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# 9.1: Introduction to Linear Higher Order Equations

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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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.