Question

Consider the given differential equation on the interval \(-\infty

Short answer

Answer: Yes

Step-by-step solution

Check if the solutions satisfy the initial conditions

Let's first check if the given functions, \(y_1(t)\) and \(y_2(t)\), satisfy the given initial conditions: For \(y_1(t)\): - \(y_1(0)=0\) - \(y_1'(0)=2\) For \(y_2(t)\): - \(y_2(0)=-1\) - \(y_2'(0)=0\) From the exercise, the initial conditions are met for both functions.

Check for linear independence

Now, we need to check if the given functions \(y_1(t)\) and \(y_2(t)\) are linearly independent. There are different ways to do this, but the most straightforward method in this situation is to use the Wronskian, which is defined as: \(W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t)\\ y_1'(t) & y_2'(t) \end{vmatrix} = y_1(t) y_2'(t) - y_1'(t) y_2(t)\) If the Wronskian is nonzero within the interval \(-\infty<t<\infty\), the functions are linearly independent. We have: \(W(y_1, y_2) = y_1(t) y_2'(t) - y_1'(t) y_2(t)\) Recall the initial conditions: \(y_1(0)=0, y_1'(0)=2, y_2(0)=-1, y_2'(0)=0\) Thus, at \(t=0\), we can compute the Wronskian as: \(W(y_1, y_2) = (0)(0) - (2)(-1) = 2\) Since the Wronskian is nonzero, the functions \(y_1(t)\) and \(y_2(t)\) are linearly independent.

Check if the functions are solutions to the differential equation

To complete our analysis, we need to confirm that both \(y_1(t)\) and \(y_2(t)\) are indeed solutions to the differential equation: \(y'' + ty = 0\) As the exercise states that the functions are solutions, we'll assume that both functions satisfy the differential equation. Thus, the given functions form a fundamental set.

Key concepts

Initial Conditions

In the context of differential equations, initial conditions are specific values that solutions of the equations must satisfy at a given point. They help to uniquely define a solution from a family of possible solutions.
For instance, in our exercise, we have two initial conditions for each function:

• For the function \(y_1(t)\), the conditions are \(y_1(0)=0\) and \(y_1'(0)=2\).
• For the function \(y_2(t)\), the conditions are \(y_2(0)=-1\) and \(y_2'(0)=0\).

This information is crucial in determining whether the solutions \(y_1(t)\) and \(y_2(t)\) are compatible with the differential equation throughout the considered interval. These initial conditions ensure that we are on the right path and our solutions are specific to the given differential problem.

Wronskian

The Wronskian is a determinant used in the study of differential equations, primarily to determine if a set of solutions is linearly independent.
It is calculated using the functions and their derivatives. For two functions \(y_1(t)\) and \(y_2(t)\), the Wronskian \(W(y_1, y_2)\) is defined as:

• \(W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t) \ y_1'(t) & y_2'(t) \end{vmatrix} = y_1(t) y_2'(t) - y_1'(t) y_2(t)\)

If this determinant is non-zero at a point or throughout the interval, the functions are linearly independent.
In our example, the Wronskian was calculated at \(t=0\) to be 2, confirming \(y_1(t)\) and \(y_2(t)\) are linearly independent, which is critical for them to form a fundamental set of solutions for the differential equation.

Linear Independence

Linear independence refers to a concept in linear algebra and differential equations, where a set of functions is considered linearly independent if no function in the set can be written as a linear combination of the others.
This property is essential when working with differential equations, as it ensures that each solution contributes uniquely to the solution set.
To test for linear independence, you can use the Wronskian, as shown in the previous section. If the Wronskian is not zero, the solutions are linearly independent. In our exercise:

• \(y_1(t)\) and \(y_2(t)\) have non-zero Wronskian at \(t=0\) which confirms their linear independence.

Because of this linear independence, they can form a comprehensive solution set that spans the solution space for the differential equation.

Solution Set

The solution set in the context of differential equations refers to the collection of functions that satisfy both the differential equation and any given initial conditions.
The completeness and uniqueness of this set often depend on the linear independence of its solutions.
In the exercise, we have two solutions \(y_1(t)\) and \(y_2(t)\), both satisfying the given initial conditions and differential equation:

• Due to the linear independence confirmed by the Wronskian, \(y_1(t)\) and \(y_2(t)\) together form a fundamental set for the differential equation.
• This fundamental set is critical as it represents the complete solution set spanning all possible solutions of the differential equation within the given interval.

Thus, ensuring both initial conditions and linear independence guarantees the accuracy and completeness of the solution set.