Question

Suppose \(p\) and \(q\) are continuous on \((a, b)\) and \(\left\\{y_{1}, y_{2}\right\\}\) is a set of solutions of $$y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0 $$ on \((a, b)\) such that either \(y_{1}\left(x_{0}\right)=y_{2}\left(x_{0}\right)=0\) or \(y_{1}^{\prime}\left(x_{0}\right)=y_{2}^{\prime}\left(x_{0}\right)=0\) for some \(x_{0}\) in \((a, b)\). Show that \(\left\\{y_{1}, y_{2}\right\\}\) is linearly dependent on \((a, b)\).

Short answer

Answer: Yes, the solutions y1 and y2 are linearly dependent on the interval (a, b) if either the function values or the first derivative values are the same at the point x0.

Step-by-step solution

Define the Wronskian and the main result

For a set of solutions \(\{y_1, y_2\}\) of the given second-order linear differential equation, define their Wronskian as \[W(x) = \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix} = y_1(x)y_2'(x) - y_2(x)y_1'(x).\] The key result for this exercise is that if the Wronskian is identically zero, i.e., \(W(x)=0\) for all \(x\) in \((a, b)\), then the set \(\{y_1, y_2\}\) is linearly dependent on \((a, b)\). This can be shown by computing the Wronskian for a linearly independent set, then showing it cannot be zero.

Derive the Wronskian of the given differential equation

Now, we will derive the Wronskian of the solutions of the given differential equation in terms of \(y_1'\), \(y_2'\), \(p(x)\), and \(q(x)\). By multiplying the first solution by the second solution's derivative, subtract the second solution by the first solution's derivative and apply the differential equation, we obtain: \[W'(x) = y_1'(x)y_2'(x) - y_2'(x)y_1'(x) = -p(x)W(x).\]

Solve the differential equation for the Wronskian

The above equation is a first-order linear homogeneous differential equation for the Wronskian. The general solution is given by: \[W(x) = W(x_0)e^{-\int_{x_0}^x p(t) \ \mathrm{d}t},\] where \(W(x_0)\) is the value of the Wronskian at \(x_0\).

Apply the conditions given in the problem

Now, we have two cases: either \(y_{1}(x_{0})=y_{2}(x_{0})=0\) or \(y_{1}^{\prime}(x_{0})=y_{2}^{\prime}(x_{0})=0\). Case 1: If \(y_{1}(x_{0})=y_{2}(x_{0})=0\), then by the definition of the Wronskian at that point, \[W(x_0) = y_1(x_0)y_2'(x_0) - y_2(x_0)y_1'(x_0) = 0.\] Hence, the Wronskian is identically zero on \((a, b)\). Case 2: If \(y_{1}^{\prime}(x_{0})=y_{2}^{\prime}(x_{0})=0\), then by the definition of the Wronskian at that point, \[W(x_0) = y_1(x_0)y_2'(x_0) - y_2(x_0)y_1'(x_0) = 0.\] Again, the Wronskian is identically zero on \((a, b)\).

Conclude that the solutions are linearly dependent

In both cases, the Wronskian of the solutions \(y_1\) and \(y_2\) is identically zero on \((a, b)\). Therefore, by the main result, the set of solutions \(\{y_1, y_2\}\) is linearly dependent on the interval \((a, b)\).

Key concepts

Wronskian

The Wronskian is a crucial tool for determining the linear independence of a set of functions, particularly solutions to differential equations. Let's take a closer look at two solutions to a differential equation, denoted as \(y_1(x)\) and \(y_2(x)\). To compute their Wronskian, we arrange these functions and their first derivatives into a matrix and find its determinant:
\[ W(x) = \begin{vmatrix} y_1(x) & y_2(x) \ y_1'(x) & y_2'((x) \end{vmatrix} = y_1(x)y_2'(x) - y_2(x)y_1'(x). \]
If our computation results in the Wronskian being zero for every point in an interval, this implies that the functions are linearly dependent in that interval. In other words, one function is a multiple of the other, and they do not provide unique information. This concept is especially useful in the study of differential equations, as it helps us understand the structure of their solution space.

Second-Order Linear Differential Equation

Differential equations are equations that relate a function with its derivatives. When dealing with a second-order linear differential equation, we are looking for a function \(y(x)\) that satisfies an equation of the form:
\[ y'' + p(x)y' + q(x)y = 0, \]
where \(y''\) is the second derivative of \(y\) with respect to \(x\), and \(p(x)\) and \(q(x)\) are continuous functions on a given interval. These types of equations often arise in physics and engineering, modeling systems with acceleration, such as oscillations. The solutions to these equations can be used to predict future behavior of systems, making them invaluable tools for scientists and engineers.

Homogeneous Differential Equation

In the context of differential equations, 'homogeneous' has a specific meaning: it describes an equation where every term is a function of the dependent variable \(y\) and its derivatives. Essentially, there's no 'free term' or constant term included. For our second-order linear differential equation, \(y''+p(x)y'+q(x)y=0\) is homogeneous because there is no term without dependency on \(y\).
A crucial property of homogeneous linear differential equations is the principle of superposition, which says that if \(y_1(x)\) and \(y_2(x)\) are solutions, then so is any linear combination of these solutions: \(c_1y_1(x) + c_2y_2(x)\), where \(c_1\) and \(c_2\) are constants. This principle guides us in understanding the structure of the solution space and forms the basis for our analysis of linear dependence and independence through the Wronskian.

Continuous Functions

When tackling differential equations, it's essential to consider the continuity of the functions involved. A continuous function is one where small changes in the input (within its domain) lead to small changes in the output; there are no abrupt jumps or breaks.
Our problem requires the functions \(p(x)\) and \(q(x)\) to be continuous on an interval \((a, b)\). Continuity is crucial because it ensures the behavior of the solutions to the differential equation can be well predicted and understood within that interval. In the context of the Wronskian, continuous coefficients in the differential equation allow us to conclude that a zero Wronskian over some interval implies linear dependence of solutions over the entire interval.