Problem 58 - Advanced Engineering Mathematics

Question

In Problems 49-58 find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} \cos x+c_{2} \sin x+c_{3} \cos 3 x+c_{4} \sin 3 x $$

Step-by-Step Solution

Answer

The differential equation is \(y'''' + 10y'' + 9y = 0\).

Step 1: Identify the Form of the Solution

The general solution is given as \( y = c_{1} \cos x + c_{2} \sin x + c_{3} \cos 3x + c_{4} \sin 3x \). This indicates that the solution consists of a combination of trigonometric functions with arguments \(x\) and \(3x\).

Step 2: Determine the Characteristic Equation

Each trigonometric term corresponds to a complex root of the form \(r = \pm i, \pm 3i\). Therefore, the characteristic equation of the differential equation is \((r^2 + 1)(r^2 + 9) = 0\).

Step 3: Expand the Characteristic Equation

Expand \((r^2 + 1)(r^2 + 9) = 0\) to get the characteristic polynomial: \(r^4 + 10r^2 + 9 = 0\). This polynomial represents the characteristic equation of the differential equation.

Step 4: Write the Homogeneous Differential Equation

Based on the characteristic polynomial \(r^4 + 10r^2 + 9 = 0\), the corresponding homogeneous linear differential equation with constant coefficients is \(y'''' + 10y'' + 9y = 0\). This differential equation will have the given general solution.

Key Concepts

Characteristic Equation

When dealing with linear differential equations, the characteristic equation is a crucial concept. It's derived from the differential equation itself but in a simplified algebraic form. From the original step-by-step solution, we see that the characteristic equation is linked to the trigonometric functions in the general solution. Specifically, for each term in the solution that involves trigonometric functions like \(\cos x\) or \(\sin x\), there are corresponding roots. These roots are complex numbers, and in our case, they are \(\pm i\) and \(\pm 3i\).

These roots form the basis of the characteristic equation. From the exercise, we know that the form \((r^2 + 1)(r^2 + 9) = 0\) needs to be expanded. When it's expanded, we end up with a polynomial: \(r^4 + 10r^2 + 9 = 0\). This equation is significant because it connects directly to the formation of the differential equation, expressing it in terms of a single polynomial equation.

Trigonometric Functions

Trigonometric functions like \(\cos x\) and \(\sin x\) are often seen in solutions for differential equations. These functions, when included in a solution, indicate that the associated differential equation has complex roots. In the exercise solution, the given form \(y = c_{1} \cos x + c_{2} \sin x + c_{3} \cos 3x + c_{4} \sin 3x\) shows how these periodic functions combine with coefficients \(c_1, c_2, c_3,\) and \(c_4\).

Each pair of trigonometric functions refers to specific frequencies, \(x\) and \(3x\), which indicate cycles and periodicism in the differential equation's solutions. These pairs correlate to complex roots in the characteristic equation, with \(x\) corresponding to roots \(\pm i\) and \(3x\) to \(\pm 3i\).

The presence of such functions modifies how we perceive solutions, making them periodic, and helps in mapping these solutions back to real-world physical phenomena, like oscillations or waves.

Constant Coefficients

Linear differential equations with constant coefficients are simpler since their coefficients do not change with \(x\). These equations are expressed in constant forms such as \(y'''' + 10y'' + 9y = 0\). With constant coefficients, the root-finding process becomes more systematic, allowing solutions to be deduced using the characteristic equation.

In the original exercise, once we derived the characteristic polynomial \(r^4 + 10r^2 + 9 = 0\), it directly led to forming the corresponding differential equation. This indicates that the coefficients of the polynomial are directly tied to those in the differential equation.

Understanding this concept helps in solving and analyzing differential equations efficiently, as constant coefficients make it feasible to use algebraic methods to predict the behavior of the solutions. This predictability is very valuable in applications across physics and engineering.