We develop a technique for solving homogeneous linear differential equations.

Homogeneous Linear Equations

A second order differential equation is said to be linear if it can be written as

We call the function on the right a forcing function, since in physical applications it’s often related to a force acting on some system modeled by the differential equation. We say that (eq:5.1.1) is homogeneous if or nonhomogeneous if . Since these definitions are like the corresponding definitions in Module linearFirstOrderDiffEq for the linear first order equation it’s natural to expect similarities between methods of solving (eq:5.1.1) and (eq:5.1.2). However, solving (eq:5.1.1) is more difficult than solving (eq:5.1.2). For example, while Theorem thmtype:2.1.1 gives a formula for the general solution of (eq:5.1.2) in the case where and Theorem thmtype:2.1.2 gives a formula for the case where , there are no formulas for the general solution of (eq:5.1.1) in either case. Therefore we must be content to solve linear second order equations of special forms.

In Module linearFirstOrderDiffEq we considered the homogeneous equation first, and then used a nontrivial solution of this equation to find the general solution of the nonhomogeneous equation . Although the progression from the homogeneous to the nonhomogeneous case isn’t that simple for the linear second order equation, it’s still necessary to solve the homogeneous equation

in order to solve the nonhomogeneous equation (eq:5.1.1). This section is devoted to (eq:5.1.3).

The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for (eq:5.1.3). We omit the proof.

Since is obviously a solution of (eq:5.1.3) we call it the trivial solution. Any other solution is nontrivial. Under the assumptions of Theorem thmtype:5.1.1, the only solution of the initial value problem on is the trivial solution.

The next three examples illustrate concepts that we’ll develop later in this section. You shouldn’t be concerned with how to find the given solutions of the equations in these examples. This will be explained in later sections.

Theorem thmtype:5.1.1 implies that if and are arbitrary real numbers then the initial value problem

has a unique solution on an interval that contains , provided that , , and are continuous and has no zeros on . To see this, we rewrite the differential equation in (eq:5.1.12) as and apply Theorem thmtype:5.1.1 with and .

Although the formulas for the solutions of (eq:5.1.14) and (eq:5.1.15) are both , you should not conclude that these two initial value problems have the same solution. Remember that a solution of an initial value problem is defined on an interval that contains the initial point; therefore, the solution of (eq:5.1.14) is on the interval , which contains the initial point , while the solution of (eq:5.1.15) is on the interval , which contains the initial point .

The General Solution of a Homogeneous Linear Second Order Equation

If and are defined on an interval and and are constants, then is a linear combination of and . For example, is a linear combination of and , with and .

The next theorem states a fact that we’ve already verified in Examples example:5.1.1, example:5.1.2, and example:5.1.3.

Proof
If then Therefore
since and are solutions of (eq:5.1.18).

We say that is a fundamental set of solutions of on if every solution of (eq:5.1.18) on can be written as a linear combination of and as in (eq:5.1.19). In this case we say that (eq:5.1.19) is general solution of on .

Linear Independence

We need a way to determine whether a given set of solutions of (eq:5.1.18) is a fundamental set. The next definition will enable us to state necessary and sufficient conditions for this.

We say that two functions and defined on an interval are linearly independent on if neither is a constant multiple of the other on . (In particular, this means that neither can be the trivial solution of (eq:5.1.18), since, for example, if we could write .) We’ll also say that the set is linearly independent on .

We’ll present the proof of Theorem thmtype:5.1.3 in steps worth regarding as theorems in their own right. However, let’s first interpret Theorem thmtype:5.1.3 in terms of Examples example:5.1.1, example:5.1.2, and example:5.1.3.

The Wronskian and Abel’s Formula

To motivate a result that we need in order to prove Theorem thmtype:5.1.3, let’s see what is required to prove that is a fundamental set of solutions of (eq:5.1.20) on . Let be an arbitrary point in , and suppose is an arbitrary solution of (eq:5.1.20) on . Then is the unique solution of the initial value problem

that is, and are the numbers obtained by evaluating and at . Moreover, and can be any real numbers, since Theorem thmtype:5.1.1 implies that (eq:5.1.21) has a solution no matter how and are chosen. Therefore is a fundamental set of solutions of (eq:5.1.20) on if and only if it’s possible to write the solution of an arbitrary initial value problem (eq:5.1.21) as . This is equivalent to requiring that the system has a solution for every choice of . Let’s try to solve (eq:5.1.22).

Multiplying the first equation in (eq:5.1.22) by and the second by yields

and subtracting the second equation here from the first yields Multiplying the first equation in (eq:5.1.22) by and the second by yields
and subtracting the first equation here from the second yields If it’s impossible to satisfy (eq:5.1.23) and (eq:5.1.24) (and therefore (eq:5.1.22)) unless and happen to satisfy
On the other hand, if we can divide (eq:5.1.23) and (eq:5.1.24) through by the quantity on the left to obtain no matter how and are chosen. This motivates us to consider conditions on and that imply (eq:5.1.25).

Proof
Differentiating (eq:5.1.28) yields Since and both satisfy (eq:5.1.27), Substituting these into (eq:5.1.30) yields
Therefore ; that is, is the solution of the initial value problem We leave it to you to verify by separation of variables that this implies (eq:5.1.29). If , (eq:5.1.29) implies that has no zeros in , since an exponential is never zero. On the other hand, if , (eq:5.1.29) implies that for all in .

The function defined in (eq:5.1.28) is the Wronskian of . Formula (eq:5.1.29) is Abel’s formula.

The Wronskian of is usually written as the determinant The expressions in (eq:5.1.26) for and can be written in terms of determinants as If you’ve taken linear algebra you may recognize this as Cramer’s rule.

The next theorem will enable us to complete the proof of Theorem thmtype:5.1.3.

Proof
We first show that if for some in , then and are linearly dependent on . Let be a subinterval of on which has no zeros. (If there’s no such subinterval, on , so and are linearly independent, and we’re finished with this part of the proof.) Then is defined on , and However, if , Theorem thmtype:5.1.4 implies that on . Therefore (eq:5.1.33) implies that , so (constant) on . This shows that for all in . However, we want to show that for all in . Let . Then is a solution of (eq:5.1.32) on such that on , and therefore on . Consequently, if is chosen arbitrarily in then is a solution of the initial value problem which implies that on , by the paragraph following Theorem thmtype:5.1.1. Hence, on , which implies that and are not linearly independent on .

Now suppose has no zeros on . Then can’t be identically zero on (why not?), and therefore there is a subinterval of on which has no zeros. Since (eq:5.1.33) implies that is nonconstant on , isn’t a constant multiple of on . A similar argument shows that isn’t a constant multiple of on , since on any subinterval of where has no zeros.

We can now complete the proof of Theorem thmtype:5.1.3.

Proof of Theorem thmtype:5.1.3:
From Theorem thmtype:5.1.5, two solutions and of (eq:5.1.32) are linearly independent on if and only if has no zeros on . From Theorem thmtype:5.1.4 and the motivating comments preceding it, is a fundamental set of solutions of (eq:5.1.32) if and only if has no zeros on . Therefore is a fundamental set for (eq:5.1.32) on if and only if is linearly independent on .

The next theorem summarizes the relationships among the concepts discussed in this section.

We can apply this theorem to an equation written as on an interval where , , and are continuous and has no zeros.

Proof
Since and are not both zero, (eq:5.1.35) implies that and Theorem thmtype:5.1.6 implies the stated conclusion.

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)

https://digitalcommons.trinity.edu/mono/8/