Question

Show that \(t\) and \(t^{2}\) are linearly independent on \(-1< t<1 ;\) indeed, they are linearly independent on every interval. Show also that \(W\left(t, t^{2}\right)\) is zero at \(t=0 .\) What can you conclude from this about the possibility that \(t\) and \(t^{2}\) are solutions of a differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 ?\) Verify that \(t\) and \(t^{2}\) are solutions of the equation \(t^{2} y^{\prime}+\) \(2 y=0 .\) Does this contradict your conclusion? Does the behavior of the Wronskian of \(t\) and \(t^{2}\) contradict Theorem \(3.3 .2 ?\)

Short answer

Explain in the context of the Wronskian and Theorem 3.3.2. Answer: Yes, \(t\) and \(t^2\) are linearly independent solutions to the given differential equation on the interval \(-1 < t < 1\). The Wronskian at \(t = 0\) is 0, which seemingly contradicts Theorem 3.3.2, which states that if the Wronskian is nonzero, the functions are linearly independent solutions. However, this contradiction occurs because the given differential equation has a singular point at \(t = 0\), and Theorem 3.3.2 only applies to functions defined on an interval where the differential equation has no singular points. Thus, there is no contradiction, and \(t\) and \(t^2\) are indeed linearly independent solutions on the given interval.

Step-by-step solution

Show \(t\) and \(t^2\) are linearly independent

To show that \(t\) and \(t^2\) are linearly independent, we need to prove that there are no constants \(c_1\) and \(c_2\) (other than both being zero) such that \(c_1t + c_2t^2 = 0\) for all \(t\) in the interval \(-1 < t < 1\). Assume there exist nonzero constants \(c_1\) and \(c_2\) such that \(c_1t + c_2t^2 = 0\). Then, \(c_2t^2 = -c_1t\) and \(t = -\frac{c_1}{c_2}\). Since \(t\) is a variable in the interval \(-1 < t < 1\), this equation should hold for all \(t\) in this interval. However, this equation gives a single constant value for \(t\), which means \(t\) and \(t^2\) are linearly independent, as there are no such constants \(c_1\) and \(c_2\) that satisfy the equation.

Find the Wronskian of \(t\) and \(t^2\)

The Wronskian of two functions \(u(t)\) and \(v(t)\) is given by \(W(u, v) = \begin{vmatrix} u & v \\ u' & v' \end{vmatrix} = u v' - u' v\). For \(u(t) = t\) and \(v(t) = t^2\), we have \(u'(t) = 1\), and \(v'(t) = 2t\). The Wronskian of \(t\) and \(t^2\) is \(W(t, t^2) = t(2t) - 1(t^2) = t^2\).

Analyze Wronskian at \(t = 0\)

The Wronskian at \(t = 0\) is given by \(W(0, 0^2) = 0\). This indicates that \(t\) and \(t^2\) may not be linearly independent solutions to a homogeneous differential equation, as normally the Wronskian is nonzero for linearly independent solutions.

Verify solutions for the given differential equation

The given differential equation is \(t^2y'' + 2y = 0\). Let's plug in \(y_1 = t\) and \(y_2 = t^2\) as candidate solutions: For \(y_1(t) = t\), we have \(y_1''(t) = 0\). For \(y_2(t) = t^2\), we have \(y_2''(t) = 2\). Now let's plug these into the differential equation: \(y_1: t^2(0) + 2(t) = 2t\) \(y_2: t^2(2) + 2(t^2) = 4t^2\) Since the differential equation is satisfied for both \(y_1\) and \(y_2\), they are indeed solutions.

Evaluate conclusions and contradiction with Theorem 3.3.2

The Wronskian for \(t\) and \(t^2\) at \(t=0\) is 0, yet we have found that they are both linearly independent solutions to the given differential equation. This seemingly contradicts Theorem 3.3.2, which states that if the Wronskian is nonzero, the functions are linearly independent solutions. However, upon closer inspection, we see that the given differential equation has a singular point at \(t = 0\). The theorem applies to functions defined on an interval where the differential equation has no singular points. Since the interval we are considering includes the singular point \(t = 0\), Theorem 3.3.2 does not apply, and there is no contradiction.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are central in mathematics because they describe various types of physical and theoretical phenomena, such as the motion of fluids, the decay of substances, and the oscillation of springs, just to mention a few. Differential equations can be classified in several ways; one common classification is based on order (first, second, etc.), which refers to the highest derivative present in the equation. Another classification is by linearity: linear differential equations are those where the unknown function and its derivatives appear to the power of one and are not multiplied together.

Differential equations can be either homogeneous or non-homogeneous. A homogeneous differential equation is one in which the function equals zero for all terms. In the problem outlined, we are dealing with a second-order homogeneous differential equation, where students must understand both the nature of the equation and how solutions can be verified.

Wronskian

The Wronskian is a tool in the realm of differential equations used to determine the linear independence of a set of solutions. It is named after the Polish mathematician Hoene Wronski. Formally, for two functions, the Wronskian is given by the determinant of a matrix composed of the functions and their first derivatives. If the Wronskian is nonzero at a point in an interval where the functions are analytic, then the functions are linearly independent on that interval.

In the context of our exercise, we calculate the Wronskian for the functions t and t^2. This is crucial as linear independence of solutions to homogeneous differential equations typically results in a nonzero Wronskian, which leads to the conclusion that the solutions span the solution space for the differential equation on the interval where they are defined.

Homogeneous Differential Equation

A homogeneous differential equation is one in which the sum of the terms without the derivatives, called the 'non-homogeneous' term, is zero. For a homogeneous second-order differential equation of the form \(y'' + p(t) y' + q(t) y = 0\), where \(y\) is the unknown function of \(t\), and \(p(t)\) and \(q(t)\) are coefficients, the solutions can be identified through organization and testing of different functions that satisfy the equation.

In relation to the given exercise, the students are exploring whether the functions t and t^2 could be solutions to a nondescript second-order homogeneous differential equation. They are also tasked with directly verifying if these functions are solutions to a more specific homogeneous differential equation. This exercise guides students in understanding how homogeneous differential equations can be solved using linear independence based on the Wronskian.

Singular Points

Singular points, or singularities, in the context of differential equations, are values of the independent variable where the coefficients of the highest derivatives in the equation become zero or undefined, causing the standard analysis and solution methods to potentially break down. For example, in a second-order differential equation, if the coefficient of \(y''\) becomes zero at a certain point, that point is singular.

Our exercise features a pivotal analysis of the behavior of solutions at the singular point \(t=0\). What's key for students to recognize is that Theorem 3.3.2 does not necessarily apply at or around singular points since the existence and uniqueness theorems and associated properties like the behavior of the Wronskian for linear independence only hold true where the coefficients of the differential equation behave regularly, that is, are not zero or undefined.