Consider the secondorder linear homogeneous equation Ly y

Consider the second-order linear homogeneous equation L[y] = y + p(t)y + q(t)y = 0, where p, q are continuous. Suppose that y1 and y2 are fundamental solutions of this equation on an interval (a, b). (a) (1 point) Prove that y1 and y2 cannot both be zero at any point in (a, b). (b) (2 points) Using any method you like, show that y1(t) = t 1 and y2(t) = (t 1)^2 cannot form a fundamental set of solutions to any second-order linear homogeneous equation on the interval (2, 2). (c) (2 points) Using any method you like, show that y1(t) = esin(t) + 3 2 and arctan(t) cannot form a fundamental set of solutions to any second-order linear homogeneous equation on the interval (1, 1).

Solution

For the most part, we will only learn how to solve second order linear equation with constant coefficients (that is, when p(t) and q(t) are constants). Since a homogeneous equation is easier to solve compares to its nonhomogeneous counterpart, we start with second order linear homogeneous equations that contain constant coefficients only: a y + b y + c y = 0. Where a, b, and c are constants, a 0. A very simple instance of such type of equations is y y = 0. The equation’s solution is any function satisfying the equality y = y. Obviously y1 = e t is a solution, and so is any constant multiple of it, C1 e t . Not as obvious, but still easy to see, is that y2 = e t is another solution (and so is any function of the form C2 e t). It can be easily verified that any function of the form y = C1 e t + C2 e t will satisfy the equation. In fact, this is the general solution of the above differential equation.

We have seen that the general solution of a second order homogeneous linear equation is in the form of y = C1 y1 + C2 y2 § , where y1 and y2 are two “distinct” functions both satisfying the given equation (as a result, y1 and y2 are themselves particular solutions of the equation). Now we will examine the circumstance under which two arbitrary solutions y1 and y2 could give us a general solution. Suppose y1 and y2 are two solutions of some second order homogeneous linear equation such that their linear combinations y = C1 y1 + C2 y2 give a general solution of the equation. Then, according to the Existence and Uniqueness Theorem, for any pair of initial conditions y(t0) = y0 and y(t0) = y0 there must exist uniquely a corresponding pair of coefficients C1 and C2 that satisfies the system of (algebraic) equations ( ) ( ) ( ) ( ) 0 1 1 0 2 2 0 0 1 1 0 2 2 0 y C y t C y t y C y t C y t = + = + From linear algebra, we know that for the above system to always have a unique solution (C1, C2) for any initial values y0 and y0, the coefficient matrix of the system must be invertible, or, equivalently, the determinant of the coefficient matrix must be nonzero**. That is ( ) ( ) ( ) ( ) 0 ( ) ( ) ( ) ( ) det 1 0 2 0 1 0 2 0 1 0 2 0 1 0 2 0 = y t y t y t y t y t y t y t y t This determinant above is called the Wronskian or the Wronskian determinant. It is a function of t as well, denoted W(y1, y2)(t), and is given by the expression W(y1, y2)(t) = y1 y 2 y 1 y2.