Question

In each of Problems 1 through 8 determine whether the given pair of functions is linearly independent or linearly dependent. \(f(t)=e^{\lambda t} \cos \mu t, \quad g(t)=e^{\lambda t} \sin \mu t, \quad \mu \neq 0\)

Short answer

Answer: The functions \(f(t)\) and \(g(t)\) are linearly independent.

Step-by-step solution

Define the functions

Given functions are: \(f(t) = e^{\lambda t} \cos \mu t\) \(g(t) = e^{\lambda t} \sin \mu t\) where \(\mu \neq 0\)

Find the first derivatives

Differentiate both functions with respect to \(t\): \(f'(t) = \lambda e^{\lambda t} \cos \mu t - \mu e^{\lambda t} \sin \mu t\) \(g'(t) = \lambda e^{\lambda t} \sin \mu t + \mu e^{\lambda t} \cos \mu t\)

Calculate the Wronskian

The Wronskian of two functions \(f(t)\) and \(g(t)\) is defined as: \(W(f,g)(t) = \begin{vmatrix} f(t) & g(t) \\ f'(t) & g'(t) \end{vmatrix}\) Now, substitute the functions and their derivatives: $W(f,g)(t) = \begin{vmatrix} e^{\lambda t} \cos \mu t & e^{\lambda t} \sin \mu t \\ \lambda e^{\lambda t} \cos \mu t - \mu e^{\lambda t} \sin \mu t & \lambda e^{\lambda t} \sin \mu t + \mu e^{\lambda t} \cos \mu t \end{vmatrix}$

Evaluate the determinant

Compute the determinant of the Wronskian: \(W(f,g)(t) = (e^{\lambda t} \cos \mu t)(\lambda e^{\lambda t} \sin \mu t + \mu e^{\lambda t} \cos \mu t) - (e^{\lambda t} \sin \mu t)(\lambda e^{\lambda t} \cos \mu t - \mu e^{\lambda t} \sin \mu t)\) After simplifying, \(W(f,g)(t) = \lambda \mu e^{2 \lambda t} \cos^2 \mu t + \mu^2 e^{2 \lambda t} \cos^2 \mu t - \lambda \mu e^{2 \lambda t} \sin^2 \mu t + \mu^2 e^{2 \lambda t} \sin^2 \mu t\) Factor out the exponential term and the \(\mu\) value: \(W(f,g)(t) = \mu e^{2 \lambda t}(\lambda \cos^2 \mu t + \mu \cos^2 \mu t - \lambda \sin^2 \mu t + \mu \sin^2 \mu t)\)

Determine the linear independence or dependence

Now, consider the Wronskian we've calculated. Notice that: \(\lambda \cos^2 \mu t + \mu \cos^2 \mu t - \lambda \sin^2 \mu t + \mu \sin^2 \mu t = \mu (\cos^2 \mu t + \sin^2 \mu t)\) From the trigonometric identity, \(\cos^2 \mu t + \sin^2 \mu t = 1\) Combining the results, we have: \(W(f,g)(t) = \mu e^{2 \lambda t} (1)\) Since \(\mu \neq 0\) and \(e^{2 \lambda t} > 0\), we see that \(W(f,g)(t) \neq 0\) for all \(t\). Thus, the functions \(f(t)\) and \(g(t)\) are linearly independent.

Key concepts

Differential Equations

Differential equations are mathematical relations that involve an unknown function and its derivatives. They come in handy for modeling various phenomena in engineering, physics, and other sciences. For example, they can describe how populations change, gauge financial models, or even predict weather patterns.
A differential equation's solutions help us understand the behavior of dynamic systems. A key aspect of dealing with differential equations is determining the nature of involved functions—whether they are linearly independent or not. This determination is essential because linearly independent solutions can form the general solution of a differential equation with given conditions.
In this context, we take a set of solutions and use tools like the Wronskian determinant to explore these dependencies. A nonzero Wronskian indicates linear independence; thus, solutions are distinct and contribute uniquely to describing the system.

Wronskian

The Wronskian is a powerful tool in mathematics, particularly in solving differential equations and determining the independence of functions. Named after the mathematician Józef Wroński, it involves calculating a specific determinant from functions and their derivatives.
For two functions, say \(f(t)\) and \(g(t)\), the Wronskian is given by the determinant:

• \(W(f,g)(t) = \begin{vmatrix} f(t) & g(t) \ f'(t) & g'(t) \end{vmatrix}\)

If this determinant is nonzero for any interval of time, the functions are linearly independent. This means each function contributes uniquely to solutions of differential equations that the functions may be involved with.
In our example, calculating the Wronskian clearly showed a non-zero result, affirming the linear independence of the given trigonometric exponential functions.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics and numerous applications, primarily because they model periodic phenomena. The functions \(\sin(x)\) and \(\cos(x)\) describe wave-like patterns and are essential in fields like acoustics, optics, and electrical engineering.
These functions have unique properties, such as periodicity, meaning they repeat values in a regular cycle. This quality makes them useful for representing oscillatory behavior, like sound waves or the path of a pendulum.
In the problem, \(\cos \mu t\) and \(\sin \mu t\) are combined with an exponential term. When paired with exponential functions, they illustrate how waves can decay or grow over time, which is crucial for understanding phenomena such as damping and wave propagation.

Exponential Functions

Exponential functions, like \(e^{\lambda t}\), are essential in describing quickly changing systems. Whether it's modeling population growth, radioactive decay, or interest computations, exponential functions are an invaluable tool.
These functions exhibit rapid changes—growth or decay—due to their inherent nature where the rate of change is proportional to the function's current value. This results in exponential growth or decay, a pattern seen in many natural processes.
In our example, the exponential term \(e^{\lambda t}\) multiplies the trigonometric functions. This combination allows the modeling of situations where wave patterns may intensify or diminish over time, illustrating concepts such as resonance or damping.
Understanding how exponential and trigonometric functions combine is central to analyzing systems that respond dynamically, enhancing our ability to predict and manipulate real-world situations.