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The Wronskian of two functions is \(W(t)=t^{2}-4 .\) Are the functions linearly independent or linearly dependent? Why?

Short Answer

Expert verified
Answer: The functions are linearly independent.

Step by step solution

01

Check if the Wronskian is non-zero for any value of t.

We are given the Wronskian W(t) = t^2 - 4. To check if this is non-zero for any value of t, we can try to find the roots (if any) of the equation t^2 - 4 = 0.
02

Solve the equation t^2 - 4 = 0 for t.

To find the roots, we can use the quadratic formula or factoring. In this case, factoring is easier since this is a simple equation: t^2 - 4 = (t - 2)(t + 2) The equation has two roots, t = 2 and t = -2.
03

Determine if the Wronskian is non-zero for any value of t.

We found that the Wronskian has two roots, t = 2 and t = -2. This means that the Wronskian is zero at these points. However, for any other value of t, the Wronskian will be non-zero (for example, for t = 0, W(0) = 0^2 - 4 = -4).
04

Determine if the functions are linearly independent or dependent.

Since the Wronskian is non-zero for at least one point in its domain (t = 0), the functions are linearly independent. So, we conclude that the functions 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 concept that helps determine whether a set of functions or vectors are dependent on each other. Imagine you have a set of functions; they are linearly independent if none of them can be expressed as a combination of the others. In other words, no function in the set can be "made" using the others.
When dealing with the Wronskian, it serves as a handy tool for checking linear independence in the context of differential equations.
  • If the Wronskian is non-zero for any point in the domain, the functions are linearly independent.
  • If the Wronskian is zero everywhere, the functions can potentially be linearly dependent.
In the given exercise, the Wronskian was non-zero at points like \(t = 0\), indicating that the functions are indeed linearly independent.
differential equations
Differential equations involve unknown functions and their derivatives. They are equations that describe how functions change and are pivotal in modeling real-world phenomena.
The Wronskian pops up often when discussing differential equations, especially when determining the relationship between solutions.
  • Solutions to differential equations can be expressed as a linear combination of functions.
  • The Wronskian helps check if these solutions are linearly independent—important for understanding the behavior of solutions over time.
For instance, in our exercise, the Wronskian of two solutions being non-zero at least somewhere ensures these solutions are linearly independent, an important criterion in solving linear differential equations.
roots of equations
Finding the roots of an equation means solving for the values that make the equation equal to zero. For a quadratic equation, like in the exercise \(t^2 - 4 = 0\), this involves finding the values of \(t\) that satisfy the equation.
There are multiple ways to find roots:
  • Factoring: breaking down the equation, as seen with \((t - 2)(t + 2) = 0\), yielding roots \(t = 2\) and \(t = -2\).
  • Quadratic Formula: another reliable method, especially for more complex quadratics.
Knowing where the roots of the Wronskian are is crucial, as these points can hint at whether functions related to it are linearly dependent, which helps in understanding the nature of solutions to differential equations further. In the exercise, identifying the roots showed where the Wronskian becomes zero, helping deduce the linear independence of the functions involved.

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Most popular questions from this chapter

Find the solution of the initial value problem $$ u^{\prime \prime}+u=F(t), \quad u(0)=0, \quad u^{\prime}(0)=0 $$ where $$ F(t)=\left\\{\begin{array}{ll}{F_{0}(2 \pi-t),} & {0 \leq t \leq \pi} \\ {-0} & {(2 \pi-t),} & {\pi

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).

In each of Problems 13 through 18 find the solution of the given initial value problem. $$ y^{\prime \prime}+y^{\prime}-2 y=2 t, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

In each of Problems 1 through 12 find the general solution of the given differential equation. $$ y^{\prime \prime}-2 y^{\prime}-3 y=3 e^{2 x} $$

In many physical problems the nonhomogencous term may be specified by different formulas in different time periods. As an example, determine the solution \(y=\phi(t)\) of $$ y^{\prime \prime}+y=\left\\{\begin{array}{ll}{t,} & {0 \leq t \leq \pi} \\\ {\pi e^{x-t},} & {t>\pi}\end{array}\right. $$ $$ \begin{array}{l}{\text { satisfying the initial conditions } y(0)=0 \text { and } y^{\prime}(0)=1 . \text { Assume that } y \text { and } y^{\prime} \text { are also }} \\ {\text { continuous at } t=\pi \text { . Plot the nonhomogencous term and the solution as functions of time. }} \\ {\text { Hint: First solve the initial value problem for } t \leq \pi \text { ; then solve for } t>\pi \text { , determining the }} \\ {\text { constants in the latter solution from the continuity conditions at } t=\pi \text { . }}\end{array} $$

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