§14. Some Theoretical Considerations

The concept of linearity is explored in the context of differential equations.

One of the more interesting developments in modern mathematics is the number of situations which have been identified in which concepts from linear algebra play a significant role. Here the interaction between linear algebra and differential equations will be briefly explored.

The phrase ‘linear algebra’ is associated with two central concepts: vectors and linear transformations. In order to make the connection between linear algebra and differential equations apparent, vectors and linear transformations must be identified in the differential equations setting.

The important properties of vectors are that the sum of two vectors is again a vector, and the product of a number (scalar) and a vector is again a vector. The key realization is that functions also have these two properties. The sum of two functions is again a function; the product of a number and a function is again a function. Functions can therefore be viewed as vectors (points) in some space.

A linear trasformation L is a function which maps vectors into vectors and obeys the linearity property L(v + w) = L(v) + L(w). In this last equation read the phrase ‘the derivative of’ in place of L; differentiation can be viewed as a linear transformation on the newly found set of vectors (which used to be thought of as functions). The connection between differential equations and linear algebra is now almost complete.

Example 14–1. Define a linear transformation L on the space of functions of the

variable t by the formula L(f (t)) = dtd f (t) − 7f (t). The kernel of L is then the set of functions f (t) for which L(f (t)) = 0. Notice that the kernel of L is also the set of

solutions of the differential equation dtd f (t) − 7f (t) = 0. This sort of reasoning can be used to show that the solution set is one dimensional; this is why there is only one arbitrary constant in the solution of first order linear differential equations with constant coefficients.

Example 14–2. As a second example the annihilator method of solving second order linear non-homogeneous equations is developed. The method of undetermined coefficients can sometimes be difficult to apply. The annihilator method is often easier to use. Suppose the equation to be solved is

d2

dt2 x(t) + 4x(t) = sin 2t.

The general solution of the corresponding homogeneous equation is C1 cos 2t + C2 sin 2t. The method of undetermined coefficients will fail here if applied directly

Copyright 2002 Jerry Alan Veeh. All rights reserved.

method.

§14: Some Theoretical Considerations 87

Solutions to Problems

Problem 14–1. The function e2t satisfies the differential equation dtd x(t) = 2x(t). The characteristic equation for the annihilator method is then (m−2)(m2 −4) = 0. The general soltution is then C1e−2t + e2t(C2 + C3t). The constant C3 is then determined by substitution.

Problem 14–2. The function t2 satisfies the differential equation dtd33 x(t) = 0. The characteristic equation for the annihilator method is then m3(m2 − 4) = 0. The general solution is C1 cos 2t + C2 sin 2t + C3t2 + C4t + C5. The constants C3, C4, and C5 are then determined by substitution.

§14: Some Theoretical Considerations 88

Solutions to Exercises

Exercise 14–1. Hint: The constants C1 and C2 can be taken to be zero since cos 2t and sin 2t solve the homogeneous equation. Then x′′(t) + 4x(t) = −4C3 sin 2t + 4C4 cos 2t. This is equal to sin 2t if C3 = −1/4 and C4 = 0. The general solution to the non-homogeneous equation is x(t) = C1 cos 2t+C2 sin 2t− t/4 cos 2t.

Exercise 14–2. The equations for the coefficients are 0 = C1 and 1 = 2C2 − 1/4.