Problem 8 - 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 \(L\). $$y^{\prime \prime \prime}+4 y^{\prime \prime}+3 y^{\prime}=x^{2} \cos x-3 x$$

Step-by-Step Solution

Answer

The operator is \(L = D(D+1)(D+3)\) and \(g(x) = x^2 \cos x - 3x\).

Step 1: Identify the Differential Operator

The given differential equation is \(y''' + 4y'' + 3y' = x^2 \cos x - 3x\). Here, we need to identify the linear operator \(L(y)\) applied to \(y\), which corresponds to the left side of the equation.

Step 2: Write the Differential Operator

From the left side of the equation, we have the operator \(L(y) = D^3(y) + 4D^2(y) + 3D(y)\), where \(D^n\) denotes the nth derivative of function \(y\). This defines the linear differential operator \(L\).

Step 3: Express the Equation in Operator Form

Now, rewrite the given differential equation using the linear operator \(L\): \(L(y) = y''' + 4y'' + 3y' = x^2 \cos x - 3x\). Therefore, it is already in the desired form, \(L(y) = g(x)\), where \(g(x) = x^2 \cos x - 3x\).

Step 4: Factor the Differential Operator

Consider factoring the linear operator \(L = D^3 + 4D^2 + 3D\). Factor out a common factor \(D\): \(L = D(D^2 + 4D + 3)\). Next, factor the quadratic \(D^2 + 4D + 3\) as \((D+1)(D+3)\). Thus, the factored form of \(L\) is \(D(D+1)(D+3)\).

Key Concepts

Differential Operator

A differential operator is a tool in mathematics used primarily for handling differentiation. To understand it, think of it as a symbolic representation of taking derivatives. In our context, it helps us express operations like taking the first, second, and higher-order derivatives of a function. For the linear differential equation given, the differential operator is represented as \( L(y) \), which includes all the derivatives on the left-hand side.

In the example we're looking at, the differential operator \(L\) is applied to the function \(y\), and is composed of terms like \(D^3(y)\), \(4D^2(y)\), and \(3D(y)\). Here, \(D^n(y)\) simply means \(\frac{d^n y}{dx^n}\), or the \(n\)-th derivative of \(y\). This notation helps simplify complex differential equations by concentrating on the structure rather than the details of each derivative.

Hence, the operator \(L\) in this exercise captures all the ways \(y\) is differentiated, forming the equation \(L(y) = g(x)\). This shorthand is especially useful when analyzing and applying techniques to solve differential equations.

Constant Coefficients

Linear differential equations often have what are called constant coefficients. These coefficients are the parameters in the operator that don't change.

In the differential equation \(y''' + 4y'' + 3y' = x^2 \cos x - 3x\), the coefficients associated with each derivative term are \(1\) for \(D^3\), \(4\) for \(4D^2\), and \(3\) for \(3D\). These numbers are constant, meaning they don't change with respect to \(x\) or with the function \(y\).

Constant coefficients simplify the process of solving differential equations significantly. Thanks to them, techniques like factoring and applying the characteristic equation become more straightforward and predictable. They ensure that the solutions are stable, as they enforce a consistency in how each derivative interacts within the equation.

Factoring Differential Equations

Factoring a differential equation involves breaking it down into simpler components. This is analogous to factoring algebraic equations by finding common factors and expressing them in a multiplicative form.

In our given exercise, the operator \(L = D^3 + 4D^2 + 3D\) is factored by first recognizing a common term, \(D\), which represents a basic derivative. By factoring out \(D\), we transform \(L\) into \(D(D^2 + 4D + 3)\).

The next step is to factor the remaining quadratic expression \(D^2 + 4D + 3\). We can find its roots, \(-1\) and \(-3\), which allows us to express the quadratic in its factored form as \((D + 1)(D + 3)\). As a result, the entire operator can be expressed as the product \(D(D+1)(D+3)\).

Factoring is powerful because it transforms differential equations into simpler pieces. Each factored part corresponds to a simpler type of derivative operation, making it easier to solve the equation by considering these components individually.