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Mathematics LibreTexts

9.2: Higher Order Constant Coefficient Homogeneous Equations

  • Page ID
    9447
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    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 \(\PageIndex{1}\)

    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

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

    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

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

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

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

    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 \(\PageIndex{1}\)).

    imageedit_3_9167602326.png
    Figure \(\PageIndex{1}\): \(y=4e^x-e^{2x}+e^{3x}\)

    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 \(\PageIndex{2}\)

    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),

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

    This implies that

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

    From (b),

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

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

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

    Example \(\PageIndex{3}\)

    Use the result of Example \(\PageIndex{2}\) 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

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

    Since

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

    \(\{\cos x,\sin x,e^x\}\) 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 \(\PageIndex{4}\)

    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 \(\PageIndex{2}\) and \(\PageIndex{4}\), 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

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

    Since

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

    \(\{e^{2x},e^{-2x},\cos2x,\sin2x\}\) 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 \(\PageIndex{1}\)

    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

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

    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 \(c_1=c_2=\cdots=c_n=0\).

    Example \(\PageIndex{5}\)

    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 \(\PageIndex{1}\) 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 \(\PageIndex{2}\)

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

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

    is a fundamental set of solutions of

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

    Example \(\PageIndex{6}\)

    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 \(\PageIndex{2}\) 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 \(\PageIndex{7}\)

    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

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

    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

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

    Example \(\PageIndex{8}\)

    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}.