Problem 4 - Advanced Engineering Mathematics

Question

In Problems 3 and 4 , fill in the blank and then write this result as a linear second-order differential equation that is free of the symbols \(c_{1}\) and \(c_{2}\) and has the form \(F\left(y, y^{\prime \prime}\right)=0\). The symbols \(c_{1}, c_{2}\), and \(k\) represent constants. \(\frac{d^{2}}{d x^{2}}\left(c_{1} \cosh k x+c_{2} \sinh k x\right)=\)

Step-by-Step Solution

Answer

The linear second-order differential equation is \(y'' - k^2 y = 0\).

Step 1: Differentiate the function once

First, differentiate the function inside the parentheses with respect to \(x\). The function given is \(c_1 \cosh(kx) + c_2 \sinh(kx)\). Therefore, the first derivative is\[\frac{d}{dx}\left(c_1 \cosh(kx) + c_2 \sinh(kx)\right) = c_1 k \sinh(kx) + c_2 k \cosh(kx).\]

Step 2: Differentiate the function a second time

Now, differentiate the result from Step 1 again with respect to \(x\):\[\frac{d^2}{dx^2}\left(c_1 \cosh(kx) + c_2 \sinh(kx)\right) = \frac{d}{dx}(c_1 k \sinh(kx) + c_2 k \cosh(kx))\]This results in:\[c_1 k^2 \cosh(kx) + c_2 k^2 \sinh(kx).\]

Step 3: Use the hyperbolic identity

To eliminate \(c_1\) and \(c_2\) from the equation, consider the original function and squared terms. Use the identity of the derivatives applied backwards:\[\text{Original function: } c_1 \cosh(kx) + c_2 \sinh(kx)\].

Step 4: Setting up the differential equation

Recognize that the equation obtained in Step 2,\[y'' = c_1 k^2 \cosh(kx) + c_2 k^2 \sinh(kx),\]is equal to \(k^2( c_1 \cosh(kx) + c_2 \sinh(kx))\) and can be simplified:\[ y'' = k^2 y.\]Rewrite this equation free from constants \(c_1\) and \(c_2\), forming the linear second-order differential equation:\[ y'' - k^2 y = 0.\]

Key Concepts

Linear Differential Equations

Linear differential equations are a type of differential equation where the unknown function and its derivatives appear to the power one, and they are multiplied by functions of only the independent variables. In simpler terms, these equations do not involve powers or products of the solution or its derivatives.
Linear second-order differential equations, like the one in our problem, take the general form:

• \[ a(x)y'' + b(x)y' + c(x)y = f(x) \]

In our specific example, we deal with a homogeneous linear second-order differential equation where the right side equals zero. This equation is seen in many physical systems, especially in mechanics and electrical circuits.
A linear differential equation like \[ y'' - k^2 y = 0 \] indicates a relationship between a function \( y \) and its second derivative. Understanding how to form and solve these equations helps model various real-world phenomena.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. Similar to how sine and cosine help describe circular motion, \( \sinh(x) \) and \( \cosh(x) \) help describe hyperbolic shapes and growth patterns. These functions are defined as:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

In our exercise, we used \( \cosh(kx) \) and \( \sinh(kx) \) to express the core problem, which indicates an association of the solution \( y \) with hyperbolic functions. Hyperbolic functions often appear in problems involving differential equations due to their natural relationship with exponential functions.
They also have useful properties such as:

• Derivatives that mirror trigonometric functions (\[ \frac{d}{dx}(\sinh(kx)) = k \cosh(kx) \text{ and } \frac{d}{dx}(\cosh(kx)) = k \sinh(kx) \])

These properties make hyperbolic functions convenient for solving second-order differential equations.

Differentiation Techniques

Differentiation is a fundamental tool in calculus, dealing with finding the rate at which a function changes. Differentiation techniques are especially crucial when working with complex functions like those involving hyperbolic functions or forming part of differential equations.
In our solution, we employed differentiation to transform the original hyperbolic function into a second-order differential equation. This process involved:

• Finding the first derivative, which describes the rate of change of the function.
• Determining the second derivative, required for the second-order differential equation.

The rules for differentiation remain consistent, including the chain rule, product rule, and standard derivative limitations for functions like \( \sinh(x) \) and \( \cosh(x) \). The ease of computing derivatives for hyperbolic functions makes them practical in many mathematical and physical applications.
Understanding how to correctly compute these derivatives ensures accurate formation and solution of differential equations, as seen when rewriting the problem in the form of \( y'' - k^2 y = 0 \).