Question

If the functions \(y_{1}\) and \(y_{2}\) are linearly independent solutions of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), prove that \(y_{3}=y_{1}+y_{2}\) and \(y_{4}=y_{1}-y_{2}\) also form a linearly independent set of solutions. Conversely, if \(y_{3}\) and \(y_{4}\) are linearly independent solutions of the differential equation, show that \(y_{1}\) and \(y_{2}\) are also.

Short answer

Answer: Yes, if \(y_1\) and \(y_2\) are linearly independent solutions, then \(y_3 = y_1 + y_2\) and \(y_4 = y_1 - y_2\) are also linearly independent solutions. The converse is also true: if \(y_3\) and \(y_4\) are linearly independent solutions, then \(y_1\) and \(y_2\) are also linearly independent solutions.

Step-by-step solution

Calculate the Wronskian of \(y_1\) and \(y_2\)

Given that \(y_1\) and \(y_2\) are linearly independent solutions of the differential equation, we can calculate their Wronskian to ensure non-zero value. The Wronskian is defined as \(W(y_{1},y_{2})=\begin{vmatrix}y_{1} & y_{2}\\ y_{1}' & y_{2}'\end{vmatrix}=y_{1}y_{2}'-y_{1}'y_{2}\). Since they are linearly independent, \(W(y_{1},y_{2})\neq 0\).

Calculate the Wronskian of \(y_3\) and \(y_4\)

To show that \(y_3 = y_1+y_2\) and \(y_4=y_1-y_2\) are linearly independent, we will calculate their Wronskian. First, find the derivatives: \(y_{3}'=y_{1}'+y_{2}'\) and \(y_{4}'=y_{1}'-y_{2}'\) Now, compute the Wronskian \(W(y_{3},y_{4})\): \(W(y_{3},y_{4})=\begin{vmatrix}y_{3} & y_{4}\\ y_{3}' & y_{4}'\end{vmatrix}=\begin{vmatrix}y_{1}+y_{2} & y_{1}-y_{2}\\ y_{1}'+y_{2}' & y_{1}'-y_{2}'\end{vmatrix}=(y_{1}+y_{2})(y_{1}'-y_{2}')-(y_{1}'+y_{2}')(y_{1}-y_{2})\) Simplify the expression: \(W(y_{3},y_{4})=y_{1}y_{1}'+y_{1}(-y_{2}')-y_{2}(y_{1}')+y_{2}(-y_{2}')-(y_{1}y_{1'}+y_{1}y_{2}'-y_{2}y_{1}'-y_{2}y_{2}')\) Collect the terms related to the Wronskian of \(y_{1}\) and \(y_{2}\): \(W(y_{3},y_{4})=2(y_{1}y_{2}'-y_{1}'y_{2})=2W(y_{1},y_{2})\) Since \(W(y_{1},y_{2})\neq 0\), we can conclude that \(W(y_{3},y_{4})\neq 0\). Therefore, \(y_3\) and \(y_4\) are also linearly independent solutions.

Proving the converse

We now need to prove that if \(y_3\) and \(y_4\) are linearly independent solutions, then \(y_1\) and \(y_2\) are also linearly independent. As before, we will use the Wronskian. Given that \(y_3\) and \(y_4\) are linearly independent, we have \(W(y_3, y_4) \neq 0\). We have already calculated that \(W(y_{3},y_{4})=2W(y_{1},y_{2})\). So, if \(W(y_{3},y_{4})\neq 0\), then it implies that \(W(y_{1},y_{2})\neq 0\). Therefore, \(y_1\) and \(y_2\) are also linearly independent solutions. This completes the proof.

Key concepts

Wronskian

In differential equations, the Wronskian is an important determinant that helps determine the linear independence of a set of solutions. Imagine you are dealing with functions like \(y_1\) and \(y_2\). You can find their Wronskian by setting up a 2x2 determinant as follows:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2 \]The essence of the Wronskian is straightforward: if it is not zero, the functions are linearly independent. Knowing this can help solve complex differential equations since linearly independent functions serve as a fundamental set of solutions.For example, in our exercise, the fact that \(W(y_1, y_2) eq 0\) confirmed that \(y_1\) and \(y_2\) are indeed linearly independent solutions. This foundational understanding paved the way for proving new combinations of these functions as independent as well.

Homogeneous Linear Differential Equations

Homogeneous linear differential equations are a common type of differential equation. They typically have the form \[y'' + p(t) y' + q(t) y = 0\]where \(p(t)\) and \(q(t)\) are functions of \(t\), often simplified to constant coefficients. These equations are called 'homogeneous' since the right-hand side is zero. This is a specific property that significantly impacts the type of solutions they possess.Why are these equations important? Because they often describe natural phenomena, like harmonic oscillations, where the result (solution) solely depends on the initial conditions and inherent system properties. These equations only involve terms of the function and its derivatives, making them simpler to work with in certain contexts.The key to solving these equations lies in finding a fundamental set - two or more linearly independent solutions. This characteristic determines the general solution to the homogeneous equation.

Linearly Independent Solutions

Linearly independent solutions are crucial in understanding and solving differential equations. A set of solutions, such as \(y_1\) and \(y_2\), are linearly independent if none of the solutions can be expressed as a linear combination of the others. Mathematically, this relates to the Wronskian being non-zero.Why do we need linearly independent solutions? They provide a complete basis for the solution space of a differential equation. For a second-order homogeneous linear differential equation, having two linearly independent solutions means you can express any solution as a combination of these two.In the given problem, when \(y_1\) and \(y_2\) are transformed into \(y_3 = y_1 + y_2\) and \(y_4 = y_1 - y_2\), the challenge was to show that these new solutions are also independent. Using the Wronskian, the problem evolves into proving that if \(W(y_3, y_4) eq 0\), then \(W(y_1, y_2) eq 0\), which means independence holds in both sets. Essentially, having linearly independent solutions ensures that we have a broad understanding and control over the equations we aim to solve.