Question

Show that \(\cosh (x)\) and \(\sinh (x)\) satisfy \(y^{\prime \prime}=y\).

Short answer

Both \(\cosh(x)\) and \(\sinh(x)\) satisfy \(y'' = y\).

Step-by-step solution

Recall the definitions of hyperbolic functions

The hyperbolic cosine and sine functions are defined as follows:\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]

Differentiate \\(\cosh(x)\\) twice

First, find the first derivative of \(\cosh(x)\):\[ \cosh'(x) = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh(x) \]Now, differentiate \(\cosh'(x) = \sinh(x)\) to get the second derivative:\[ \cosh''(x) = \frac{d}{dx}(\sinh(x)) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh(x) \]Thus, we have shown \(\cosh''(x) = \cosh(x)\).

Differentiate \\(\sinh(x)\\) twice

First, find the first derivative of \(\sinh(x)\):\[ \sinh'(x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh(x) \]Now, differentiate \(\sinh'(x) = \cosh(x)\) to get the second derivative:\[ \sinh''(x) = \frac{d}{dx}(\cosh(x)) = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh(x) \]Thus, we have shown \(\sinh''(x) = \sinh(x)\).

Key concepts

Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, help us describe and analyze various phenomena, especially in mathematics and physics. Two of the most important hyperbolic functions are the hyperbolic cosine, denoted as \(\cosh(x)\), and the hyperbolic sine, denoted as \(\sinh(x)\). These functions are defined using exponentials and exhibit behaviors that are similar to the cos and sin functions in trigonometry, but are theoretically distinct.

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

These definitions help explain why hyperbolic functions are so useful, as they blend exponential behavior with symmetry properties. Notably, \(\cosh(x)\) is even, meaning \(\cosh(-x) = \cosh(x)\), while \(\sinh(x)\) is odd, meaning \(\sinh(-x) = -\sinh(x)\). Recognizing these properties can provide greater insights when solving differential equations involving hyperbolic functions.

Second Derivative

In calculus, the derivative of a function provides critical information about its behavior and characteristics like slope and curvature. The second derivative, essentially the derivative of the derivative, offers further valuable insights such as concavity and acceleration. For the hyperbolic functions, calculations of their second derivatives reveal beautiful symmetrical properties.
We start by differentiating \(\cosh(x)\):

• First derivative: \(\cosh'(x) = \sinh(x)\)
• Second derivative: \(\cosh''(x) = \cosh(x)\)

and for \(\sinh(x)\):

• First derivative: \(\sinh'(x) = \cosh(x)\)
• Second derivative: \(\sinh''(x) = \sinh(x)\)

These results demonstrate that both \(\cosh(x)\) and \(\sinh(x)\) satisfy their own differential equation of the form \(y'' = y\). Such properties are immensely helpful in various fields, including engineering and physics, due to their predictable behaviors.

Mathematical Proof

Proofs in mathematics are crucial for establishing the truth of statements, allowing us to extend knowledge and explore deeper truths. For the problem of showing \(\cosh(x)\) and \(\sinh(x)\) satisfy the differential equation \(y'' = y\), each step is a part of the logical argument that confirms this proposition.
The first step is recalling the exponential definitions of the hyperbolic functions, which are expressed with \(e^x\) terms. By systematically differentiating these forms twice, we identify that their second derivatives return the original functions:

• \(\cosh''(x) = \cosh(x)\)
• \(\sinh''(x) = \sinh(x)\)

This mathematical proof is simple yet profound as it confirms consistent behavior across a variety of hyperbolic functions. It not only underscores the natural symmetry within mathematics but also is preparatory groundwork for more complex equations and models in applied sciences.