Question

find the Wronskian of the given pair of functions. $$ e^{2 t}, \quad e^{-3 t / 2} $$

Short answer

Show your work. Answer: The functions \(f_1(t) = e^{2t}\) and \(f_2(t) = e^{-\frac{3}{2}t}\) are linearly independent. To show this, we calculated the Wronskian determinant of the two functions, which came out to be \(W(f_1,f_2) = -\frac{7}{2}e^{-t}\). Since this expression is nonzero for all t, the functions are linearly independent.

Step-by-step solution

Differentiate the functions#wtag_content#First, we need to find the first derivatives of each function. Let's denote the functions as $$ f_1(t) = e^{2t} \quad\text{and}\quad f_2(t) = e^{-\frac{3}{2}t}. $$ Now, we find their first derivatives: $$ \frac{d}{dt}f_1(t) = \frac{d}{dt}(e^{2t}) = 2e^{2t} $$ and $$ \frac{d}{dt}f_2(t) = \frac{d}{dt}(e^{-\frac{3}{2}t}) = -\frac{3}{2}e^{-\frac{3}{2}t} $$

Step 2: Calculate the Wronskian determinant#wtag_content# Now that we have the first derivatives, we can calculate the Wronskian determinant, denoted by W, as follows: $$ W(f_1,f_2) = \begin{vmatrix} f_1(t) & f_2(t) \\ \frac{d}{dt}f_1(t) & \frac{d}{dt}f_2(t) \end{vmatrix}= \begin{vmatrix} e^{2t} & e^{-\frac{3}{2}t} \\ 2e^{2t} & -\frac{3}{2}e^{-\frac{3}{2}t} \end{vmatrix}. $$

Find the value of the determinant#wtag_content#Finally, we find the determinant of the matrix: $$ W(f_1, f_2) = e^{2t} \left(-\frac{3}{2}e^{-\frac{3}{2}t}\right) - e^{-\frac{3}{2}t}(2e^{2t}). $$ Simplify this expression to get the Wronskian: $$ W(f_1 , f_2) = -\frac{3}{2}e^{-t} - 2e^{-t} = -\frac{7}{2}e^{-t}. $$

Step 4: Determine linear independence/wtag_content#Since the Wronskian \(W(f_1,f_2) = -\frac{7}{2}e^{-t}\) is nonzero for all t, the functions \(f_1(t) = e^{2t}\) and \(f_2(t) = e^{-\frac{3}{2}t}\) are linearly independent.

Key concepts

Linear Independence

Linear independence is a fundamental concept in linear algebra and differential equations. When we say a set of functions is linearly independent, we mean that no function in the set can be written as a linear combination of the others. This concept is crucial because it ensures that the functions offer distinct solutions for equation sets or systems.
The Wronskian is a tool used to determine if a pair or set of functions is linearly independent. If the Wronskian of a set of functions is non-zero for some interval, the functions are said to be linearly independent over that interval. In our example, the functions \( f_1(t) = e^{2t} \) and \( f_2(t) = e^{-rac{3}{2}t} \) have a Wronskian of \(-\frac{7}{2} e^{-t}\), which is non-zero for all values of \( t \). Therefore, they are linearly independent.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how the function changes as its input changes. In the context of the Wronskian, differentiation is needed because the Wronskian determinant involves the original functions and their first derivatives.
For the given functions \( f_1(t) = e^{2t} \) and \( f_2(t) = e^{-\frac{3}{2}t} \), we differentiate them to find their derivatives. The derivative of \( f_1(t) \) with respect to \( t \) is \( \frac{d}{dt}(e^{2t}) = 2e^{2t} \), and the derivative of \( f_2(t) \) is \( \frac{d}{dt}(e^{-\frac{3}{2}t}) = -\frac{3}{2}e^{-\frac{3}{2}t} \).
These derivatives are used in the 2x2 matrix to calculate the Wronskian determinant. Differentiation is thus a key step in determining whether the function pairs are linearly independent.

Determinant

The determinant is a scalar value that can be computed from a square matrix and provides important information about the matrix. In the case of the Wronskian, we calculate the determinant of a 2x2 matrix formed by the functions and their first derivatives. This determinant helps us assess linear independence of functions.
For our function pairs, the matrix is:

• First row: \( f_1(t) = e^{2t} \), \( f_2(t) = e^{-\frac{3}{2}t} \)
• Second row: \( \frac{d}{dt}f_1(t) = 2e^{2t} \), \( \frac{d}{dt}f_2(t) = -\frac{3}{2}e^{-\frac{3}{2}t} \)

The determinant of this matrix is calculated as:\[ W(f_1, f_2) = e^{2t}\left(-\frac{3}{2}e^{-\frac{3}{2}t}\right) - e^{-\frac{3}{2}t}(2e^{2t}) \]Simplifying this yields \( W(f_1, f_2) = -\frac{7}{2}e^{-t} \). The non-zero value indicates linear independence, thereby confirming that the determinant is a powerful indicator in this context.

Function Pairs

When dealing with pairs of functions in the context of the Wronskian, we specifically examine how these functions behave together. The relationship between them can tell us a lot about the solutions of differential equations.
In our exercise, the function pair \( f_1(t) = e^{2t} \) and \( f_2(t) = e^{-rac{3}{2}t} \) are exponential functions. Exponential functions are particularly interesting because they do not cross zero, making them useful for examining growth and decay processes.
By computing the Wronskian for these function pairs, we confirm that they are linearly independent. This means in solving differential equations, each function contributes uniquely to the general solution, providing valuable individual solutions without redundancy. Function pairs, therefore, are central in understanding the diversity or similarity of solutions they provide in problem-solving scenarios.