Question

If the functions \(y_{1}\) and \(y_{2}\) are linearly independent solutions of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0,\) determine under what conditions the functions \(y_{3}=a_{1} y_{1}+a_{2} y_{2}\) and \(y_{4}=b_{1} y_{1}+b_{2} y_{2}\) also form a linearly independent set of solutions.

Short answer

Answer: The functions \(y_3\) and \(y_4\) form a linearly independent set of solutions if the following conditions hold: 1. \(a_1b_2 - a_2b_1 \neq 0\) 2. \(y_1y_2' - y_2y_1' \neq 0\), which is implied by the linear independence of \(y_1\) and \(y_2\).

Step-by-step solution

Compute the first and second derivatives of \(y_3\) and \(y_4\)

To find the derivatives, we'll use the given differential equation: \(y'' + p(t)y' + q(t)y = 0\). Since \(y_1\) and \(y_2\) are solutions, then replacing y with \(y_1\) or \(y_2\) will satisfy the equation. Now, we compute the derivatives of \(y_3\) and \(y_4\): $$y_3 = a_1y_1 + a_2y_2$$ $$y_3' = a_1y_1' + a_2y_2'$$ $$y_3'' = a_1y_1'' + a_2y_2''$$ Similarly, $$y_4 = b_1y_1 + b_2y_2$$ $$y_4' = b_1y_1' + b_2y_2'$$ $$y_4'' = b_1y_1'' + b_2y_2''$$

Calculate the Wronskian determinant

The Wronskian determinant is given by: $$\begin{vmatrix} y_3 & y_4\\ y_3' & y_4' \end{vmatrix} = \begin{vmatrix} a_1y_1 + a_2y_2 & b_1y_1 + b_2y_2\\ a_1y_1' + a_2y_2' & b_1y_1' + b_2y_2' \end{vmatrix}$$

Evaluate the determinant

Expand the determinant using the definition of the determinant of a 2x2 matrix: $$W = (a_1y_1 + a_2y_2)(b_1y_1' + b_2y_2') - (b_1y_1 + b_2y_2)(a_1y_1' + a_2y_2')$$

Simplify the expression

Distribute the terms and collect like terms: $$W = a_1b_1y_1y_1' + a_1b_2y_1y_2' + a_2b_1y_2y_1' + a_2b_2y_2y_2' - a_1b_1y_1y_1' - a_1b_2y_2y_1' - a_2b_1y_1y_2' - a_2b_2y_2y_2' $$ Now some terms cancel out: $$W = a_1b_2y_1y_2' + a_2b_1y_2y_1' - a_1b_2y_2y_1' - a_2b_1y_1y_2'$$ Rearrange and factor: $$W = (a_1b_2 - a_2b_1)(y_1y_2' - y_2y_1')$$

Determine the conditions for linear independence

The functions \(y_3\) and \(y_4\) will be linearly independent if their Wronskian determinant is nonzero. Thus, we have: $$W = (a_1b_2 - a_2b_1)(y_1y_2' - y_2y_1') \neq 0$$ Therefore, \(y_3\) and \(y_4\) are linearly independent if \((a_1b_2 - a_2b_1)(y_1y_2' - y_2y_1')\) is nonzero. The product will be nonzero if both of its factors are nonzero. So, \(y_3\) and \(y_4\) form a linearly independent set of solutions under the following conditions: 1. \(a_1b_2 - a_2b_1 \neq 0\) 2. \(y_1y_2' - y_2y_1' \neq 0\), which is implied by the linear independence of \(y_1\) and \(y_2\).

Key concepts

Wronskian determinant

The Wronskian determinant is a key tool in analyzing whether two functions, say \( y_1 \) and \( y_2 \), are linearly independent solutions of a differential equation. The determinant itself is a specific value calculated using the derivatives of these functions. For two functions, the Wronskian \( W(y_1, y_2) \) is calculated as follows:

• \( W(y_1, y_2) = y_1 y_2' - y_2 y_1' \)

This expression involves the functions themselves, \( y_1 \) and \( y_2 \), as well as their first derivatives, \( y_1' \) and \( y_2' \).
If the Wronskian determinant is non-zero at some point \( t_0 \), then the functions \( y_1 \) and \( y_2 \) are linearly independent. In simpler terms, it means that neither function can be expressed as a linear combination of the other. However, if the Wronskian is zero for all values of \( t \), it indicates the possibility of a linear dependence between the functions, meaning that one can be expressed as a multiple of the other.
The concept of the Wronskian is invaluable because it provides a straightforward method to validate the independence of solutions, which is essential in solving differential equations.

second-order linear differential equations

Second-order linear differential equations are a type of equation that involve a second derivative, along with possibly a first derivative and the function itself. These equations generally have the form:

• \( y'' + p(t)y' + q(t)y = 0 \)

where \( y'' \) denotes the second derivative of \( y \), \( y' \) is the first derivative, and \( p(t) \) and \( q(t) \) are potentially variable coefficients dependent on \( t \).
Solving these equations involves finding a function, or functions, \( y \), such that when plugged back into the equation, it satisfies the equation for all \( t \).
These types of equations are crucial in many physical applications, like mechanical vibrations and electrical circuits, where the behavior of systems is often modeled by the relationships between measurements like position, velocity, and acceleration.
Identifying solutions to these equations often involves finding two linearly independent solutions, and verifying their independence with the Wronskian determinant as previously discussed. Each independent solution allows the general solution to be expressed as a linear combination of these solutions.

linear combinations of solutions

A linear combination of solutions is a method used in solving differential equations, particularly when dealing with multiple solutions. If \( y_1 \) and \( y_2 \) are solutions to a differential equation, a new function can be constructed as:

• \( y = a_1y_1 + a_2y_2 \)

where \( a_1 \) and \( a_2 \) are constants. This new function \( y \) represents a linear combination of \( y_1 \) and \( y_2 \). Linear combinations are useful because they harness the independence of functions to create more complex solutions while still respecting the initial conditions of the system being analyzed.
The ability to form linear combinations is what allows us to write the general solution of a second-order linear differential equation. For instance, considering the linear independence conditions derived through the Wronskian:

• If \( a_1b_2 - a_2b_1 eq 0 \), then distinct solutions form an independent set.

Linear combinations are at the core of building solutions tailored to specific initial conditions or constraints, which is a foundational aspect of differential equations in the modeling of real-life phenomena.