equation with constant coefficients.

Step 2: Solve the Homogeneous Equation

First, solve the homogeneous part, \( y'' - 4y = 0 \). The characteristic equation is \( r^2 - 4 = 0 \), which factors to \( (r-2)(r+2) = 0 \). The roots are \( r = 2 \) and \( r = -2 \). Thus, the general solution of the homogeneous equation is \( y_h = C_1 e^{2x} + C_2 e^{-2x} \).

Step 3: Form the Particular Solution Guess

For the non-homogeneous part \((x^2 - 3)\sin(2x)\), use the method of undetermined coefficients. We will try \( y_p = (Ax^2 + Bx + C)\cos(2x) + (Dx^2 + Ex + F)\sin(2x) \) as a particular solution.

Step 4: Compute the Derivatives of the Particular Solution Guess

Calculate the first and second derivatives of \( y_p \). This is straightforward but will involve the product and chain rule for derivatives of trigonometric functions, yielding quite lengthy expressions.

Step 5: Plug the Guess into the Differential Equation

Substitute \( y_p, y_p', \) and \( y_p'' \) into the original differential equation \( y'' - 4y = (x^2 - 3)\sin(2x) \). Collect terms according to powers of \( x \) and trig functions.

Step 6: Simplify and Compare Coefficients

Equate the coefficients of corresponding like terms on both sides of the equation. This will yield a system of equations in terms of \( A, B, C, D, E, \) and \( F \).

Step 7: Solve the System of Equations

Solve the system of equations obtained in Step 6 to find the values of \( A, B, C, D, E, \) and \( F \).

Step 8: Write the General Solution

Combine the particular solution \( y_p \) with the homogeneous solution to get the general solution: \( y = y_h + y_p = C_1 e^{2x} + C_2 e^{-2x} + y_p \).

Step 9: Verify the Solution

Check the solution by substituting it back into the original differential equation to ensure that both sides are equal.

Key Concepts

Non-homogeneous Differential Equation

A non-homogeneous differential equation is one that contains a term that is not a result of the function or its derivatives. In contrast to homogeneous differential equations, which have the form \( y'' + py' + qy = 0 \), non-homogeneous equations include an additional function, often denoted as \( g(x) \), on the right-hand side.
For example, the equation \( y'' - 4y = (x^2 - 3)\sin(2x) \) is non-homogeneous because of the \( (x^2 - 3)\sin(2x) \) term, which is independent of \( y \) and its derivatives.

Key characteristics:

• The presence of a non-zero function \( g(x) \) that makes it non-homogeneous.
• Solutions typically involve finding both a particular solution for the non-homogeneous part and the general solution for the homogeneous part.
• The overall solution consists of the sum of both the homogeneous and particular solutions.

Linear Differential Equations

Linear differential equations, as the name suggests, involve derivatives of a function and exhibit linearity. This means that each term in the equation is either a constant or a derivative of the function multiplied by a constant.
For example, in \( y'' - 4y = (x^2 - 3)\sin(2x) \), the left-hand side is linear as it involves \( y'' \) and \( -4y \), both of which are linear in \( y \).

Characteristics of linear differential equations:

• They do not have products or powers of the unknown function and its derivatives.
• The coefficients of \( y \), \( y' \), \( y'' \), etc., are constants or functions of the independent variable.
• They exhibit superposition, meaning the algebraic sum of any two solutions is also a solution.

Particular Solution

In the context of non-homogeneous differential equations, the particular solution is one that accounts for the non-homogeneous part or the specific input of the system.
Using the method of undetermined coefficients, we guess a suitable form for this particular solution based on the type of \( g(x) \). For the equation \( y'' - 4y = (x^2 - 3)\sin(2x) \), we used:\[ y_p = (Ax^2 + Bx + C)\cos(2x) + (Dx^2 + Ex + F)\sin(2x) \]

Step-by-step approach:

• Construct a trial solution that has similar form as \( g(x) \) and includes unknown coefficients.
• Differentiate the trial solution to find its derivatives.
• Substitute these derivatives and the particular solution back into the differential equation.
• Equate coefficients for like terms to solve for the unknown coefficients.

Characteristic Equation

The characteristic equation is used to find solutions to the homogeneous part of a linear differential equation with constant coefficients. By solving this polynomial equation, we determine the roots that guide us in forming the homogeneous solution.
For the homogeneous equation \( y'' - 4y = 0 \), the characteristic equation derived is:\[ r^2 - 4 = 0 \]which factors to \((r-2)(r+2) = 0\) resulting in roots \( r = 2 \) and \( r = -2 \).

Importance of characteristic equation:

• Solutions are formed based on the nature of the roots (real and distinct, real and repeated, or complex).
• Helps build the general solution for the homogeneous equation through terms like \( e^{rx} \).
• Provides a foundational part of the complete solution by establishing the behavior of the system without external influence.