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

Short Answer

Expert verified
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

01

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: $$ W(f_1,f_2) = \begin{vmatrix} f_1(t) & f_2(t) \\ \frac{d}{dt}f_1(t) & \frac{d}{dt}f_2(t) \end{vmatrix}= \begin{vmatrix} e^{2t} & e^{-\frac{3}{2}t} \\ 2e^{2t} & -\frac{3}{2}e^{-\frac{3}{2}t} \end{vmatrix}. $$
02

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Determine a suitable form for \(Y(t)\) if the method of undetermined coefficients is to be used. (b) Use a computer algebra system to find a particular solution of the given equation. $$ y^{\prime \prime}+4 y=t^{2} \sin 2 t+(6 t+7) \cos 2 t $$

Consider the initial value problem $$ u^{\prime \prime}+\gamma u^{\prime}+u=0, \quad u(0)=2, \quad u^{\prime}(0)=0 $$ We wish to explore how long a time interval is required for the solution to become "negligible" and how this interval depends on the damping coefficient \(\gamma\). To be more precise, let us seek the time \(\tau\) such that \(|u(t)|<0.01\) for all \(t>\tau .\) Note that critical damping for this problem occurs for \(\gamma=2\) (a) Let \(\gamma=0.25\) and determine \(\tau,\) or at least estimate it fairly accurately from a plot of the solution. (b) Repeat part (a) for several other values of \(\gamma\) in the interval \(0<\gamma<1.5 .\) Note that \(\tau\) steadily decreases as \(\gamma\) increases for \(\gamma\) in this range. (c) Obtain a graph of \(\tau\) versus \(\gamma\) by plotting the pairs of values found in parts (a) and (b). Is the graph a smooth curve? (d) Repeat part (b) for values of \(\gamma\) between 1.5 and \(2 .\) Show that \(\tau\) continues to decrease until \(\gamma\) reaches a certain critical value \(\gamma_{0}\), after which \(\tau\) increases. Find \(\gamma_{0}\) and the corresponding minimum value of \(\tau\) to two decimal places. (e) Another way to proceed is to write the solution of the initial value problem in the form (26). Neglect the cosine factor and consider only the exponential factor and the amplitude \(R\). Then find an expression for \(\tau\) as a function of \(\gamma\). Compare the approximate results obtained in this way with the values determined in parts (a), (b), and (d).

Find the general solution of the given differential equation. $$ y^{\prime \prime}-y^{\prime}-2 y=\cosh 2 t \quad \text { Hint } \cosh t=\left(e^{\prime}+e^{-t}\right) / 2 $$

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}-(x-0.1875) y=0, \quad x>0 ; \quad y_{1}(x)=x^{1 / 4} e^{2 \sqrt{x}}\)

A spring is stretched 6 in. by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant of \(0.25 \mathrm{lb}-\) sec/ft and is acted on by an external force of \(4 \cos 2 t\) lb. (a) Determine the steady-state response of this system. (b) If the given mass is replaced by a mass \(m,\) determine the value of \(m\) for which the amplitude of the steady-state response is maximum.

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