Problem 11 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given system of differential equations by systematic elimination. $$\begin{array}{l} \left(D^{2}-1\right) x-y=0 \\ (D-1) x+D y=0 \end{array}$$

Step-by-Step Solution

Answer

Solve the characteristic polynomial for \(x\), then use \(x = C_1 e^{m_1 t} + C_2 e^{m_2 t} + C_3 e^{m_3 t}\) and \(y = (D^2 - 1)x\).

Step 1: Express in Operator Form

First, rewrite the system of differential equations using operator notation. The given equations are:\[(D^2 - 1)x - y = 0\]\[(D - 1)x + Dy = 0\] Where \(D\) represents the differential operator \(\frac{d}{dt}\).

Step 2: Solve for y in terms of x

From the first equation, express \(y\) in terms of \(x\):\[y = (D^2 - 1)x\] We'll use this expression to substitute \(y\) in the second equation.

Step 3: Substitute for y

Substitute \(y = (D^2 - 1)x\) into the second equation:\[(D - 1)x + D((D^2 - 1)x) = 0\]Which simplifies to:\[(D - 1)x + (D^3 - D)x = 0\]

Step 4: Simplify the x Equation

Combine like terms in the x equation:\[(D^3 + D - 1)x = 0\]This is now a differential equation solely in terms of \(x\).

Step 5: Solve the Characteristic Equation for x

Solve the characteristic equation of the differential operator:\[m^3 + m - 1 = 0\]Factor or find roots. A logical approach is using trial and error for simple roots like ±1 or polynomial factorization if necessary.

Step 6: Determine the General Solution for x

Suppose the roots found are \(m_1, m_2, m_3\). The general solution for \(x\) will take the form:\[x = C_1 e^{m_1 t} + C_2 e^{m_2 t} + C_3 e^{m_3 t}\]Where \(C_1, C_2, C_3\) are constants.

Step 7: Solve for y using x's solution

Using the relation \(y = (D^2 - 1)x\), substitute the general solution for \(x\) back to find a corresponding solution for \(y\):\[y = (D^2 - 1)(C_1 e^{m_1 t} + C_2 e^{m_2 t} + C_3 e^{m_3 t})\]Compute the derivatives to determine \(y\).

Key Concepts

Differential Operator

In the world of differential equations, a differential operator plays a pivotal role. It's a symbolic representation that simplifies the operation of differentiation. For instance, in the given exercise, the operator \(D\) is used to represent \(\frac{d}{dt}\), making the equations more compact.

By using the differential operator, it becomes easier to manipulate the terms in differential equations as if they are algebraic equations. This simplifies complex systems and allows us to systematically eliminate variables or isolate terms. In our example, expressions like \((D^2 - 1)x - y = 0\) show how operators can concisely handle derivatives.

Using the differential operator helps streamline the process of solving differential equations. It transforms the problem into a more manageable form and is invaluable in systematic elimination methods.

Characteristic Equation

Once we have our differential equation neatly packaged using operators, the next step is often to derive its characteristic equation. This involves substituting potential solutions in the form of exponential functions \(e^{mt}\) into the differential equation.

The characteristic equation springs from this substitution because when you apply the differential operator to an exponential function, it scales the function by the exponent. For example, \(D e^{mt} = m e^{mt}\). In our exercise, the characteristic equation for \((D^3 + D - 1)x = 0\) is found by replacing \(D\) with \(m\).

We then get a polynomial equation like \(m^3 + m - 1 = 0\). Solving this polynomial gives us the roots, which indicate the possible values of \(m\) and form the basis of the equation's general solution. The characteristic equation transforms the problem from calculus into algebra, which is typically much easier to handle.

General Solution

The general solution of a differential equation represents a family of possible solutions that satisfy the equation. This solution incorporates constants that can be determined if initial conditions are provided.

From the earlier step of solving the characteristic equation \(m^3 + m - 1 = 0\), suppose we found roots \(m_1, m_2, m_3\). Each root corresponds to a term in the general solution for \(x\). The form of the general solution for \(x\) in our case will be:

• \(x = C_1 e^{m_1 t} + C_2 e^{m_2 t} + C_3 e^{m_3 t}\)

where \(C_1, C_2,\) and \(C_3\) are arbitary constants.

The exponential form \(e^{mt}\) comes from the property of differential operators acting on exponentials, as these handle varying frequencies and growth rates possible in dynamic systems. This general solution indicates all possible behaviors of the system described by the differential equations.

Elimination Method

The elimination method is a clever way to solve systems of differential equations by reducing the number of variables. It's somewhat akin to elimination methods used in algebraic equations.

In our exercise, we have two equations and two unknowns, \(x\) and \(y\). By expressing \(y\) in terms of \(x\) using one of the equations, we eliminate \(y\) from the system, reducing it to one equation that only involves \(x\).

The substitution \(y = (D^2 - 1)x\) into the second equation results in a single differential equation in \(x\).

• This step simplifies solving the problem because it narrows down the complexity to a single equation.
• You handle one unknown at a time instead of juggling multiple variables simultaneously.

Post elimination, we're left with an equation that can be tackled using our standard techniques like finding the characteristic equation and determining the general solution. This method is fundamental in solving larger systems where handling multiple variables simultaneously can become unwieldy.