Problem 46 - Differential Equations and Linear Algebra

Question

For all constants \(a\) and \(b,\) prove that the set of functions \(\left\\{e^{a x} \cos b x, e^{a x} \sin b x, x e^{a x} \cos b x, x e^{a x} \sin b x\right\\}\) is linearly independent on \((-\infty, \infty).\)

Step-by-Step Solution

Answer

The set of functions \(\left\{e^{ax} \cos bx, e^{ax} \sin bx, xe^{ax} \cos bx, xe^{ax} \sin bx\right\}\) is linearly independent on the interval \((-∞, ∞)\). This is proven by first assuming the functions are linearly dependent, then differentiating the linear combination, simplifying it, and solving the resulting system of linear equations. We end up with the trivial solution for the constants, which contradicts our assumption, proving that these functions are linearly independent.

Step 1: Assume the Set of Functions is Linearly Dependent

Assume that the set of functions \(\left\{e^{ax} \cos bx, e^{ax} \sin bx, xe^{ax} \cos bx, xe^{ax} \sin bx\right\}\) is linearly dependent. If this is true, then there exists scalars \(c_1, c_2, c_3, c_4\) not all equal to zero such that: \(c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx + c_3 xe^{ax} \cos bx + c_4 xe^{ax} \sin bx = 0\) for any \(x \in (-\infty, \infty)\).

Step 2: Derivative

It could be helpful to analyze the derivative of this equation. So let's apply the differentiation for the left side of the equation: \(\frac{d}{dx}(c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx + c_3 xe^{ax} \cos bx + c_4 xe^{ax} \sin bx) = 0\) After applying the product and chain rules, we get: \(c_1 ae^{ax} \cos bx - c_1 b e^{ax} \sin bx + c_2 a e^{ax} \sin bx + c_2 b e^{ax} \cos bx + c_3 e^{ax} \cos bx - c_3 b x e^{ax} \sin bx + c_4 e^{ax} \sin bx + c_4 b x e^{ax} \cos bx = 0\)

Step 3: Group

To simplify this equation, group the terms with the same function: \((c_1 a + c_2 b + c_3) e^{ax} \cos bx + (-c_1 b + c_2 a + c_4) e^{ax} \sin bx + (-c_3 b + c_4 b) xe^{ax} \cos bx + (c_3 a - c_4 a) xe^{ax} \sin bx = 0\) Now notice that the functions \(e^{ax} \cos bx, e^{ax} \sin bx, xe^{ax} \cos bx, xe^{ax} \sin bx\) are linearly independent, so the coefficients in front of each function must be equal to zero: \[ \begin{cases} c_1 a + c_2 b + c_3 = 0 \\ -c_1 b + c_2 a + c_4 = 0 \\ -c_3 b + c_4 b = 0 \\ c_3 a - c_4 a = 0 \end{cases} \]

Step 4: Solve for Constants

Now we have a system of linear equations: From the third and fourth equations, we get: \[ \begin{cases} c_4 = c_3 \\ c_3 = c_4 \end{cases} \] Plugging these back into the first two equations: \[ \begin{cases} c_1 a + c_2 b + c_3 = 0 \\ -c_1 b + c_2 a + c_3 = 0 \end{cases} \] Add the two equations to get: \(c_1 a + c_2 b + c_3 - c_1 b + c_2 a + c_3 = 0\) \((c_1 + c_2)(a + b) + 2c_3 = 0\) If \(a = -b\), then both \(c_1\) and \(c_2\) can be any nonzero constant such that their sum is zero, but this will contradict our assumption that at least one of the constants is nonzero. If \(a \neq -b\), then \(c_1 = -c_2\) and \(c_3 = 0\). Thus, the only solution for the constants is the trivial one, meaning our assumption was incorrect.

Step 5: Conclusion

Since our assumption was incorrect, we conclude that the set of functions \(\left\{e^{ax} \cos bx, e^{ax} \sin bx, xe^{ax} \cos bx, xe^{ax} \sin bx\right\}\) is linearly independent on the interval \(-\infty, \infty\).

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, these functions usually represent physical quantities, and the derivatives represent their rates of change. Differential equations play a pivotal role in engineering, physics, economics, and other disciplines.

For example, in our exercise, derivatives help establish the linear independence of a set of functions, which are potential solutions to differential equations. By taking derivatives and setting up conditions from the characteristics of a zero function, students can infer critical information about the solutions to a differential equation. This strategy is used in our solution steps, using differentiation to help determine linear independence.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations in vector spaces and through matrices. When we talk about the independence of functions, we are often placing these functions within a vector space framework.

In the context of our set of functions, they can be thought of as vectors in a function space. For vectors (or functions) to be linearly independent, no one in the set can be written as a linear combination of the others. This is analogous to the concept that one cannot be reached by scaling and adding the other vectors. The step-by-step solution goes through a process much like solving a system of linear equations to verify the linear independence of the set.

Eigenfunctions

Eigenfunctions are significant in both differential equations and linear algebra. They are solutions to a type of non-trivial homogenous linear differential equation called an eigenvalue problem. In quantum mechanics, they are the functions that describe the state of quantum systems.

While our exercise does not deal directly with eigenfunctions, understanding the concept of eigenfunctions is helpful because they illustrate linear independence within the context of differential equations. The set of functions given in our exercise could serve in constructing eigenfunctions for certain differential operators.

System of Linear Equations

A system of linear equations consists of two or more linear equations made up of two or more variables. The point where these equations intersect is the solution to the system. Systems of linear equations can have one solution, infinite solutions, or no solution.

When we talk about a set of functions being linearly independent, we're referring to the absence of non-trivial solutions to a particular system of linear equations. The solution provided for our exercise involves creating and solving a system of linear equations to prove that no non-zero constants can turn our set of functions into a zero function, hence proving linear independence.