Question

Suppose it is known that Eq. (1) has as solutions the functions \(t\) and \(\sin t\). Find the order \(n\) of the equation.

Short answer

Answer: The order of the given differential equation, Eq. (1), is 4.

Step-by-step solution

Differentiate the given solutions

Start by differentiating the solutions \(t\) and \(\sin t\) with respect to \(t\). For \(t\): \begin{align} y_1(t) &= t \\ y_1'(t) &= 1 \\ y_1''(t) &= 0 \\ \end{align} Continue differentiating if needed. For \(\sin t\): \begin{align} y_2(t) &= \sin t \\ y_2'(t) &= \cos t \\ y_2''(t) &= -\sin t \\ y_2'''(t) &= -\cos t \\ y_2^{(4)}(t) &= \sin t \\ \end{align}

Look for a common linear differential equation

We are looking for a linear differential equation with constant coefficients that is satisfied by both solutions. One possible differential equation is formed using the second and fourth derivatives of the given functions. \begin{equation} a_2 y''(t) + a_4 y^{(4)}(t) = 0 \end{equation} where \(a_2\) and \(a_4\) are constants.

Verify that the given solutions satisfy the found linear differential equation

Plug the given solutions and their derivatives into the equation (2) to verify that they satisfy the equation. For \(t\): \begin{equation} a_2 y_1''(t) + a_4 y_1^{(4)}(t) = a_2(0) + 0 = 0 \end{equation} For \(\sin t\): \begin{equation} a_2 y_2''(t) + a_4 y_2^{(4)}(t) = a_2(-\sin t) + a_4(\sin t) = 0 \end{equation} For the equation to be satisfied, we need \(a_4 = a_2\). As long as \(a_2\) and \(a_4\) are non-zero constants, both solutions satisfy the found differential equation.

Determine the order of the equation

Since the highest derivative involved in our equation (2) is the fourth derivative, the order \(n\) of the equation is 4. Therefore, the order \(n\) of the differential equation Eq. (1) is 4.

Key concepts

Linear Differential Equation

A linear differential equation is a crucial concept in the field of differential equations and appears frequently in various applications across engineering, physics, and other sciences. It is characterized by the equation involving derivatives of an unknown function that appear in a linear fashion. This means that the unknown function and its derivatives are not multiplied together, raised to a power other than one, or composed with other functions.

Such differential equations can be written in the general form \[ a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + \ldots + a_1(t)y' + a_0(t)y = g(t) \],where \(y\) is the unknown function of the independent variable \(t\), the primes denote derivatives with respect to \(t\), \(a_n(t)\), \(a_{n-1}(t)\), ..., \(a_0(t)\) are coefficient functions, and \(g(t)\) is a given function. In the case where all the coefficients are constants, we have a special subcategory called linear differential equations with constant coefficients.

In our given problem, we use this principle to identify a linear differential equation that satisfies both proposed solutions, \(t\) and \(\sin t\). By observing the derived equations for these functions, we can construct a common linear differential equation.

Derivative of Trigonometric Functions

Understanding the derivatives of trigonometric functions is fundamental to solving many problems in calculus. Calculating the derivatives of functions like \(\sin t\), \(\cos t\), and others is a common task. The basic derivatives that one must know are:

• The derivative of \(\sin t\) is \(\cos t\).
• The derivative of \(\cos t\) is \(-\sin t\).

These derivatives follow a repetitive pattern, which is particularly useful when dealing with higher-order derivatives, such as those required in our exercise. The process of differentiation often leads to a cyclical pattern every \(\pi\) radians, which is a valuable characteristic when trying to find a common differential equation to fit a set of functions. The exercise illustrates this with the higher derivatives of \(\sin t\), showing this cyclical behavior that simplifies the process of constructing our differential equation.

Constant Coefficients

When we talk about differential equations with constant coefficients, we refer to equations where the coefficients \(a_n\), \(a_{n-1}\), ..., \(a_0\) are constants rather than functions of the independent variable \(t\). This is a significant simplification because it means that the behavior of the differential equation is uniform across any interval of the independent variable. Constant coefficients are often seen in physical systems where they represent some consistent property, such as mass or spring stiffness.

In our specific example, finding a linear differential equation with constant coefficients that both \(t\) and \(\sin t\) satisfy is the goal. The equation we construct, \(a_2 y'' + a_4 y^{(4)} = 0\), illustrates a situation where the coefficients are constant. Once we have established that the order of the highest derivative is four, and the constant coefficients \(a_2\) and \(a_4\) are non-zero and equal, we conclude that our differential equation is of the fourth order with constant coefficients, satisfying both proposed solutions.