19.2: Higher Order Constant Coefficient Homogeneous Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
If a0, a1, …, an are constants and a0≠0, then
a0y(n)+a1y(n−1)+⋯+any=F(x)
is said to be a constant coefficient equation. In this section we consider the homogeneous constant coefficient equation
a0y(n)+a1y(n−1)+⋯+any=0.
Since Equation ??? is normal on (−∞,∞), the theorems in Section 9.1 all apply with (a,b)=(−∞,∞).
As in Section 5.2, we call
p(r)=a0rn+a1rn−1+⋯+an
the characteristic polynomial of Equation ???. We saw in Section 5.2 that when n=2 the solutions of Equation ??? 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 Dk stand for the k-th derivative operator; that is
Dky=y(k).
If
q(r)=b0rm+b1rm−1+⋯+bm
is an arbitrary polynomial, define the operator
q(D)=b0Dm+b1Dm−1+⋯+bm
such that
q(D)y=(b0Dm+b1Dm−1+⋯+bm)y=b0y(m)+b1y(m−1)+⋯+bmy
whenever y is a function with m derivatives. We call q(D) a polynomial operator.
With p as in Equation ???,
p(D)=a0Dn+a1Dn−1+⋯+an,
so Equation ??? can be written as p(D)y=0. If r is a constant then
p(D)erx=(a0Dnerx+a1Dn−1erx+⋯+anerx)=(a0rn+a1rn−1+⋯+an)erx;
that is
p(D)(erx)=p(r)erx.
This shows that y=erx is a solution of Equation ??? if p(r)=0. In the simplest case, where p has n distinct real zeros r1, r2,…, rn, this argument yields n solutions
y1=er1x,y2=er2x,…,yn=ernx.
It can be shown (Exercise 9.2.39) that the Wronskian of {er1x,er2x,…,ernx} is nonzero if r1, r2, …, rn are distinct; hence, {er1x,er2x,…,ernx} is a fundamental set of solutions of p(D)y=0 in this case.
- Find the general solution of y‴−6y″+11y′−6y=0.
- and solve the initial value problem y‴−6y″+11y′−6y=0,y(0)=4,y′(0)=5,y″(0)=9.
Solution a
The characteristic polynomial of Equation ??? is
p(r)=r3−6r2+11r−6=(r−1)(r−2)(r−3).
Therefore {ex,e2x,e3x} is a set of solutions of Equation ???. It is a fundamental set, since its Wronskian is
W(x)=|exe2xe3xex2e2x3e3xex4e2x9e3x|=e6x|111123149|=2e6x≠0.
Therefore the general solution of Equation ??? is
y=c1ex+c2e2x+c3e3x.
Solution b
We must determine c1, c2 and c3 in Equation ??? so that y satisfies the initial conditions in Equation ???. Differentiating Equation ??? twice yields
y′=c1ex+2c2e2x+3c3e3xy″=c1ex+4c2e2x+9c3e3x.
Setting x=0 in Equation ??? and Equation ??? and imposing the initial conditions yields
c1+2c2+3c3=4c1+2c2+3c3=5c1+4c2+9c3=9.
The solution of this system is c1=4, c2=−1, c3=1. Therefore the solution of Equation ??? is
y=4ex−e2x+e3x
(Figure 19.2.1 ).

Now we consider the case where the characteristic polynomial Equation ??? 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.
Consider the polynomial
p(r)=r3−r2+r−1
and the associated polynomial operator
p(D)=D3−D2+D−1.
Since p(r) can be factored as
p(r)=(r−1)(r2+1)=(r2+1)(r−1),
it is reasonable to expect that p(D) can be factored as
p(D)=(D−1)(D2+1)=(D2+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 ???. 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 ??? is this: if y is any three-times differentiable function then
- (D−1)(D2+1)y is the function obtained by first applying D2+1 to y and then applying D−1 to the resulting function
- (D2+1)(D−1)y is the function obtained by first applying D−1 to y and then applying D2+1 to the resulting function.
From (a),
(D−1)(D2+1)y=(D−1)[(D2+1)y]=(D−1)(y″+y)=D(y″+y)−(y″+y)=(y‴+y′)−(y″+y)=y‴−y″+y′−y=(D3−D2+D−1)y.
This implies that
(D−1)(D2+1)=(D3−D2+D−1).
From (b),
(D2+1)(D−1)y=(D2+1)[(D−1)y]=(D2+1)(y′−y)=D2(y′−y)+(y′−y)=(y‴−y″)+(y′−y)=y‴−y″+y′−y=(D3−D2+D−1)y,
(D2+1)(D−1)=(D3−D2+D−1),
which completes the justification of Equation ???.
Use the result of Example 19.2.2 to find the general solution of
y‴−y″+y′−y=0.
Solution
From Equation ???, we can rewrite Equation ??? as
(D−1)(D2+1)y=0,
which implies that any solution of (D2+1)y=0 is a solution of Equation ???. Therefore y1=cosx and y2=sinx are solutions of Equation ???.
From Equation ???, we can rewrite Equation ??? as
(D2+1)(D−1)y=0,
which implies that any solution of (D−1)y=0 is a solution of Equation ???. Therefore y3=ex is solution of Equation ???.
The Wronskian of {ex,cosx,sinx} is
W(x)=|cosxsinxex−sinxcosxex−cosx−sinxex|.
Since
W(0)=|101011−101|=2,
{cosx,sinx,ex} is linearly independent and
y=c1cosx+c2sinx+c3ex
is the general solution of Equation ???.
Find the general solution of
y(4)−16y=0.
Solution
The characteristic polynomial of Equation ??? is
p(r)=r4−16=(r2−4)(r2+4)=(r−2)(r+2)(r2+4).
By arguments similar to those used in Examples 19.2.2 and 19.2.4 , it can be shown that Equation ??? can be written as
(D2+4)(D+2)(D−2)y=0
or
(D2+4)(D−2)(D+2)y=0
or
(D−2)(D+2)(D2+4)y=0.
Therefore y is a solution of Equation ??? if it is a solution of any of the three equations
(D−2)y=0,(D+2)y=0,(D2+4)y=0.
Hence, {e2x,e−2x,cos2x,sin2x} is a set of solutions of Equation ???. The Wronskian of this set is
W(x)=|e2xe−2xcos2xsin2x2e2x−2e−2x−2sin2x2cos2x4e2x4e−2x−4cos2x−4sin2x8e2x−8e−2x8sin2x−8cos2x|.
Since
W(0)=|11102−20244−408−80−8|=−512,
{e2x,e−2x,cos2x,sin2x} is linearly independent, and
y1=c1e2x+c2e−2x+c3cos2x+c4sin2x
is the general solution of Equation ???.
It is known from algebra that every polynomial
p(r)=a0rn+a1rn−1+⋯+an
with real coefficients can be factored as
p(r)=a0p1(r)p2(r)⋯pk(r),
where no pair of the polynomials p1, p2, …, pk has a common factor and each is either of the form
pj(r)=(r−rj)mj,
where rj is real and mj is a positive integer, or
pj(r)=[(r−λj)2+ω2j]mj,
where λj and ωj are real, ωj≠0, and mj is a positive integer. If Equation ??? holds then rj is a real zero of p, while if Equation ??? holds then λ+iω and λ−iω are complex conjugate zeros of p. In either case, mj is the multiplicity of the zero(s).
By arguments similar to those used in our examples, it can be shown that
p(D)=a0p1(D)p2(D)⋯pk(D)
and that the order of the factors on the right can be chosen arbitrarily. Therefore, if pj(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.
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 $g=f$)}\\ (D-a)^2e^{ax}f&=& (D-a)e^{ax}f'=e^{ax}f'' &\mbox{(from \eqref{eq:9.2.17} with $g=f'$)}\\ (D-a)^3e^{ax}f&=& (D-a)e^{ax}f''=e^{ax}f''' &\mbox{(from \eqref{eq:9.2.17} with $g=f''$)}\\ &\vdots&\\ (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.
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 19.2.1 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.
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,\\ 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
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 19.2.2 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
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\\ &=r(r^3+4r^2+6r+4)\\ &=r(r+2)(r^2+2r+2)\\ &=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
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}.