Question

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ y^{\prime \prime \prime}+y^{\prime}=0 ; \quad 1, \quad \cos t, \quad \sin t $$

Short answer

Based on the analysis and solution steps, it was found that the function \(1\) is a solution to the given differential equation, while \(\cos(t)\) and \(\sin(t)\) are not solutions. The Wronskian of the given functions was calculated to be 1, which indicates that the functions are linearly independent.

Step-by-step solution

Calculate the first and third derivatives of the given functions

To do this, we will differentiate each function with respect to t once and then differentiate the result twice to get the third derivative. 1. For \(1\): - \(y'(t) = \frac{d}{dt}(1)= 0\) - \(y'''(t)= \frac{d^3}{dt^3}(1) = 0\) 2. For \(\cos(t)\): - \(y'(t) = \frac{d}{dt}(\cos(t))=-\sin(t)\) - \(y'''(t) = \frac{d^3}{dt^3}(\cos(t))=-\cos(t)\) 3. For \(\sin(t)\): - \(y'(t)=\frac{d}{dt}(\sin(t))=\cos(t)\) - \(y'''(t)=\frac{d^3}{dt^3} (\sin(t)) =-\sin(t)\)

Substitute the calculated derivatives into the differential equation

We need to substitute the first and third derivatives into the equation \(y^{\prime\prime\prime} + y^{\prime} = 0\) and see if it holds true. For \(1\): - \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = 0 + 0 = 0\) - \(1\) is a solution. For \(\cos(t)\): - \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = -\cos(t) - \sin(t)\) - This does not equal zero, so \(\cos(t)\) is not a solution. For \(\sin(t)\): - \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = -\sin(t) + \cos(t)\) - This does not equal zero, so \(\sin(t)\) is not a solution.

Calculate the Wronskian of the given functions

To calculate the Wronskian, we will set up a determinant using the original functions and their first and second derivatives. The Wronskian of a set of functions is defined as: $$ W(y_1, y_2, y_3) = \begin{vmatrix} y_1 & y_2 & y_3\\ y'_1 & y'_2 & y'_3\\ y''_1 & y''_2 & y''_3 \end{vmatrix} $$ where in this case, \(y_1 = 1, y_2 = \cos(t), y_3 = \sin(t)\). After calculating the first and second derivatives in Step 1, we can create the determinant like so: $$ W(1, \cos(t), \sin(t)) = \begin{vmatrix} 1 & \cos(t) & \sin(t)\\ 0 & -\sin(t) & \cos(t)\\ 0 & -\cos(t) & -\sin(t) \end{vmatrix} $$

Evaluate the determinant

To find the Wronskian, we need to evaluate this determinant. We can do this using the rule for expanding a determinant along the first row: $$ W(1, \cos(t), \sin(t)) = 1 \begin{vmatrix} -\sin (t) & \cos (t) \\ -\cos (t) & -\sin (t) \end{vmatrix} - \cos(t) \begin{vmatrix} 0 &\cos(t) \\ 0 & -\sin(t) \end{vmatrix} + \sin(t) \begin{vmatrix} 0 & -\sin(t) \\ 0 & -\cos(t) \end{vmatrix} $$ Now, we can evaluate each of the remaining 2x2 determinants: $$ W(1, \cos(t), \sin(t)) = 1((- \sin (t))(-\sin(t)) - (\cos (t))(-\cos (t))) - 0 \sin(t) + 0 \cos(t) $$ Simplifying the expression, we find: $$ W(1, \cos(t), \sin(t)) = \sin^2(t) + \cos^2(t) = 1 $$ The Wronskian of the given functions is 1.

Key concepts

Differential Equations

In the given exercise, verifying that the functions are actual solutions involves checking whether they satisfy the differential equation when their derivatives are substituted into it. Steps 1 and 2 in the solutions involve calculating and then substituting these derivatives, verifying which functions are indeed solutions.

Understanding differential equations and the principles behind them is a vital skill for students in various fields such as engineering, physics, economics, and beyond. They provide a clear path from an abstract mathematical description to real-world applications that can be observed and measured.

Derivatives of Functions

In calculus, derivatives represent the rate at which a function is changing at any point. Formally, the derivative of a function at a certain point is the slope of the tangent line to the function's graph at that point. It's an instantaneous rate of change, which can be thought of as how quickly or slowly the function's output (value) is changing in comparison to its input.

When dealing with differential equations, you encounter derivatives of various orders. The first derivative represents the primary rate of change, the second derivative can represent acceleration or the curvature of the function's graph, and higher-order derivatives continue to reveal deeper aspects of the function's behavior.

In the context of our exercise, calculating the Wronskian requires knowing the derivatives up to the third order. The solution steps detail precisely how to compute these derivatives for the functions given. It is important to follow these derivative rules correctly to ensure the accuracy of the Wronskian and the conclusions drawn from it.

Determinants

Determinants are a property of square matrices that provide useful information in various areas including linear algebra, differential equations, and analysis. They are used to determine the solvability of system of linear equations, calculate the area or volume of parallelograms and parallelepipeds defined by vectors, and to understand the behavior of linear transformations.

The determinant of a matrix is a single number that can tell us many things about the matrix. For instance, if the determinant of a matrix is zero, the matrix is said to be singular, and it does not have an inverse. Determinants are especially relevant to the study of differential equations because of their role in the Wronskian, which helps determine if a set of solutions is linearly independent.

In this exercise, the solution provided shows how to calculate the Wronskian, which involves setting up and evaluating the determinant of a matrix built from the given functions and their derivatives. The capability of evaluating determinants efficiently is essential for verifying the linear independence of solutions to differential equations. Accurate calculation of the Wronskian, through determinants, leads to a profound understanding of the behavior of the function system being analyzed.