Theory for Higher Order Linear Homogeneous Differential Equations

We call differential equation described in the form of Eqn. (1) the nth-order linear homogeneous ODE, where a_i(x) are functions of x, for i = 0,1,…,n.

Note All theories about nth-order linear ODEs are analogous to those about the second-order linear ODEs. For example,

Theorem 1
Let y₁, y₂, …, y_m be the m solutions of L[y]=0.
Then c₁y₁ + c₂y₂ + … + c_my_m is a solution of L[y]=0, for any set of c_i.

Theorem 2
Let y₁, y₂, …, y_n be the n independent solutions of L[y] = 0.
If φ(x) is another solution of L[y] = 0, then there exists a set of constants c₁, c₂, …, c_n, such that φ(x) = c₁y₁ + c₂y₂ + … + c_ny_n.

Definition 1
If y₁, y₂, …, y_n are the n independent solutions of L[y] = 0,
then they are called the fundamental set of solutions of L[y] = 0, and c₁y₁ + c₂y₂ + … + c_ny_n is called the general solution of L[y]=0.

Definition 2
Let f₁, f₂, …, f_n be of class (n-1) and be defined on an interval (a, b). The Wronskian of f₁, f₂, …, f_n is defined by

Theorem 3
Let y₁, y₂, …, y_n be the n solutions of L[y] = 0, and all be defined on an open interval (α, β).
Then, either: (i) W(y₁, y₂, …, y_n)(x) = 0 for all x ∈ (α, β), or
(ii) W(y₁, y₂, …, y_n)(x) ≠ 0 for all x ∈ (α, β).

The Wroskian provides a quick method to examine whether a set of functions are linearly independent.

Theorem 4
Let y₁, y₂, …, y_n be the n solutions of L[y] = 0, and all be defined on an open interval (α, β).
Then, W(y₁, y₂, …, y_n)(x) ≠ 0 for some x ∈ (α, β),
if and only if y₁, y₂, …, y_n are linearly independent.

Note
In order to make Theorem 4 hold, y₁, y₂, …, y_n need to be solutions of SAME equation.

Example 1

Consider the function sets (a) and (b):

Assume they are, individually, solutions of some linear homogeneous equations with constant coefficients.
Examine if they are linearly independent.

Solution (a)
We may choose that c₁= 1, c₂ = −2, c₃ = 0, so that

Thus, they are linearly dependent. ❏

Solution (b)
We can use Theorem 4 and calculate the Wronskian:

Because for any x ∈ ℝ, W(x) ≠ 0, the functions of (b) are linearly independent. ❏

Note
Actually the function set (b) is the fundamental set of solutions of the equation

or, in terms of differential operator D ≡ d/dx,

Notice that exp(−3x) is the solution of (D+3)y=0 ; cos(2x) and sin(2x) form the fundamental set of solution of (D² + 4)y = 0.

The General Solution of L[y] = 0 with Constant Coefficients

Now, consider the nth-order linear homogeneous ODEs with constant coefficients

where a_i, i = 0,1,…,n, are constants.

Inspired by the fact that exponential functions are the very solutions of D[y] = 0, we specify the function

to be a solution of Eqn. (3). Substituting Eqn. (4) into Eqn. (3), one will obtain the characteristic equation

Eqn. (5) is an algebraic equation rather than differential equation. According to the fundamental theorem of algebra, there exists, counted with multiplicity, exact n (complex) roots of Eqn. (5).

Discussion

Case 1 (only distinct real roots)

Suppose the roots of Eqn. (5) are n distinct real roots:

then the corresponding solution set of Eqn. (3) is

and the general solution of Eqn. (3) is

It can be proved that the Wronskian of Eqn. (7) is a non-vanishing function of x, and thus those solutions are linearly independent.

Case 2 (complex conjugate roots)

Suppose the roots of Eqn. (5) are (n – 2) distinct real roots and one pair of complex conjugates:

where r_j (j = 1,2,…,k), a and b are real,then the corresponding solution set of Eqn. (3) is

One can write the general solution of Eqn. (3) based on the linear combination of elements of solution set (10). If there are two pairs of complex conjugates,

where r_ℓ (ℓ = 1,2,…,n − 2k), a_j, and b_j (j = 1,2,…,k) are real, then the corresponding solution set of Eqn. (3) is

Case 3 (repeated roots)

Suppose one of the roots of Eqn. (5) r = r_k has multiplicity p, that is, r = r_k is repeated p times. If r_k is real, then it corresponds to the solution set

which is a subset of the solution set of Eqn. (3). And if r_k is a pair of complex conjugates, and can be written a ± bi, where a and b are real, then the root r_k corresponds to the solution set

Example 2

Find the general solution of the following equation:

Solution
The characteristic equation of Eqn. (15) is

which can be factorized into

This leads to the roots r = 0, −5, 2.
Therefore, the general solution of Eqn. (15) is

❏

Example 3

Find the general solution of the following equation:

Solution
The characteristic equation of Eqn. (16) is

its roots are

Thus, the general solution of Eqn. (16) is

Otherwise, if we write down each solution of r,

then the general solution of Eqn. (16) will be

❏

Example 4

Find the general solution of the following equation:

Solution
The characteristic equation of Eqn. (17) is

which can be easily solved if we factorize it into

The solution is r = −1, −1, −1. Thus, the general solution of Eqn. (17) is

❏

Example 5

Find the general solution of the following equation:

Solution
The characteristic equation of Eqn. (1) is

It is not easy to find the factors of Eqn. (19) directly, but we may assume that it has rational roots: ±1, ±2, ±4, ±8. By the factor theorem, if Eqn. (19) has a root r = a, then P(a) = 0, where P(r) stands for the polynomial in Eqn. (19). We find that P(−1) = 1 + 5 + 6 − 4 − 8 = 0, so we can factorize Eqn. (19) and obtain

And we find that P(2) = 16 − 40 +24 + 8 − 8 = 0, so we factorize again :

Finally,

The roots of Eqn. (19) are −1, 2, 2, 2. Therefore, the general solution of Eqn. (18) is

❏