Question

In each exercise, (a) Verify that the given functions form a fundamental set of solutions. (b) Solve the initial value problem. \(t y^{\prime \prime \prime}+3 y^{\prime \prime}=0, \quad t>0 ; \quad y(2)=\frac{1}{2}, \quad y^{\prime}(2)=-\frac{5}{4}, \quad y^{\prime \prime}(2)=\frac{1}{4}\) \(y_{1}(t)=1, \quad y_{2}(t)=t, \quad y_{3}(t)=t^{-1}\)

Short answer

Question: Do the given functions \(y_{1}(t)=1, y_{2}(t)=t, y_{3}(t)=t^{-1}\) form a fundamental set of solutions, and if so, find the particular solution for \(y(2)=\frac{1}{2}, y^{\prime}(2)=-\frac{5}{4}, y^{\prime \prime}(2)=\frac{1}{4}\). Answer: Yes, the given functions form a fundamental set of solutions, as the Wronskian is nonzero for \(t>0\). The particular solution is \(y(t)=\frac{9}{4} - t + \frac{1}{4}t^{-1}\).

Step-by-step solution

Verifying the given functions form a fundamental set of solutions

First, we need to find the Wronskian of the given functions \(y_{1}(t)=1, y_{2}(t)=t, y_{3}(t)=t^{-1}\). Recall that the Wronskian is defined as: $$W(t) = \begin{vmatrix} y_{1}(t) & y_{2}(t) & y_{3}(t) \\ y_{1}^\prime(t) & y_{2}^\prime(t) & y_{3}^\prime(t) \\ y_{1}^{\prime \prime}(t) & y_{2}^{\prime \prime}(t) & y_{3}^{\prime \prime}(t) \end{vmatrix}$$ Now we need to compute the derivatives of the given functions: $$y_{1}^\prime(t)=0, \quad y_{1}^{\prime \prime}(t)=0$$ $$y_{2}^\prime(t)=1, \quad y_{2}^{\prime \prime}(t)=0$$ $$y_{3}^\prime(t)=-t^{-2}, \quad y_{3}^{\prime \prime}(t)=2t^{-3}$$ Now we compute the determinant of the matrix formed by these derivatives: $$W(t) = \begin{vmatrix} 1 & t & t^{-1} \\ 0 & 1 & -t^{-2} \\ 0 & 0 & 2t^{-3} \end{vmatrix} = \left( 1\times \det \begin{vmatrix} 1 & -t^{-2} \\ 0 & 2t^{-3} \end{vmatrix} \right) = 2t^{-1}$$ The Wronskian is nonzero for \(t>0\). Thus, the given functions form a fundamental set of solutions.

Solving the initial value problem

To solve the initial value problem, we assume the solution is in the form: $$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+c_{3}y_{3}(t)=c_{1}+c_{2}t+c_{3}t^{-1}$$ It's given that \(y(2)=\frac{1}{2}, y^{\prime}(2)=-\frac{5}{4}, y^{\prime \prime}(2)=\frac{1}{4}\). We can differentiate our general solution with respect to \(t\) to obtain the required derivatives. $$y^{\prime}(t)=c_{2}-c_{3}t^{-2}$$ $$y^{\prime \prime}(t)=2c_{3}t^{-3}$$ Now use the initial values to form a system of equations: $$y(2)=\frac{1}{2}=c_{1}+2c_{2}+\frac{1}{2}c_{3}$$ $$y^{\prime}(2)=-\frac{5}{4}=c_{2}-\frac{1}{4}c_{3}$$ $$y^{\prime \prime}(2)=\frac{1}{4}=c_{3}$$ The last equation immediately gives \(c_{3}=\frac{1}{4}\). Plug this into the second equation to get: $$-\frac{5}{4}=c_{2}-\frac{1}{4} \cdot \frac{1}{4} \implies c_{2}=-1$$ Finally, plug these values into the first equation: $$\frac{1}{2}=c_{1} - 2 + \frac{1}{8} \implies c_{1}=\frac{9}{4}$$ Now we have found the constants, and our particular solution is: $$y(t)=\frac{9}{4} - t + \frac{1}{4}t^{-1}$$

Key concepts

Wronskian Determinant

The Wronskian determinant is a powerful tool used to determine if a set of functions are linearly independent, which is key to solving differential equations. When dealing with a differential equation, finding a fundamental set of solutions that are linearly independent is essential. The Wronskian is a specific kind of determinant that, when not equal to zero, confirms the functions' linear independence within an interval.

In the context of the given exercise, we calculate the Wronskian using three functions, which are proposed solutions to a third-order differential equation. By differentiating these functions and arranging them into a matrix, with each row being the successive derivatives of the functions, we form the matrix required to calculate the Wronskian. As long as the Wronskian is not zero, it implies that these functions do indeed form a fundamental set of solutions for the interval in question, in this case, for any positive value of t.

Fundamental Set of Solutions

A fundamental set of solutions consists of functions that solve a differential equation and are linearly independent. For a nth-order differential equation, we need n such solutions to form the fundamental set. This concept is vital because it lays the groundwork for the general solution of the differential equation.

In our problem, we verify that the set of functions given is indeed fundamental by using the Wronskian. Verification is a two-step process where first, we ensure that we have the correct number of functions (in this case, three for a third-order equation), and second, we confirm they are linearly independent through their Wronskian. Once a fundamental set has been established, any linear combination of these solutions is also a solution to the differential equation.

Third-Order Differential Equation

A third-order differential equation is a relationship involving an unknown function and its derivatives up to the third order. The order of a differential equation indicates the highest derivative it contains.

For instance, the exercise we're considering features a third-order differential equation that includes a term with the third derivative of y, denoted as y'''. These types of equations require three linearly independent solutions to construct a general solution, as we have seen with our fundamental set. Third-order equations can model various real-world phenomena, including certain motions in physics and engineering problems. Solving these requires not only finding the general solution but also applying initial conditions to find a particular solution.

Initial Conditions

The term initial conditions refers to the values of the function and its derivatives at a specific point. They are absolutely necessary to solve an initial value problem (IVP), which aims to find a particular solution to a differential equation that satisfies these conditions.

Our differential equation comes with three initial conditions, each providing vital information about the function and its derivatives at t = 2. With these conditions, we can determine the exact values of the coefficients in the general solution, converting the general solution into a specific one that fits the given constraints. These initial conditions help pinpoint the exact trajectory among the infinite solutions represented by the general solution.