9.2: Higher Order Constant Coefficient Homogeneous Equations

If $a_0$, $a_1$, …, $a_n$ are constants and $a_0\ne0$, then

$$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=F(x)\nonumber$$

is said to be a constant coefficient equation. In this section we consider the homogeneous constant coefficient equation

$$\label{eq:9.2.1} a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0.$$

Since Equation $\ref{eq:9.2.1}$ is normal on $(-\infty,\infty)$, the theorems in Section 9.1 all apply with $(a,b)=(-\infty,\infty)$.

As in Section 5.2, we call

$$\label{eq:9.2.2} p(r)=a_0r^n+a_1r^{n-1}+\cdots+a_n$$

the characteristic polynomial of Equation $\ref{eq:9.2.1}$. We saw in Section 5.2 that when $n=2$ the solutions of Equation $\ref{eq:9.2.1}$ are determined by the zeros of the characteristic polynomial. This is also true when $n>2$, but the situation is more complicated in this case. Consequently, we take a different approach.

If $k$ is a positive integer, let $D^k$ stand for the $k$-th derivative operator; that is

$$D^ky=y^{(k)}. \nonumber$$

If

$$q(r)=b_0r^m+b_1r^{m-1}+\cdots+b_m \nonumber$$

is an arbitrary polynomial, define the operator

$$q(D)=b_0D^m+b_1D^{m-1}+\cdots+b_m \nonumber$$

such that

$$q(D)y=(b_0D^m+b_1D^{m-1}+\cdots+b_m)y=b_0y^{(m)}+b_1y^{(m-1)}+\cdots+ b_my \nonumber$$

whenever $y$ is a function with $m$ derivatives. We call $q(D)$ a polynomial operator.

With $p$ as in Equation $\ref{eq:9.2.2}$,

$$p(D)=a_0D^n+a_1D^{n-1}+\cdots+a_n, \nonumber$$

so Equation $\ref{eq:9.2.1}$ can be written as $p(D)y=0$. If $r$ is a constant then

$$\begin{align*} p(D)e^{rx}&= \left(a_0D^ne^{rx}+a_1D^{n-1}e^{rx}+\cdots+a_ne^{rx}\right)\\[4pt] &= (a_0r^n+a_1r^{n-1}+\cdots+a_n)e^{rx};\end{align*}\nonumber$$

that is

$$p(D)(e^{rx})=p(r)e^{rx}. \nonumber$$

This shows that $y=e^{rx}$ is a solution of Equation $\ref{eq:9.2.1}$ if $p(r)=0$. In the simplest case, where $p$ has $n$ distinct real zeros $r_1$, $r_2$,…, $r_n$, this argument yields $n$ solutions

$$y_1=e^{r_1x},\quad y_2=e^{r_2x},\dots,\quad y_n=e^{r_nx}. \nonumber$$

It can be shown (Exercise 9.2.39) that the Wronskian of $\{e^{r_1x},e^{r_2x},\dots,e^{r_nx}\}$ is nonzero if $r_1$, $r_2$, …, $r_n$ are distinct; hence, $\{e^{r_1x},e^{r_2x},\dots,e^{r_nx}\}$ is a fundamental set of solutions of $p(D)y=0$ in this case.

Example 9.2.1

1. Find the general solution of $$\label{eq:9.2.3} y'''-6y''+11y'-6y=0.$$
2. and solve the initial value problem $$\label{eq:9.2.4} y'''-6y''+11y'-6y=0, \quad y(0)=4,\quad y'(0)=5,\quad y''(0)=9.$$

Solution a

The characteristic polynomial of Equation $\ref{eq:9.2.3}$ is

$$p(r)=r^3-6r^2+11r-6=(r-1)(r-2)(r-3). \nonumber$$

Therefore $\{e^x,e^{2x},e^{3x}\}$ is a set of solutions of Equation $\ref{eq:9.2.3}$. It is a fundamental set, since its Wronskian is

$$W(x)=\left|\begin{array}{rrr}e^x&e^{2x}&e^{3x}\\[4pt]e^x&2e^{2x}& 3e^{3x}\\[4pt]e^x&4e^{2x}&9e^{3x}\end{array}\right|= e^{6x}\left|\begin{array}{rrr}1&1&1\\[4pt]1&2& 3\\[4pt]1&4&9\end{array}\right|=2e^{6x}\ne0. \nonumber$$

Therefore the general solution of Equation $\ref{eq:9.2.3}$ is

$$\label{eq:9.2.5} y=c_1e^{x}+c_2e^{2x}+c_3e^{3x}.$$

Solution b

We must determine $c_1$, $c_2$ and $c_3$ in Equation $\ref{eq:9.2.5}$ so that $y$ satisfies the initial conditions in Equation $\ref{eq:9.2.4}$. Differentiating Equation $\ref{eq:9.2.5}$ twice yields

$$\label{eq:9.2.6} \begin{array}{rcl} y'&=&c_1e^{x}+2c_2e^{2x}+3c_3e^{3x}\\[4pt] y''&=&c_1e^{x}+4c_2e^{2x}+9c_3e^{3x}. \end{array}$$

Setting $x=0$ in Equation $\ref{eq:9.2.5}$ and Equation $\ref{eq:9.2.6}$ and imposing the initial conditions yields

$$\begin{array}{rcl} c_1+\phantom{2}c_2+\phantom{3}c_3&=&4\\[4pt] c_1+2c_2+3c_3&=&5\\[4pt] c_1+4c_2+9c_3&=&9. \end{array}\nonumber$$

The solution of this system is $c_1=4$, $c_2=-1$, $c_3=1$. Therefore the solution of Equation $\ref{eq:9.2.4}$ is

$$y=4e^x-e^{2x}+e^{3x} \nonumber$$

(Figure 9.2.1 ).

Figure 9.2.1 : $y=4e^x-e^{2x}+e^{3x}$

Now we consider the case where the characteristic polynomial Equation $\ref{eq:9.2.2}$ does not have $n$ distinct real zeros. For this purpose it is useful to define what we mean by a factorization of a polynomial operator. We begin with an example.

Example 9.2.2

Consider the polynomial

$$p(r)=r^3-r^2+r-1 \nonumber$$

and the associated polynomial operator

$$p(D)=D^3-D^2+D-1. \nonumber$$

Since $p(r)$ can be factored as

$$p(r)=(r-1)(r^2+1)=(r^2+1)(r-1), \nonumber$$

it is reasonable to expect that p(D) can be factored as

$$\label{eq:9.2.7} p(D)=(D-1)(D^2+1)=(D^2+1)(D-1).$$

However, before we can make this assertion we must define what we mean by saying that two operators are equal, and what we mean by the products of operators in Equation $\ref{eq:9.2.7}$. We say that two operators are equal if they apply to the same functions and always produce the same result. The definitions of the products in Equation $\ref{eq:9.2.7}$ is this: if $y$ is any three-times differentiable function then

1. $(D-1)(D^2+1)y$ is the function obtained by first applying $D^2+1$ to $y$ and then applying $D-1$ to the resulting function
2. $(D^2+1)(D-1)y$ is the function obtained by first applying $D-1$ to $y$ and then applying $D^2+1$ to the resulting function.

From (a):

$$\label{eq:9.2.8} \begin{array}{rcl} (D-1)(D^2+1)y&=&(D-1)[(D^2+1)y]\\[4pt] &=&(D-1)(y''+y)=D(y''+y)-(y''+y)\\[4pt]&=&(y'''+y')-(y''+y)\\[4pt] &=&y'''-y''+y'-y=(D^3-D^2+D-1)y. \end{array}$$

This implies that

$$(D-1)(D^2+1)=(D^3-D^2+D-1). \nonumber$$

From (b):

$$\label{eq:9.2.9} \begin{array}{rcl} (D^2+1)(D-1)y&=&(D^2+1)[(D-1)y]\\[4pt] &=&(D^2+1)(y'-y)=D^2(y'-y)+(y'-y)\\[4pt]&=&(y'''-y'')+(y'-y)\\[4pt] &=&y'''-y''+y'-y=(D^3-D^2+D-1)y, \end{array}$$

$$(D^2+1)(D-1)=(D^3-D^2+D-1), \nonumber$$

which completes the justification of Equation $\ref{eq:9.2.7}$.

Example 9.2.3

Use the result of Example 9.2.2 to find the general solution of

$$\label{eq:9.2.10} y'''-y''+y'-y=0.$$

Solution

From Equation $\ref{eq:9.2.8}$, we can rewrite Equation $\ref{eq:9.2.10}$ as

$$(D-1)(D^2+1)y=0, \nonumber$$

which implies that any solution of $(D^2+1)y=0$ is a solution of Equation $\ref{eq:9.2.10}$. Therefore $y_1=\cos x$ and $y_2=\sin x$ are solutions of Equation $\ref{eq:9.2.10}$.

From Equation $\ref{eq:9.2.9}$, we can rewrite Equation $\ref{eq:9.2.10}$ as

$$(D^2+1)(D-1)y=0, \nonumber$$

which implies that any solution of $(D-1)y=0$ is a solution of Equation $\ref{eq:9.2.10}$. Therefore $y_3=e^x$ is solution of Equation $\ref{eq:9.2.10}$.

The Wronskian of $\{e^x,\cos x,\sin x\}$ is

$$W(x)=\left|\begin{array}{rrr}\cos x&\sin x&e^x\\[4pt]-\sin x&\cos x&e^x\\[4pt] -\cos x&-\sin x&e^x\end{array}\right|. \nonumber$$

Since

$$W(0)=\left|\begin{array}{rrr}1&0&1\\[4pt]0&1&1\\[4pt] -1&0&1\end{array}\right|=2, \nonumber$$

$\{\cos x,\sin x,e^x\}$ is linearly independent and

$$y=c_1\cos x+c_2\sin x+c_3e^x \nonumber$$

is the general solution of Equation $\ref{eq:9.2.10}$.

Example 9.2.4

Find the general solution of

$$\label{eq:9.2.11} y^{(4)}-16y=0.$$

Solution

The characteristic polynomial of Equation $\ref{eq:9.2.11}$ is

$$\begin{align*} p(r) &=r^4-16 \\[4pt] &=(r^2-4)(r^2+4) \\[4pt] &=(r-2)(r+2)(r^2+4). \end{align*}\nonumber$$

By arguments similar to those used in Examples 9.2.2 and 9.2.4 , it can be shown that Equation $\ref{eq:9.2.11}$ can be written as

$$(D^2+4)(D+2)(D-2)y=0 \nonumber$$

or

$$(D^2+4)(D-2)(D+2)y=0 \nonumber$$

or

$$(D-2)(D+2)(D^2+4)y=0. \nonumber$$

Therefore $y$ is a solution of Equation $\ref{eq:9.2.11}$ if it is a solution of any of the three equations

$$(D-2)y=0,\quad (D+2)y=0, \quad(D^2+4)y=0. \nonumber$$

Hence, $\{e^{2x},e^{-2x},\cos2x,\sin2x\}$ is a set of solutions of Equation $\ref{eq:9.2.11}$. The Wronskian of this set is

$$W(x)=\left|\begin{array}{rrrr} e^{2x}&e^{-2x}&\cos2x&\sin2x\\[4pt] 2e^{2x}&-2e^{-2x}&-2\sin2x&2\cos2x\\[4pt] 4e^{2x}&4e^{-2x}&-4\cos2x&-4\sin2x\\[4pt] 8e^{2x}&-8e^{-2x}&8\sin2x&-8\cos2x\\[4pt] \end{array}\right|. \nonumber$$

Since

$$W(0)=\left|\begin{array}{rrrr} 1&1&1&0\\[4pt] 2&-2&0&2\\[4pt] 4&4&-4&0\\[4pt] 8&-8&0&-8\\[4pt] \end{array}\right|=-512, \nonumber$$

$\{e^{2x},e^{-2x},\cos2x,\sin2x\}$ is linearly independent, and

$$y_1=c_1e^{2x}+c_2e^{-2x}+c_3\cos2x+c_4\sin2x \nonumber$$

is the general solution of Equation $\ref{eq:9.2.11}$.

It is known from algebra that every polynomial

$$p(r)=a_0r^n+a_1r^{n-1}+\cdots+a_n \nonumber$$

with real coefficients can be factored as

$$p(r)=a_0p_1(r)p_2(r)\cdots p_k(r), \nonumber$$

where no pair of the polynomials $p_1$, $p_2$, …, $p_k$ has a commom factor and each is either of the form

$$\label{eq:9.2.12} p_j(r)=(r-r_j)^{m_j},$$

where $r_j$ is real and $m_j$ is a positive integer, or

$$\label{eq:9.2.13} p_j(r)=\left[(r-\lambda_j)^2+\omega_j^2\right]^{m_j},$$

where $\lambda_j$ and $\omega_j$ are real, $\omega_j\ne0$, and $m_j$ is a positive integer. If Equation $\ref{eq:9.2.12}$ holds then $r_j$ is a real zero of $p$, while if Equation $\ref{eq:9.2.13}$ holds then $\lambda+i\omega$ and $\lambda-i\omega$ are complex conjugate zeros of $p$. In either case, $m_j$ is the multiplicity of the zero(s).

By arguments similar to those used in our examples, it can be shown that

$$\label{eq:9.2.14} p(D)=a_0p_1(D)p_2(D)\cdots p_k(D)$$

and that the order of the factors on the right can be chosen arbitrarily. Therefore, if $p_j(D)y=0$ for some $j$ then $p(D)y=0$. To see this, we simply rewrite Equation $\ref{eq:9.2.14}$ so that $p_j(D)$ is applied first. Therefore the problem of finding solutions of $p(D)y=0$ with $p$ as in Equation $\ref{eq:9.2.14}$ reduces to finding solutions of each of these equations

$$p_j(D)y=0,\quad 1\le j\le k, \nonumber$$

where $p_j$ is a power of a first degree term or of an irreducible quadratic. To find a fundamental set of solutions $\{y_1,y_2,\dots,y_n\}$ of $p(D)y=0$, we find fundamental set of solutions of each of the equations and take $\{y_1,y_2,\dots,y_n\}$ to be the set of all functions in these separate fundamental sets. In Exercise 9.2.40 we sketch the proof that $\{y_1,y_2,\dots,y_n\}$ is linearly independent, and therefore a fundamental set of solutions of $p(D)y=0$.

To apply this procedure to general homogeneous constant coefficient equations, we must be able to find fundamental sets of solutions of equations of the form

$$(D-a)^my=0 \nonumber$$

and

$$\left[(D-\lambda)^2+\omega^2\right]^my=0, \nonumber$$

where $m$ is an arbitrary positive integer. The next two theorems show how to do this.

Theorem 9.2.1

If $m$ is a positive integer, then

$$\label{eq:9.2.15} \{e^{ax}, xe^{ax},\dots, x^{m-1}e^{ax}\}$$

is a fundamental set of solutions of

$$\label{eq:9.2.16} (D-a)^my=0.$$

Proof

We’ll show that if

$$f(x)=c_1+c_2x+\cdots+c_mx^{m-1} \nonumber$$

is an arbitrary polynomial of degree $\le m-1$, then $y=e^{ax}f$ is a solution of Equation $\ref{eq:9.2.16}$. First note that if $g$ is any differentiable function then

$$(D-a)e^{ax}g=De^{ax}g-ae^{ax}g=ae^{ax}g+e^{ax}g'-ae^{ax}g, \nonumber$$

so

$$\label{eq:9.2.17} (D-a)e^{ax}g=e^{ax}g'.$$

Therefore

$$\begin{array}{lcll} (D-a)e^{ax}f&=&e^{ax}f'&\mbox{(from \eqref{eq:9.2.17} with $g=f$)}\\[4pt] (D-a)^2e^{ax}f&=& (D-a)e^{ax}f'=e^{ax}f'' &\mbox{(from \eqref{eq:9.2.17} with $g=f'$)}\\[4pt] (D-a)^3e^{ax}f&=& (D-a)e^{ax}f''=e^{ax}f''' &\mbox{(from \eqref{eq:9.2.17} with $g=f''$)}\\[4pt] &\vdots&\\[4pt] (D-a)^me^{ax}f &=&(D-a)e^{ax}f^{(m-1)}=e^{ax}f^{(m)} &\mbox{(from \eqref{eq:9.2.17} with $g=f^{(m-1)}$)}. \end{array}\nonumber$$

Since $f^{(m)}=0$, the last equation implies that $y=e^{ax}f$ is a solution of Equation $\ref{eq:9.2.16}$ if $f$ is any polynomial of degree $\le m-1$. In particular, each function in Equation $\ref{eq:9.2.15}$ is a solution of Equation $\ref{eq:9.2.16}$. To see that Equation $\ref{eq:9.2.15}$ is linearly independent (and therefore a fundamental set of solutions of Equation $\ref{eq:9.2.16}$), note that if

$$c_1e^{ax}+c_2xe^{ax}+c\dots+c_{m-1}x^{m-1}e^{ax}=0 \nonumber$$

for all $x$ in some interval $(a,b)$, then

$$c_1+c_2x+c\dots+c_{m-1}x^{m-1}=0 \nonumber$$

for all $x$ in $(a,b)$. However, we know from algebra that if this polynomial has more than $m-1$ zeros then $c_1=c_2=\cdots=c_n=0$.

Example 9.2.5

Find the general solution of

$$\label{eq:9.2.18} y'''+3y''+3y'+y=0.$$

Solution

The characteristic polynomial of Equation $\ref{eq:9.2.18}$ is

$$p(r)=r^3+3r^2+3r+1=(r+1)^3. \nonumber$$

Therefore Equation $\ref{eq:9.2.18}$ can be written as

$$(D+1)^3y=0, \nonumber$$

so Theorem 9.2.1 implies that the general solution of Equation $\ref{eq:9.2.18}$ is

$$y=e^{-x}(c_1+c_2x+c_3x^2). \nonumber$$

The proof of the next theorem is sketched in Exercise 9.2.41.

Theorem 9.2.2

If $\omega \ne0$ and $m$ is a positive integer, then

$$\begin{array}{rl} \{e^{\lambda x}\cos\omega x, xe^{\lambda x}\cos\omega x, &\dots, x^{m-1}e^{\lambda x}\cos\omega x,\\[4pt] e^{\lambda x}\sin\omega x, xe^{\lambda x}\sin\omega x,& \dots, x^{m-1}e^{\lambda x}\sin\omega x\} \end{array}\nonumber$$

is a fundamental set of solutions of

$$[(D-\lambda)^2+\omega^2]^my=0. \nonumber$$

Example 9.2.6

Find the general solution of

$$\label{eq:9.2.19} (D^2+4D+13)^3y=0.$$

Solution

The characteristic polynomial of Equation $\ref{eq:9.2.19}$ is

$$p(r)=(r^2+4r+13)^3=\left((r+2)^2+9\right)^3. \nonumber$$

Therefore Equation $\ref{eq:9.2.19}$ can be be written as

$$[(D+2)^2+9]^3y=0, \nonumber$$

so Theorem 9.2.2 implies that the general solution of Equation $\ref{eq:9.2.19}$ is

$$y=(a_1+a_2x+a_3x^2)e^{-2x}\cos3x +(b_1+b_2x+b_3x^2)e^{-2x}\sin3x. \nonumber$$

Example 9.2.7

Find the general solution of

$$\label{eq:9.2.20} y^{(4)}+4y'''+6y''+4y'=0.$$

Solution

The characteristic polynomial of Equation $\ref{eq:9.2.20}$ is

$$\begin{aligned} p(r)&=r^4+4r^3+6r^2+4r\\[4pt] &=r(r^3+4r^2+6r+4)\\[4pt] &=r(r+2)(r^2+2r+2)\\[4pt] &=r(r+2)[(r+1)^2+1].\end{aligned}\nonumber$$

Therefore Equation $\ref{eq:9.2.20}$ can be written as

$$[(D+1)^2+1](D+2)Dy=0. \nonumber$$

Fundamental sets of solutions of

$$\left[(D+1)^2+1\right] y=0,\quad (D+2) y=0,\quad \text{and} \quad Dy=0. \nonumber$$

are given by

$$\{e^{-x}\cos x,e^{-x}\sin x\},\quad \{e^{-2x}\},\quad \text{and} \quad \{1\}, \nonumber$$

respectively. Therefore the general solution of Equation $\ref{eq:9.2.20}$ is

$$y=e^{-x}(c_1\cos x+c_2\sin x)+c_3e^{-2x}+c_4. \nonumber$$

Example 9.2.8

Find a fundamental set of solutions of

$$\label{eq:9.2.21} [(D+1)^2+1]^2(D-1)^3(D+1)D^2y=0.$$

Solution

A fundamental set of solutions of Equation $\ref{eq:9.2.21}$ can be obtained by combining fundamental sets of solutions of

• $\left[(D+1)^{2}+1\right]^{2} y=0$
• $(D-1)^{3} y=0$
• $(D+1) y=0$
• $D^{2} y=0$

Fundamental sets of solutions of these equations are given by

• $\{e^{-x} \cos x, x e^{-x} \cos x, e^{-x} \sin x, x e^{-x} \sin x\}$
• $\left\{e^{x}, x e^{x}, x^{2} e^{x}\right\}$
• $\left\{e^{-x}\right\}$,
• $\{1, x\}$

respectively. These ten functions form a fundamental set of solutions of Equation $\ref{eq:9.2.21}$.