Problem 69 - Calculus Single Variable

Question

Show that \(y=\sinh (\omega x)\) and \(y=\cosh (\omega x)\) are solutions of the differential equation \(y^{\prime \prime}(x)-\omega^{2} y(x)=0\). Let \(a\) be a positive constant, and set $$ T(x)=a \cdot \sec h^{-1}\left(\frac{x}{a}\right)-\sqrt{a^{2}-x^{2}} $$

Step-by-Step Solution

Answer

Both \(y = \sinh(\omega x)\) and \(y = \cosh(\omega x)\) solve the differential equation. \(T(x)\) requires simplification of inverse hyperbolic functions.

Step 1: Verify Solution for \(y = \sinh(\omega x)\)

First, find the first and second derivatives of \(y = \sinh(\omega x)\). The first derivative is \( y' = \omega \cosh(\omega x) \) and the second derivative is \( y'' = \omega^2 \sinh(\omega x) \). Substitute these into the given differential equation \( y'' - \omega^2 y = 0 \): \[y'' - \omega^2 y = \omega^2 \sinh(\omega x) - \omega^2 \sinh(\omega x) = 0.\]This verifies that \( y = \sinh(\omega x) \) is a solution of the differential equation.

Step 2: Verify Solution for \(y = \cosh(\omega x)\)

Next, check if \(y = \cosh(\omega x)\) satisfies the differential equation. Compute its first derivative: \( y' = \omega \sinh(\omega x) \), and its second derivative: \( y'' = \omega^2 \cosh(\omega x) \). Substitute these into the differential equation:\[y'' - \omega^2 y = \omega^2 \cosh(\omega x) - \omega^2 \cosh(\omega x) = 0.\]Hence, \( y = \cosh(\omega x) \) is also a solution.

Step 3: Analyze Function \(T(x)\)

Let's analyze the given function \(T(x) = a \cdot \sec h^{-1}\left(\frac{x}{a}\right) - \sqrt{a^2 - x^2}\). Begin by recognizing that \( \sec h^{-1}(z) \) is the inverse hyperbolic secant, defined in terms of logarithms and \( \cosh^{-1} \). Formally, \( \sec h^{-1}(z) = \cosh^{-1}\left(\frac{1}{z}\right) \), which connects back to hyperbolic trig functions and their inverses.

Step 4: Simplify and Relate \(T(x)\) to Known Solutions

Convert the terms involving \( \sec h^{-1} \) into simpler forms via properties of inverse hyperbolic functions. Since \(T(x)\) seems unrelated to the differential equation directly through the original statement, re-express as:\[\cosh^{-1}\left(\frac{1}{x/a}\right) = \ln\left(\frac{1+\sqrt{1-(x/a)^2}}{x/a}\right).\]Subtracting \(\sqrt{a^2 - x^2}\) involves simplification relying on conditional identities of hyperbolic functions.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are a fascinating part of mathematics, often coming into play when dealing with differential equations. They are analogs of the trigonometric functions but defined using hyperbolas instead of circles. There are a few key hyperbolic functions you should know:

• Hyperbolic sine, denoted as \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• Hyperbolic cosine, denoted as \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

Hyperbolic sine and cosine have properties similar to their trigonometric counterparts, but they aren't periodic.
They also satisfy identity relations similar to trigonometric identities, such as \( \cosh^2(x) - \sinh^2(x) = 1 \).
In the context of differential equations, hyperbolic functions often appear as solutions to linear second-order equations.
They provide simple forms that can satisfy the conditions of the equations, just as sine and cosine do in some problems.

Solution Verification

Verifying that a function is a solution to a differential equation means showing that the function satisfies the equation when substituted into it.
Let's look at the equation used in our exercise:
\( y'' - \omega^2 y = 0.\)
We need to check if our proposed solutions, \( y = \sinh(\omega x) \) and \( y = \cosh(\omega x) \), satisfy this equation.For \( y = \sinh(\omega x) \):

• First derivative: \( y' = \omega \cosh(\omega x) \).
• Second derivative: \( y'' = \omega^2 \sinh(\omega x) \).

Substituting these into the differential equation yields zero, verifying it.
The same process applies to \( y = \cosh(\omega x) \), and again the equation balances.
This step is crucial as it ensures that the proposed solutions genuinely satisfy the differential equation under consideration.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are those that undo hyperbolic functions, similar to how inverse trigonometric functions work for sine, cosine, etc.
Important inverse hyperbolic functions include:

• Inverse hyperbolic sine, \( \sinh^{-1}(x) \), and
• Inverse hyperbolic cosine, \( \cosh^{-1}(x) \).

These functions are defined in terms of logarithms:

• \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \)
• \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \) for \( x \geq 1 \)

In the exercise, \( \sec h^{-1} \) appears, which is less common and defined as \( \cosh^{-1}\left(\frac{1}{x}\right) \).
Such transformations are useful in simplifying complex expressions to forms that are more manageable or meaningful.
Understanding these inverse functions provides a deeper insight into the relationships and behaviors of hyperbolic functions in solving various mathematical problems.