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Mathematics LibreTexts

9.1: Introduction to Linear Higher Order Equations

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    9446
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    An \(n\)th order differential equation is said to be linear if it can be written in the form

    \[\label{eq:9.1.1} y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=f(x).\]

    We considered equations of this form with \(n=1\) in Section 2.1 and with \(n=2\) in Chapter 5. In this chapter \(n\) is an arbitrary positive integer. In this section we sketch the general theory of linear \(n\)th order equations. Since this theory has already been discussed for \(n=2\) in Sections 5.1 and 5.3, we’ll omit proofs.

    For convenience, we consider linear differential equations written as

    \[\label{eq:9.1.2} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),\]

    which can be rewritten as Equation \ref{eq:9.1.1} on any interval on which \(P_0\) has no zeros, with \(p_1=P_1/P_0\), …, \(p_n=P_n/P_0\) and \(f=F/P_0\). For simplicity, throughout this chapter we’ll abbreviate the left side of Equation \ref{eq:9.1.2} by \(Ly\); that is,

    \[Ly=P_0y^{(n)}+P_1y^{(n-1)}+\cdots+P_ny.\nonumber \]

    We say that the equation \(Ly=F\) is normal on \((a,b)\) if \(P_0\), \(P_1\), …, \(P_n\) and \(F\) are continuous on \((a,b)\) and \(P_0\) has no zeros on \((a,b)\). If this is so then \(Ly=F\) can be written as Equation \ref{eq:9.1.1} with \(p_1\), …, \(p_n\) and \(f\) continuous on \((a,b)\).

    The next theorem is analogous to Theorem 5.3.1.

    Theorem \(\PageIndex{1}\)

    Suppose \(Ly=F\) is normal on \((a,b)\), let \(x_0\) be a point in \((a,b),\) and let \(k_0\), \(k_1\), …, \(k_{n-1}\) be arbitrary real numbers\(.\) Then the initial value problem

    \[Ly=F, \quad y(x_0)=k_0,\quad y'(x_0)=k_1,\dots,\quad y^{(n-1)}(x_0)=k_{n-1}\nonumber \]

    has a unique solution on \((a,b)\).

    Homogeneous Equations

    Equation \ref{eq:9.1.2} is said to be homogeneous if \(F\equiv0\) and nonhomogeneous otherwise. Since \(y\equiv0\) is obviously a solution of \(Ly=0\), we call it the trivial solution. Any other solution is nontrivial.

    If \(y_1\), \(y_2\), …, \(y_n\) are defined on \((a,b)\) and \(c_1\), \(c_2\), …, \(c_n\) are constants, then

    \[\label{eq:9.1.3} y=c_1y_1+c_2y_2+\cdots+c_ny_n\]

    is a linear combination of \(\{y_1,y_2\dots,y_n\}\). It’s easy to show that if \(y_1\), \(y_2\), …, \(y_n\) are solutions of \(Ly=0\) on \((a,b)\), then so is any linear combination of \(\{y_1,y_2,\dots,y_n\}\). (See the proof of Theorem 5.1.2.) We say that \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of \(Ly=0\) on \((a,b)\) if every solution of \(Ly=0\) on \((a,b)\) can be written as a linear combination of \(\{y_1,y_2,\dots,y_n\}\), as in Equation \ref{eq:9.1.3}. In this case we say that Equation \ref{eq:9.1.3} is the general solution of \(Ly=0\) on \((a,b)\).

    It can be shown (Exercises 9.1.14 and 9.1.15) that if the equation \(Ly=0\) is normal on \((a,b)\) then it has infinitely many fundamental sets of solutions on \((a,b)\). The next definition will help to identify fundamental sets of solutions of \(Ly=0\).

    We say that \(\{y_1,y_2,\dots,y_n\}\) is linearly independent on \((a,b)\) if the only constants \(c_1\), \(c_2\), …, \(c_n\) such that

    \[\label{eq:9.1.4} c_1y_1(x)+c_2y_2(x)+\cdots+c_ny_n(x)=0,\quad a<x<b,\]

    are \(c_1=c_2=\cdots=c_n=0\). If Equation \ref{eq:9.1.4} holds for some set of constants \(c_1\), \(c_2\), …, \(c_n\) that are not all zero, then \(\{y_1,y_2,\dots,y_n\}\) is linearly dependent on \((a,b)\)

    The next theorem is analogous to Theorem 5.1.3.

    Theorem \(\PageIndex{1}\)

    If \(Ly=0\) is normal on \((a,b)\), then a set \(\{y_1,y_2,\dots,y_n\}\) of \(n\) solutions of \(Ly=0\) on \((a,b)\) is a fundamental set if and only if it is linearly independent on \((a,b)\).

    Example \(\PageIndex{1}\)

    The equation

    \[\label{eq:9.1.5} x^3y'''-x^2y''-2xy'+6y=0\]

    is normal and has the solutions \(y_1=x^2\), \(y_2=x^3\), and \(y_3=1/x\) on \((-\infty,0)\) and \((0,\infty)\). Show that \(\{y_1,y_2,y_3\}\) is linearly independent on \((-\infty, 0)\) and \((0,\infty)\). Then find the general solution of Equation \ref{eq:9.1.5} on \((-\infty, 0)\) and \((0,\infty)\).

    Solution

    Suppose

    \[\label{eq:9.1.6} c_1x^2+c_2x^3+{c_3\over x}=0\]

    on \((0,\infty)\). We must show that \(c_1=c_2=c_3=0\). Differentiating Equation \ref{eq:9.1.6} twice yields the system

    \[\label{eq:9.1.7} \begin{array}{rr} {c_1x^2+c_2x^3+ \dfrac{c_3}{x}} &{=  0} \\ {2c_1x+3c_2x^2- \dfrac{c_3}{x^2}} &{=0} \\ {2c_1+6c_2x + \dfrac{2c_3}{x^3}} &{=0.} \end{array}\]

    If Equation \ref{eq:9.1.7} holds for all \(x\) in \((0,\infty)\), then it certainly holds at \(x=1\); therefore,

    \[\label{eq:9.1.8} \begin{array}{rr} {\phantom{2}c_1+\phantom{3}c_2+\phantom{2}c_3} &{= 0} \\ {2c_1+3c_2-\phantom{2}c_3}&{= 0}\\ {2c_1+6c_2+2c_3 }&{=0. }\end{array}\]

    By solving this system directly, you can verify that it has only the trivial solution \(c_1=c_2=c_3=0\); however, for our purposes it is more useful to recall from linear algebra that a homogeneous linear system of \(n\) equations in \(n\) unknowns has only the trivial solution if its determinant is nonzero. Since the determinant of Equation \ref{eq:9.1.8} is

    \[\left|\begin{array}{rrr}1&1&1\\2&3&-1\\2&6&2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\2&1&-3\\2&4&0\end{array}\right|=12, \nonumber\]

    it follows that Equation \ref{eq:9.1.8} has only the trivial solution, so \(\{y_1,y_2,y_3\}\) is linearly independent on \((0,\infty)\). Now Theorem \(\PageIndex{2}\) implies that

    \[y=c_1x^2+c_2x^3+{c_3\over x} \nonumber\]

    is the general solution of Equation \ref{eq:9.1.5} on \((0,\infty)\). To see that this is also true on \((-\infty,0)\), assume that Equation \ref{eq:9.1.6} holds on \((-\infty,0)\). Setting \(x=-1\) in Equation \ref{eq:9.1.7} yields

    \[\begin{aligned} \phantom{-2}c_1-\phantom{3}c_2-\phantom{2}c_3&=0\\ -2c_1+3c_2-\phantom{2}c_3&=0\\ \phantom{-}2c_1-6c_2-2c_3&=0.\end{aligned}\nonumber \]

    Since the determinant of this system is

    \[\left|\begin{array}{rrr}1&-1&-1\\-2&3&-1\\2&-6&-2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\-2&1&-3\\2&-4&0\end{array}\right|=-12, \nonumber\]

    it follows that \(c_1=c_2=c_3=0\); that is, \(\{y_1,y_2,y_3\}\) is linearly independent on \((-\infty,0)\).

    Example \(\PageIndex{2}\)

    The equation

    \[\label{eq:9.1.9} y^{(4)}+y'''-7y''-y'+6y=0\]

    is normal and has the solutions \(y_1=e^x\), \(y_2=e^{-x}\), \(y_3=e^{2x}\), and \(y_4=e^{-3x}\) on \((-\infty,\infty)\). (Verify.) Show that \(\{y_1,y_2,y_3,y_4\}\) is linearly independent on \((-\infty,\infty)\). Then find the general solution of Equation \ref{eq:9.1.9}.

    Solution

    Suppose \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are constants such that

    \[\label{eq:9.1.10} c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x}=0\]

    for all \(x\). We must show that \(c_1=c_2=c_3=c_4=0\). Differentiating Equation \ref{eq:9.1.10} three times yields the system

    \[\label{eq:9.1.11} \begin{array}{rcl} c_1e^x+c_2e^{-x}+\phantom{2}c_3e^{2x}+\phantom{27}c_4e^{-3x}&=&0\\ c_1e^x-c_2e^{-x}+2c_3e^{2x}-\phantom{2}3c_4e^{-3x}&=&0\\ c_1e^x+c_2e^{-x}+4c_3e^{2x}+\phantom{2}9c_4e^{-3x}&=&0\\ c_1e^x-c_2e^{-x}+8c_3e^{2x}-27c_4e^{-3x}&=&0. \end{array}\]

    If Equation \ref{eq:9.1.11} holds for all \(x\), then it certainly holds for \(x=0\). Therefore

    \[\begin{array}{rcl} c_1+c_2+\phantom{2}c_3+\phantom{27}c_4&=&0\\ c_1-c_2+2c_3-\phantom{2}3c_4&=&0\\ c_1+c_2+4c_3+\phantom{2}9c_4&=&0\\ c_1-c_2+8c_3-27c_4&=&0. \end{array}\nonumber \]

    The determinant of this system is

    \[\label{eq:9.1.12} \begin{array}{rcl} \left|\begin{array}{rrrr}1&1&1&1\\1&-1&2&-3\\1&1&4&9\\ 1&-1&8&-27\end{array}\right|&=& \left|\begin{array}{rrrr}1&1&1&1\\0&-2&1&-4\\0&0&3&8\\ 0&-2&7&-28\end{array}\right| =\left|\begin{array}{rrr}-2&1&-4\\0&3&8\\ -2&7&-28\end{array}\right|\\[5 pt]&=& \left|\begin{array}{rrr}-2&1&-4\\0&3&8 \\0&6&-24\end{array}\right|= -2\left|\begin{array}{rr}3&8\\6&-24\end{array}\right|=240, \end{array}\]

    so the system has only the trivial solution \(c_1=c_2=c_3=c_4=0\). Now Theorem \(\PageIndex{2}\) implies that

    \[y=c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x} \nonumber\]

    is the general solution of Equation \ref{eq:9.1.9}.

    The Wronskian

    We can use the method used in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) to test \(n\) solutions \(\{y_1,y_2,\dots,y_n\}\) of any \(n\)th order equation \(Ly=0\) for linear independence on an interval \((a,b)\) on which the equation is normal. Thus, if \(c_1\), \(c_2\) ,…, \(c_n\) are constants such that

    \[c_1y_1+c_2y_2+\cdots+c_ny_n=0,\quad a<b,\nonumber\]

    then differentiating \(n-1\) times leads to the \(n\times n\) system of equations

    \[\label{eq:9.1.13} \begin{array}{rcl} c_1y_1(x)+c_2y_2(x)+&\cdots&+c_ny_n(x)=0\\ c_1y_1'(x)+c_2y_2'(x)+&\cdots&+c_ny_n'(x)=0\\ \phantom{c_1y_1^{(n)}(x)+c_2y_2^{(n-1)}(x)}&\vdots& \phantom{\cdots+c_ny_n^{(n-1)(x)}=q}\\ c_1y_1^{(n-1)}(x)+c_2y_2^{(n-1)}(x)+&\cdots&+c_ny_n^{(n-1)}(x) =0 \end{array}\]

    for \(c_1\), \(c_2\), …, \(c_n\). For a fixed \(x\), the determinant of this system is

    \[W(x)=\left|\begin{array}{cccc} y_1(x)&y_2(x)&\cdots&y_n(x)\\[4pt] y'_1(x)&y'_2(x)&\cdots&y_n'(x)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_1^{(n-1)}(x)&y_2^{(n-1)}(x)&\cdots&y_n^{(n-1)}(x) \end{array}\right|.\nonumber\]

    We call this determinant the Wronskian of \(\{y_1,y_2,\dots,y_n\}\). If \(W(x)\ne0\) for some \(x\) in \((a,b)\) then the system Equation \ref{eq:9.1.13} has only the trivial solution \(c_1=c_2=\cdots=c_n=0\), and Theorem \(\PageIndex{2}\) implies that

    \[y=c_1y_1+c_2y_2+\cdots+c_ny_n\nonumber\]

    is the general solution of \(Ly=0\) on \((a,b)\).

    The next theorem generalizes Theorem 5.1.4. The proof is sketched in Exercises 9.1.17-9.1.20.

    Theorem \(\PageIndex{3}\) Abel's Formula

    Suppose the homogeneous linear \(n\)th order equation

    \[\label{eq:9.1.14} P_0(x)y^{(n)}+P_1(x)y^{n-1}+\cdots+P_n(x)y=0\]

    is normal on \((a,b),\) let \(y_1,\) \(y_2,\) …, \(y_n\) be solutions of Equation \ref{eq:9.1.14} on \((a,b),\) and let \(x_0\) be in \((a,b)\). Then the Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is given by

    \[\label{eq:9.1.15} W(x)= W(x_0) \exp\left\{-\int^x_{x_0}{P_1(t) \over P_0(t)} \, dt \right\},\quad a<b.\]

    Therefore, either \(W\) has no zeros in \((a,b)\) or \(W\equiv0\) on \((a,b).\)

    Formula Equation \ref{eq:9.1.15} is Abel’s formula.

    The next theorem is analogous to Theorem 5.1.6.

    Theorem \(\PageIndex{4}\) 

    Suppose \(Ly=0\) is normal on \((a,b)\) and let \(y_1\), \(y_2\), …, \(y_n\) be \(n\) solutions of \(Ly=0\) on \((a,b)\). Then the following statements are equivalent\(;\) that is, they are either all true or all false:

    1. The general solution of \(Ly=0\) on \((a,b)\) is \(y=c_1y_1+c_2y_2+\cdots+c_ny_n.\)
    2. \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of \(Ly=0\) on \((a,b).\)
    3. \(\{y_1,y_2,\dots,y_n\}\) is linearly independent on \((a,b).\)
    4. The Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is nonzero at some point in \((a,b).\)
    5. The Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is nonzero at all points in \((a,b).\)

    Example \(\PageIndex{3}\)

    In Example \(\PageIndex{1}\) we saw that the solutions \(y_1=x^2\), \(y_2=x^3\), and \(y_3=1/x\) of

    \[x^3y'''-x^2y''-2xy'+6y=0 \nonumber\]

    are linearly independent on \((-\infty,0)\) and \((0,\infty)\). Calculate the Wronskian of \(\{y_1,y_2,y_3\}\).

    Solution

    If \(x\ne0\), then

    \[W(x)=\left|\begin{array}{ccc}{x^{2}} & {x^{3}} & {\frac{1}{x}} \\ {2 x} & {3 x^{2}} & {-\frac{1}{x^{2}}} \\ {2} & {6 x} & {\frac{2}{x^{3}}}\end{array}\right|=2 x^{3}\left|\begin{array}{ccc}{1} & {x} & {\frac{1}{x^{3}}} \\ {2} & {3 x} & {-\frac{1}{x^{3}}} \\ {1} & {3 x} & {\frac{1}{x^{3}}}\end{array}\right| \nonumber\]

    where we factored \(x^2\), \(x\), and \(2\) out of the first, second, and third rows of \(W(x)\), respectively. Adding the second row of the last determinant to the first and third rows yields

    \[W(x)=2 x^{3}\left|\begin{array}{ccc}{3} & {4 x} & {0} \\ {2} & {3 x} & {-\frac{1}{x^{3}}} \\ {3} & {6 x} & {0}\end{array}\right|=2 x^{3}\left(\frac{1}{x^{3}}\right)\left|\begin{array}{cc}{3} & {4 x} \\ {3} & {6 x}\end{array}\right|=12 x\nonumber \]

    Therefore \(W(x)\ne0\) on \((-\infty,0)\) and \((0,\infty)\).

    Example \(\PageIndex{4}\)

    In Example \(\PageIndex{2}\) we saw that the solutions \(y_1=e^x\), \(y_2=e^{-x}\), \(y_3=e^{2x}\), and \(y_4=e^{-3x}\) of

    \[y^{(4)}+y'''-7y''-y'+6y=0 \nonumber\]

    are linearly independent on every open interval. Calculate the Wronskian of \(\{y_1,y_2,y_3,y_4\}\).

    Solution

    For all \(x\),

    \[W(x)=\left|\begin{array}{rrrr} e^x&e^{-x}&e^{2x}&e^{-3x}\\ e^x&-e^{-x}&2e^{2x}&-3e^{-3x}\\ e^x&e^{-x}&4e^{2x}&9e^{-3x}\\ e^x&-e^{-x}&8e^{2x}&-27e^{-3x} \end{array}\right|. \nonumber\]

    Factoring the exponential common factor from each row yields

    \[W(x)=e^{-x}\left|\begin{array}{rrrr}1&1&1&1\\1&-1&2&-3\\1&1&4&9\\ 1&-1&8&-27\end{array}\right|=240e^{-x}, \nonumber\]

    from Equation \ref{eq:9.1.12}.

    Note

    Under the assumptions of Theorem \(\PageIndex{4}\), it isn’t necessary to obtain a formula for \(W(x)\). Just evaluate \(W(x)\) at a convenient point in \((a, b)\), as we did in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\). 

    Theorem \(\PageIndex{5}\)

    Suppose \(c\) is in \((a,b)\) and \(\alpha_{1},\) \(\alpha_{2},\) …, are real numbers, not all zero. Under the assumptions of Theorem 10.3.3, suppose \(y_{1}\) and \(y_{2}\) are solutions of Equation 5.1.35 such that

    \[\label{eq:9.1.16} \alpha y_{i}(c)+ y_{i}'(c)+\cdots +y_{i}^{(n-1)}(c)=0,\quad 1\le i\le n.\]

    Then \(\{y_{1},y_{2},\dots y_{n}\}\) isn’t linearly independent on \((a,b).\)

    Proof

    Since \(\alpha_{1}\), \(\alpha_{2}\), …, \(\alpha_{n}\) are not all zero, Equation \ref{eq:9.1.14} implies that

    \[\left|\begin{array}{ccccccc} y_{1}(c)&y_{1}'(c)&\cdots&y_{1}^{(n-1)}(c)\\ y_{2}(c)&y_{2}'(c)&\cdots&y_{2}^{(n-1)}(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{n}(c)&y_{n}'(c)&\cdots&y_{n}^{(n-1)}(c)\\ \end{array}\right|=0,\nonumber \]

    so

    \[\left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)&\cdots& y_{n}(c)\\ y_{1}'(c)&y_{2}'(c)&\cdots& y_{n}'(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{1}^{(n-1)}(c)&y_{2}^{(n-1)}(c)(c)&\cdots& y_{n}^{(n-1)}(c)(c)\\ \end{array}\right|=0\nonumber \]

    and Theorem \(\PageIndex{4}\) implies the stated conclusion.

    General Solution of a Nonhomogeneous Equation

    The next theorem is analogous to Theorem 5.3.2. It shows how to find the general solution of \(Ly=F\) if we know a particular solution of \(Ly=F\) and a fundamental set of solutions of the complementary equation \(Ly=0\).

    Theorem \(\PageIndex{6}\)

    The probabilities assigned to events by a distribution function on a sample space are given by.

    Proof

    Suppose \(Ly=F\) is normal on \((a,b).\) Let \(y_p\) be a particular solution of \(Ly=F\) on \((a,b),\) and let \(\{y_1,y_2,\dots,y_n\}\) be a fundamental set of solutions of the complementary equation \(Ly=0\) on \((a,b)\). Then \(y\) is a solution of \(Ly=F\) on \((a,b)\) if and only if

    \[y=y_p+c_1y_1+c_2y_2+\cdots+c_ny_n,\nonumber \]

    where \(c_1,c_2,\dots,c_n\) are constants.

    The next theorem is analogous to Theorem 5.3.2.

    Theorem \(\PageIndex{7}\) The Principle of Superposition

    Suppose for each \(i=1,\) \(2,\) …, \(r\), the function \(y_{p_i}\) is a particular solution of \(Ly=F_i\) on \((a,b).\) Then

    \[y_p=y_{p_1}+y_{p_2}+\cdots+y_{p_r} \nonumber\]

    is a particular solution of

    \[Ly=F_1(x)+F_2(x)+\cdots+F_r(x) \nonumber\]

    on \((a,b).\)

    We’ll apply Theorems \(\PageIndex{6}\) and \(\PageIndex{7}\) throughout the rest of this chapter.