Problem 54 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The differential equation is \[ D^2 + \frac{1}{4} = 0 \].

Step 1: Analyze the General Solution Given

The given general solution is \( y = c_1 \cosh \frac{1}{2}x + c_2 \sinh \frac{1}{2}x \). This indicates that the solution involves hyperbolic functions \(\cosh\) and \(\sinh\) with an argument of \(\frac{1}{2}x\).

Step 2: Recall Derivatives of Hyperbolic Functions

Recall that \(\frac{d}{dx} \cosh(u) = \sinh(u) \cdot \frac{du}{dx}\) and \(\frac{d}{dx} \sinh(u) = \cosh(u) \cdot \frac{du}{dx}\). If \(u = \frac{1}{2}x\), then \(u' = \frac{1}{2}\). Thus: \[ \frac{d}{dx} \cosh \frac{1}{2}x = \frac{1}{2} \sinh \frac{1}{2}x \] \[ \frac{d}{dx} \sinh \frac{1}{2}x = \frac{1}{2} \cosh \frac{1}{2}x \]

Step 3: Equate to the Characteristic Equation Form

Hyperbolic functions \(\cosh(ax)\) and \(\sinh(ax)\) in solutions often arise from roots of the characteristic equation of the form \(r = \pm ai\). The presence of \(\frac{1}{2}\) in the argument suggests roots are \(r = \pm \frac{1}{2}i\).

Step 4: Construct the Differential Equation

The characteristic equation for roots \(r = \pm \frac{1}{2}i\) is \(r^2 + \frac{1}{4} = 0\). Expanding, the factorized form of this would lead to the differential operator: \[ \left(D^2 + \frac{1}{4}\right)y = 0 \] where \(D\) denotes the differentiation operator. This is the homogeneous linear differential equation with constant coefficients.

Key Concepts

general solution

A general solution of a differential equation is a solution that contains all possible particular solutions of the equation. They typically involve arbitrary constants, which denote the broad set of functions that can solve the differential equation. In our exercise, the general solution is given by:

• \( y = c_1 \cosh \frac{1}{2}x + c_2 \sinh \frac{1}{2}x \)

Here, \( c_1 \) and \( c_2 \) are arbitrary constants, and the solution features hyperbolic functions: \( \cosh \) (hyperbolic cosine) and \( \sinh \) (hyperbolic sine) functions, with the argument \( \frac{1}{2}x \).
The presence of hyperbolic functions typically hints that the roots of the characteristic equation are imaginary. This is due to hyperbolic functions being analogs of the trigonometric sine and cosine functions, often used for solutions involving imaginary number roots in differential equations.
The significance of having arbitrary constants \( c_1 \) and \( c_2 \) is that they reflect the infinite possible adjustments to fit initial or boundary conditions in physical and real-world problems.

constant coefficients

A homogeneous linear differential equation with constant coefficients means that the coefficients of the terms in the differential equation are constants. This form of the differential equation can be succinctly expressed as a polynomial of the differentiation operator. For instance, if an equation is written as

• \( a_n D^n y + a_{n-1} D^{n-1} y + \cdots + a_1 D y + a_0 y = 0 \)

The coefficients \( a_n, a_{n-1}, \ldots, a_0 \) are all constants.
This characteristic simplifies the analysis considerably because it allows the solutions to be exponential functions (or combinations of exponential functions), enabling the use of exponential properties to solve them.
In the given solution of our exercise, the constant coefficient system is exemplified by the derived differential equation:

• \( \left(D^2 + \frac{1}{4}\right)y = 0 \)

Here, the coefficient \( \frac{1}{4} \) is constant, making it easier to manipulate and factorize the operator, forming solutions like the one given in hyperbolic terms.

characteristic equation

The characteristic equation is fundamental in finding solutions to linear differential equations with constant coefficients. It arises from assuming a solution of a specific form and substituting it back into the differential equation. In this context, you assume a solution of the form \( y = e^{rx} \) for constant coefficient differential equations. Upon substitution, you derive a polynomial equation, the roots of which determine the nature of the solution to your differential equation.
In our problem, the characteristic equation comes as:

• \( r^2 + \frac{1}{4} = 0 \)

The roots here are imaginary: \( r = \pm \frac{1}{2}i \). This means our solution involves hyperbolic sine and cosine functions, \( \sinh \) and \( \cosh \) respectively, as seen in our provided general solution.
Using the characteristic equation, we can proceed to build both the general and particular solutions by combining expressions derived from the roots. It’s a powerful technique to systematically solve and understand differential equations by breaking them down to manageable algebraic analysis.