Question

The \(t\)-interval of interest is \(-\infty

Short answer

Question: Verify if the functions \(y_1(t)=e^{-t/2}\) and \(y_2(t)=t e^{-t/2}\) are solutions to the differential equation \(4y^{\prime\prime}+4y^{\prime}+y=0\), and if they form a fundamental set of solutions. If so, determine the unique solution to the initial value problem with \(y(1)=1\) and \(y'(1)=0\). Answer: The functions \(y_1(t)=e^{-t/2}\) and \(y_2(t)=t e^{-t/2}\) are indeed solutions to the given differential equation and form a fundamental set of solutions. The unique solution to the initial value problem is \(y(t) = 2e^{-t/2} - 2te^{-t/2}\).

Step-by-step solution

Verify the given functions are solutions of the differential equation.

Given the two functions \(y_1(t)=e^{-t/2}\) and \(y_2(t)=t e^{-t/2}\), we need to verify if they are solutions to the given differential equation \(4y^{\prime\prime}+4y^{\prime}+y=0\). First, let's take derivatives of \(y_1(t)\): $$ y_1^{\prime}(t) = -\frac{1}{2} e^{-t / 2}, \quad y_1^{\prime\prime}(t) = \frac{1}{4} e^{-t / 2}. $$ Next, plug these derivatives back to the equation and check if it is satisfied: $$ 4\left(\frac{1}{4} e^{-t / 2}\right) + 4\left(-\frac{1}{2} e^{-t / 2}\right) + e^{-t/2} = e^{-t/2} - 2e^{-t/2} + e^{-t/2} = 0. $$ Since we have equality, \(y_1(t)\) is a solution to the given differential equation. Now, let's do the same for \(y_2(t)\). $$ y_2^{\prime}(t) = e^{-t / 2} -\frac{1}{2} t e^{-t / 2}, \quad y_2^{\prime\prime}(t) = -e^{-t / 2} + t e^{-t / 2}. $$ Plug these derivatives back to the equation and check if it is satisfied: $$ 4\left(-e^{-t/2} + t e^{-t/2}\right) + 4\left(e^{-t/2} -\frac{1}{2}t e^{-t/2}\right) + te^{-t/2} = 0. $$ Since we also have equality, \(y_2(t)\) is a solution to the given differential equation.

Calculate the Wronskian and check for fundamental set of solutions.

To determine if \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions, we need to calculate their Wronskian. The Wronskian of two functions \(y_1(t)\) and \(y_2(t)\) is defined as: $$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_1' \\ y_2 & y_2' \end{vmatrix} = y_1y_2' - y_1'y_2. $$ In this case, our Wronskian will be: $$ W(y_1, y_2) = \left(e^{-t/2}\right)\left(e^{-t/2} -\frac{1}{2}te^{-t/2}\right) - \left(-\frac{1}{2}e^{-t/2}\right)\left(te^{-t/2}\right). $$ Simplify the expression: $$ W(y_1, y_2) = e^{-t} - \frac{1}{2}te^{-t} + \frac{1}{2}te^{-t} = e^{-t}. $$ Since the Wronskian is nonzero, the two functions \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions.

Determine the unique solution of the initial value problem.

Given that \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions for the initial value problem and the initial conditions \(y(1)=1\) and \(y'(1)=0\), we proceed to find the unique solution as a linear combination of the two functions: $$ y(t) = C_1y_1(t) + C_2y_2(t) = C_1e^{-t/2} + C_2te^{-t/2}. $$ Apply the initial conditions: $$ y(1) = 1 = C_1e^{-1/2} + C_2e^{-1/2}. $$ $$ y'(t) = -\frac{1}{2}C_1e^{-t/2} + C_2e^{-t/2} -\frac{1}{2}C_2te^{-t/2}. $$ $$ y'(1) = 0 = -\frac{1}{2}C_1e^{-1/2} + C_2e^{-1/2} -\frac{1}{2}C_2e^{-1/2}. $$ Solving the above system of equations gives \(C_1 = 2\) and \(C_2 = -2\). Therefore, the unique solution to the initial value problem is: $$ y(t) = 2e^{-t/2} - 2te^{-t/2}. $$

Key concepts

Wronskian

The Wronskian is a very important tool in the study of differential equations. It helps to determine whether a set of solutions is linearly independent, which is crucial for forming a fundamental set of solutions.

Consider two functions, \(y_1(t)\) and \(y_2(t)\). The Wronskian of these functions, denoted as \(W(y_1, y_2)\), is a determinant of their derivatives:

• \(W(y_1, y_2) = \begin{vmatrix} y_1 & y_1' \ y_2 & y_2' \end{vmatrix}\)
• This simplifier gives: \(W(y_1, y_2) = y_1y_2' - y_1'y_2\)

If the Wronskian is non-zero at some point in the interval of interest, it indicates that the functions \(y_1(t)\) and \(y_2(t)\) are linearly independent. They can form a fundamental set of solutions of the differential equation.

In our example, the Wronskian of \(y_1(t) = e^{-t/2}\) and \(y_2(t) = te^{-t/2}\) is calculated and found to be \(e^{-t}\), which is non-zero for all \(t\). Therefore, these functions are linearly independent and indeed form a fundamental set of solutions.

Fundamental Set of Solutions

The concept of a fundamental set of solutions is vital in solving linear homogeneous differential equations. A fundamental set consists of solutions that are linearly independent over an interval, meaning no solution in the set can be written as a linear combination of the others.

This set contains exactly as many solutions as the order of the differential equation. For a second-order differential equation like our example, we need two linearly independent solutions. These solutions can be combined to express any solution of the differential equation in a general form.

When \(y_1(t) = e^{-t/2}\) and \(y_2(t) = te^{-t/2}\) are determined to be solutions to the equation \(4y'' + 4y' + y = 0\), and their Wronskian is non-zero, they constitute a fundamental set of solutions. Consequently, any solution \(y(t)\) can be expressed as:

• \(y(t) = C_1y_1(t) + C_2y_2(t)\)

where \(C_1\) and \(C_2\) are constants determined by initial conditions or other circumstances.

Initial Value Problem

An Initial Value Problem (IVP) in differential equations specifies the values of the solution and possibly some of its derivatives at a particular point. This additional information lets us determine constants in the general solution.

For our differential equation \(4y'' + 4y' + y = 0\), the initial conditions are \(y(1) = 1\) and \(y'(1) = 0\). By substituting these conditions into our general solution, \(y(t) = C_1e^{-t/2} + C_2te^{-t/2}\), we can solve for the constants \(C_1\) and \(C_2\).

Plugging in the initial conditions, we solve:

• \(1 = C_1e^{-1/2} + C_2e^{-1/2}\) from \(y(1) = 1\)
• \(0 = -\frac{1}{2}C_1e^{-1/2} + C_2e^{-1/2} - \frac{1}{2}C_2e^{-1/2}\) from \(y'(1) = 0\)

Solving this system gives \(C_1 = 2\) and \(C_2 = -2\), leading us to the unique solution \(y(t) = 2e^{-t/2} - 2te^{-t/2}\).

This unique solution satisfies both the differential equation and the initial conditions, which is the hallmark of solving an IVP.