Problem 63 - A First Course in Differential Equations with Modeling Applications

Question

Verify that \(y=\sinh x-2 \cos (x+\pi / 6)\) is a particular solution of \(y^{(4)}-y=0 .\) Reconcile this particular solution with the general solution of the DE.

Step-by-Step Solution

Answer

The given function satisfies the DE, and fits into the general solution form.

Step 1: Calculate First Derivative

To get the fourth derivative, we need to calculate the first derivative: \( y = \sinh x - 2 \cos(x + \frac{\pi}{6}) \). Compute \( y' = \cosh x + 2 \sin(x + \frac{\pi}{6}) \).

Step 2: Calculate Second Derivative

Differentiate \( y' = \cosh x + 2 \sin(x + \frac{\pi}{6}) \) to get \( y'' = \sinh x + 2 \cos(x + \frac{\pi}{6}) \).

Step 3: Calculate Third Derivative

Differentiate \( y'' = \sinh x + 2 \cos(x + \frac{\pi}{6}) \) to get \( y''' = \cosh x - 2 \sin(x + \frac{\pi}{6}) \).

Step 4: Calculate Fourth Derivative

Differentiate \( y''' = \cosh x - 2 \sin(x + \frac{\pi}{6}) \) to find \( y^{(4)} = \sinh x - 2 \cos(x + \frac{\pi}{6}) \), which is the same as the original \( y \).

Step 5: Verify Solution in Differential Equation

Substitute back into the differential equation: \( y^{(4)} - y = \sinh x - 2 \cos(x + \frac{\pi}{6}) - (\sinh x - 2 \cos(x + \frac{\pi}{6})) = 0 \), showing that \( y = \sinh x - 2 \cos(x + \frac{\pi}{6}) \) is a solution.

Step 6: Identify General Solution

The differential equation \( y^{(4)} - y = 0 \) has a characteristic equation with roots \( \pm 1, \pm i \), indicating the general solution is \( y = C_1 e^x + C_2 e^{-x} + C_3 \cos x + C_4 \sin x \).

Step 7: Reconcile with Particular Solution

The particular solution \( y = \sinh x - 2 \cos(x + \frac{\pi}{6}) \) fits the general solution form when expressed as \( y = \frac{e^x-e^{-x}}{2} - 2(\cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6}) \), verifying consistency.

Key Concepts

Particular Solution

A particular solution to a differential equation is a specific solution that fits the given equation. In the context of our exercise, the function \( y = \sinh x - 2 \cos(x + \frac{\pi}{6}) \) is presented as a particular solution for the differential equation \( y^{(4)} - y = 0 \). To verify this, we calculate up to the fourth derivative of \( y \) and substitute it back into the equation. The particular solution satisfies the differential equation since when we substitute, the output simplifies to zero.

In general terms, finding a particular solution involves solving the differential equation using given initial or boundary conditions. This solution is vital for understanding how a system behaves under specific circumstances, like initial displacement in a mechanical system.

General Solution

The general solution of a differential equation encompasses all possible solutions to the equation. It includes constants to account for different initial conditions. For the equation \( y^{(4)} - y = 0 \), the general solution is expressed as: \[ y = C_1 e^x + C_2 e^{-x} + C_3 \cos x + C_4 \sin x \]This general solution is derived from solving the characteristic equation, which determines the forms of the exponential and trigonometric functions.

These constants \( C_1, C_2, C_3, \) and \( C_4 \) can be adjusted to fit particular cases. By including both exponential functions and trigonometric terms, the general solution accounts for various types of oscillatory and exponential behaviors inherent in the system described by the differential equation.

Characteristic Equation

The characteristic equation is a crucial concept in solving linear differential equations with constant coefficients. It is derived by assuming solutions of the form \( e^{rx} \), where \( r \) is a constant. For the equation \( y^{(4)} - y = 0 \), you assume \( y = e^{rx} \) and perform substitution. This leads to a polynomial equation in terms of \( r \), called the characteristic polynomial:

• \( r^4 - 1 = 0 \)

This simplifies to \((r^2 - 1)(r^2 + 1) = 0\), giving roots \( \pm 1, \pm i \). The solutions \( e^x, e^{-x}, \cos x, \) and \( \sin x \) correspond to these roots and form the basis of the general solution.

Understanding the characteristic equation helps establish the types of functions – exponential or trigonometric – that are solutions to the differential equation, guiding us in forming the general solution.

Hyperbolic Functions

Hyperbolic functions, like \( \sinh x \) and \( \cosh x \), are similar to trigonometric functions but are based on hyperbolas rather than circles. In the solution for the differential equation, the hyperbolic sine function \( \sinh x \) is used as part of the particular solution.

Defined as:

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

Hyperbolic functions appear naturally in solutions of certain differential equations, especially those involving exponential growth and decay, as they provide a rich set of behaviors similar to but distinct from sine and cosine functions.

They can be used to express various types of physical phenomena, such as the shape of a hanging cable or temperature distribution over time in a solid.