Question

Hyperbolic functions are useful in solving differential equations. Show that the functions \(y=A \sinh k x\) and \(y=B \cosh k x,\) where \(A, B,\) and \(k\) are constants, satisfy the equation \(y^{\prime \prime}(x)-k^{2} y(x)=0\).

Short answer

Answer: Yes, both functions \(y=A \sinh kx\) and \(y=B \cosh kx\) satisfy the given differential equation \(y''(x)-k^2y(x)=0\).

Step-by-step solution

Calculate first derivative

We start by finding the first derivative of \(y=A \sinh kx\) with respect to \(x\): $$ \frac{d}{dx} (y) = \frac{d}{dx} (A \sinh kx) $$ Using the chain rule, we get: $$ y'(x) = A \cdot k \cdot \cosh kx $$

Calculate second derivative

Next, we find the second derivative of \(y=A \sinh kx\) with respect to \(x\): $$ \frac{d^2}{dx^2} (y) = \frac{d^2}{dx^2} (A \cdot k \cdot \cosh kx) $$ Using the chain rule again, we get: $$ y''(x) = A \cdot k^2 \cdot \sinh kx $$ Step 2: Substitute \(y(x)\) and \(y''(x)\) in the differential equation and check if they satisfy.

Verify the differential equation for \(y=A \sinh kx\)

Now we can substitute \(y(x)\) and \(y''(x)\) in the differential equation \(y''(x)-k^2y(x)=0\): $$ (A \cdot k^2 \cdot \sinh kx) - k^2 (A \sinh kx) = 0 $$ We can factor out \(A \cdot k^2 \cdot \sinh kx\) to get: $$ A \cdot k^2 \cdot \sinh kx (1-1)=0 $$ So the function \(y=A \sinh kx\) satisfies the given differential equation. Step 3: Repeat steps 1 and 2 for the function \(y=B \cosh kx\).

Find the derivatives for \(y=B \cosh kx\)

First derivative: $$ y'(x)= B \cdot k \cdot \sinh kx $$ Second derivative: $$ y''(x) = B \cdot k^2 \cdot \cosh kx $$

Verify the differential equation for \(y=B \cosh kx\)

Now we can substitute \(y(x)\) and \(y''(x)\) in the differential equation \(y''(x)-k^2y(x)=0\): $$ (B \cdot k^2 \cdot \cosh kx) - k^2 (B \cosh kx) = 0 $$ We can factor out \(B \cdot k^2 \cdot \cosh kx\) to get: $$ B \cdot k^2 \cdot \cosh kx (1-1)=0 $$ So the function \(y=B \cosh kx\) also satisfies the given differential equation. In conclusion, both functions $$y=A \sinh kx$$ and $$y=B \cosh kx$$ satisfy the differential equation $$y''(x)-k^2y(x)=0$$.

Key concepts

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are essential in various areas of mathematics and engineering, particularly when dealing with differential equations. The two most common hyperbolic functions are the hyperbolic sine, denoted as \(\sinh(x)\), and the hyperbolic cosine, denoted as \(\cosh(x)\). These functions help describe the shape of a hanging cable or chain, known as a catenary, which naturally takes the form of a hyperbolic cosine.
For any real number \(x\), the hyperbolic sine and cosine are defined as:
\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]
\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
These formulas illustrate the basis of hyperbolic functions in the exponential function, making them different in nature but similar in form to regular sine and cosine functions. Hyperbolic functions arise in various contexts, from mathematical models involving hyperbolic geometry to descriptions of special relativity in physics.

Chain Rule

The chain rule is a fundamental technique in calculus, used for calculating the derivative of a composite function. When functions are nested inside each other, the chain rule helps us break down the derivatives by considering the outer and inner functions separately.
If we have a composite function \(y = f(g(x))\), the chain rule states that the derivative of \(y\) with respect to \(x\) is the derivative of the outer function \(f\) evaluated at the inner function \(g(x)\), multiplied by the derivative of the inner function \(g(x)\). In formula form:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
In the context of hyperbolic functions like \(y = A \sinh(kx)\) or \(y = B \cosh(kx)\), the chain rule enables us to find derivatives by treating \(kx\) as the inner function. Thus, the derivative of \(\sinh(kx)\) becomes \(k \cosh(kx)\), and the derivative of \(\cosh(kx)\) becomes \(k \sinh(kx)\). This rule is crucial for solving differential equations involving these functions.

Second Order Differential Equation

A second order differential equation is a type of differential equation that contains the second derivative of a function. The general form is \(y''(x) + p(x)y'(x) + q(x)y(x) = g(x)\). In this context, a second order differential equation has constant coefficients, which simplifies the process of finding solutions.
The particular equation given, \(y''(x) - k^2y(x) = 0\), is a homogenous second order differential equation with constant coefficients. Here:

• \(y''(x)\) is the second derivative of the unknown function \(y\).
• \(k^2y(x)\) represents the term with the constant coefficient \(k^2\).

This kind of equation is common in physics and engineering, modeling systems such as harmonic oscillators or wave propagation. Solving these involves integrating the derivatives back to a function, which fulfills the original equation. Both \(y = A \sinh(kx)\) and \(y = B \cosh(kx)\) can solve this equation, as substituting these into the equation proves equality to zero.

Satisfying Differential Equation

For a function to satisfy a differential equation, it means that when the function is substituted into the equation, the resulting value holds true, making the equation balanced or equal to zero. Verification involves deriving the function until reaching the order specified by the differential equation and then substituting back these derivatives.
The exercise shows two hyperbolic functions, \(y = A \sinh(kx)\) and \(y = B \cosh(kx)\), that satisfy the second order differential equation \(y''(x) - k^2y(x) = 0\).
For \(y = A \sinh(kx)\):

• First derivative: \(y'(x) = A \cdot k \cdot \cosh(kx)\)
• Second derivative: \(y''(x) = A \cdot k^2 \cdot \sinh(kx)\)

Substituting these into the equation, we observe that:
\[y''(x) - k^2y(x) = A \cdot k^2 \cdot \sinh(kx) - k^2 \cdot A \sinh(kx) = 0\]
This confirms the equation holds true. Similar steps show \(y = B \cosh(kx)\) satisfies the equation as well. Hence, both functions are solutions, highlighting the usefulness of hyperbolic functions in solving such differential equations.