Question

The differential equation for transverse vibrations of a string whose density increases linearly from one end to the other is \(y^{\prime \prime}+(A x+B) y=0,\) where \(A\) and \(B\) are constants. Find the general solution of this equation in terms of Bessel functions. Hint: Make the change of variable \(A x+B=A u\).

Short answer

The general solution is \[ y(x) = C_1 J_0 \left(x + \frac{B}{A}\right) + C_2 Y_0 \left(x + \frac{B}{A}\right) \].

Step-by-step solution

- Substitute the Change of Variable

Given the hint, make the substitution in the differential equation. Define a new variable as \( u = \frac{Ax + B}{A} \). Consequently, the differential equation transforms based on this new variable.

- Find the Derivatives with respect to the New Variable

Calculate the derivatives of \(y\) with respect to the new variable \(u\). Start by noting that \( x = \frac{uA - B}{A} \). Therefore, \( \frac{dx}{du} = \frac{1}{A} \), resulting in the derivatives being transformed accordingly.

- Compute the Second Derivative

Compute the second derivative by applying the chain rule. This is needed because the original equation involves the second derivative of \(y\). Thus, \( \frac{d^2y}{dx^2} = A^2 \frac{d^2y}{du^2} \), where the extra \(A^2\) comes from the second application of the chain rule.

- Substitute Back into the Original Equation

Replace \( x \), \( y'' \) in the original equation with their equivalents in terms of \(u\): \(A^2 y'' + A u y = 0 \). Simplify to obtain \(y'' + u y = 0 \).

- Recognize as a Bessel Differential Equation

Identify that \( y'' + u y = 0 \) resembles a Bessel differential equation, where the general solution can be written involving Bessel functions of the first kind \(J_0(u)\) and second kind \(Y_0(u)\). Hence, the general solution is: \[ y(u) = C_1 J_0(u) + C_2 Y_0(u) \]

- Bring Back the Original Variable

Substitute \( u \) in terms of \(x\) back into the solution. Use \[ u = \frac{Ax + B}{A} = x + \frac{B}{A} \]. Thus, the general solution in terms of \(x\) is \[ y(x) = C_1 J_0 \left(x + \frac{B}{A}\right) + C_2 Y_0 \left(x + \frac{B}{A}\right) \].

Key concepts

Transverse Vibrations

Transverse vibrations involve oscillations of a string or a similar structure in a plane perpendicular to its length.
These kinds of vibrations are common in strings of musical instruments like guitars.
The physical property causing these vibrations is elasticity.
When a force displaces part of the string, it tends to return to its equilibrium, producing waves that travel along the length.
In mathematical terms, transverse vibration problems often reduce to solving differential equations derived from the wave equation.

Linear Density Variation

Linear density variation means that the density of the string changes linearly from one end to the other.
If you have a string fixed at both ends, one end may be heavier than the other.
This results in a variable density, often described in a linear form, such as \(\rho(x) = \rho_0(1 + \beta x)\), where \(\rho_0\) and \(\beta\) are constants.
Understanding this property is crucial because it directly affects the wave equation and, consequently, the differential equation that describes the system's vibrations.

Differential Equations

A differential equation relates a function to its derivatives.
In problems involving physics, like transverse vibrations, these equations come directly from physical laws.
For instance, the equation \(\frac{d^2 y}{dx^2} + (A x + B) y = 0 \) describes the balance of forces under linear density variation.
Solving this equation provides the displacement of the string at any point, helping us understand its vibrational mode.

Variable Substitution

Variable substitution is a technique used to simplify differential equations by changing variables.
In this problem, we use the substitution \(\frac{A x + B}{A} = u\).
This transformation converts the original variable \(x\) into a new, simpler variable \(u\).
This makes the differential equation easier to solve because it may convert it into a standard form for which solutions are already known.
Following through these changes, the process generally involves re-expressing the derivatives and substituting back into the equation.

Second Derivative

The second derivative \( y'' \) represents the rate of change of the first derivative, giving the acceleration of the vibrating string.
For our transformed variable \(u\), calculating the second derivative involves applying the chain rule.
Given our substitution, we find that \( \frac{d^2 y}{dx^2} = A^2 \frac{d^2 y}{du^2} \).
Correctly computing this is crucial because the original differential equation involves the second derivative.

General Solution

Finding the general solution of a differential equation means expressing the function \(y\) in the most general form.
For equations resembling Bessel's differential equations, solutions include Bessel functions of the first kind \(J_0(u) \) and second kind \( Y_0(u) \).
In our case, after transforming and solving the differential equation, the general solution becomes:
\[ y(x) = C_1 J_0 \bigg( x + \frac{B}{A} \bigg) + C_2 Y_0 \bigg( x + \frac{B}{A} \bigg) \].
This general solution includes constants \( C_1 \) and \( C_2 \) which can be determined based on specific boundary conditions.