Problem 10 - A First Course in Differential Equations with Modeling Applications

Question

W rite the given differential equation in the form \(L(y)=g(x),\) where \(L\) is a linear differential operator with constant coefficients. If possible, factor \(L\). $$y^{(4)}-8 y^{\prime \prime}+16 y=\left(x^{3}-2 x\right) e^{4 x}$$

Step-by-Step Solution

Answer

The operator is \( L = (D^2 - 4)^2 \).

Step 1: Recognize the Linear Operator Form

Given the differential equation, we need to write it in the form \( L(y) = g(x) \), where \( L \) is a linear differential operator with constant coefficients. The left side of the given equation, \( y^{(4)} - 8y'' + 16y \), naturally fits this structure as it involves derivatives of \( y \) with constant coefficients.

Step 2: Define the Linear Operator

Define the linear differential operator \( L \) as \( L = D^4 - 8D^2 + 16 \), where \( D \) denotes differentiation with respect to \( x \). Thus, \( L(y) = D^4y - 8D^2y + 16y \).

Step 3: Write the Equation Using the Operator

The differential equation can be written using the operator \( L \) as: \[ L(y) = y^{(4)} - 8y'' + 16y = (x^3 - 2x)e^{4x} \].

Step 4: Factorize the Operator, if Possible

To factorize \( L = D^4 - 8D^2 + 16 \), notice it resembles a perfect square \((D^2 - 4)^2\). Check: \((D^2 - 4)^2 = D^4 - 8D^2 + 16\). Therefore, the operator can be rewritten as \( L = (D^2 - 4)^2 \).

Key Concepts

Differential Equations

Understanding differential equations is crucial when discussing linear differential operators. A differential equation is a mathematical equation that relates some function with its derivatives. This means it can describe how a particular quantity changes over time or space. For example, in the equation \( y^{(4)} - 8y'' + 16y = (x^3 - 2x)e^{4x} \), we see derivatives up to the fourth order.

These equations are powerful tools for modeling real-world phenomena, from physics to engineering. However, solving differential equations often involves finding functions that satisfy such relationships. Sometimes, these solutions are straightforward, while in other cases, they require more sophisticated techniques, such as transforming the equation into forms that are easier to manage, such as with linear operators.

Factorization of Operators

Factorizing operators is a useful technique when working with linear differential equations. In the exercise, the operator \( L = D^4 - 8D^2 + 16 \) is factorized into \( L = (D^2 - 4)^2 \).

This process is similar to factoring algebraic expressions in solving quadratic equations, where it simplifies the problem, making analysis and solution easier. When factorizing, we aim to express the operator as a product of simpler terms, which reveals more about the structure of the equation.

• By identifying the factorized form, we can potentially apply solutions to the simpler, sub-problems, hence reducing the complexity of finding an overall solution.
• This approach can also assist in uncovering the roots and understanding the behavior of the solutions to the differential equation.

Constant Coefficients

A distinctive feature in the given differential equation is its constant coefficients. These coefficients \( (1, -8, 16) \) in the operator \( L = D^4 - 8D^2 + 16 \) stay the same, regardless of the variable \( x \). Constant coefficients simplify analysis because they do not change with space or time, allowing for more straightforward application of mathematical tools and strategies.

Due to these constant coefficients, certain solutions, such as exponential functions, can be utilized effectively. They provide simpler analyses and are more predictable when compared to equations with variable coefficients. This stability is specifically advantageous when transforming and solving these types of differential equations using methods like factorization or transformations.