9.1: Introduction to Linear Higher Order Equations

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$trench-1995.jpg$
William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathematics) 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 9.1.1

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 9.1.1

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 9.1.1

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 9.1.2 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 9.1.2

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 9.1.2 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 9.1.1 and 9.1.2 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 9.1.2 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 9.1.3 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 9.1.4

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:

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

Example 9.1.3

In Example 9.1.1 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 9.1.4

In Example 9.1.2 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 9.1.4 , 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 9.1.1 and 9.1.2 .

Theorem 9.1.5

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

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

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

and Theorem 9.1.4 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 9.1.6

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 9.1.7 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 9.1.6 and 9.1.7 throughout the rest of this chapter.