9.1: Introduction to Linear Higher Order Equations
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( \newcommand{\kernel}{\mathrm{null}\,}\)
An nth order differential equation is said to be linear if it can be written in the form
y(n)+p1(x)y(n−1)+⋯+pn(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 nth 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
P0(x)y(n)+P1(x)y(n−1)+⋯+Pn(x)y=F(x),
which can be rewritten as Equation ??? on any interval on which P0 has no zeros, with p1=P1/P0, …, pn=Pn/P0 and f=F/P0. For simplicity, throughout this chapter we’ll abbreviate the left side of Equation ??? by Ly; that is,
Ly=P0y(n)+P1y(n−1)+⋯+Pny.
We say that the equation Ly=F is normal on (a,b) if P0, P1, …, Pn and F are continuous on (a,b) and P0 has no zeros on (a,b). If this is so then Ly=F can be written as Equation ??? with p1, …, pn 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 x0 be a point in (a,b), and let k0, k1, …, kn−1 be arbitrary real numbers. Then the initial value problem
Ly=F,y(x0)=k0,y′(x0)=k1,…,y(n−1)(x0)=kn−1
has a unique solution on (a,b).
Homogeneous Equations
Equation ??? is said to be homogeneous if F≡0 and nonhomogeneous otherwise. Since y≡0 is obviously a solution of Ly=0, we call it the trivial solution. Any other solution is nontrivial.
If y1, y2, …, yn are defined on (a,b) and c1, c2, …, cn are constants, then
y=c1y1+c2y2+⋯+cnyn
is a linear combination of {y1,y2…,yn}. It’s easy to show that if y1, y2, …, yn are solutions of Ly=0 on (a,b), then so is any linear combination of {y1,y2,…,yn}. (See the proof of Theorem 5.1.2.) We say that {y1,y2,…,yn} 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 {y1,y2,…,yn}, as in Equation ???. In this case we say that Equation ??? 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 {y1,y2,…,yn} is linearly independent on (a,b) if the only constants c1, c2, …, cn such that
c1y1(x)+c2y2(x)+⋯+cnyn(x)=0,a<x<b,
are c1=c2=⋯=cn=0. If Equation ??? holds for some set of constants c1, c2, …, cn that are not all zero, then {y1,y2,…,yn} 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 {y1,y2,…,yn} 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
x3y‴−x2y″−2xy′+6y=0
is normal and has the solutions y1=x2, y2=x3, and y3=1/x on (−∞,0) and (0,∞). Show that {y1,y2,y3} is linearly independent on (−∞,0) and (0,∞). Then find the general solution of Equation ??? on (−∞,0) and (0,∞).
Solution
Suppose
c1x2+c2x3+c3x=0
on (0,∞). We must show that c1=c2=c3=0. Differentiating Equation ??? twice yields the system
c1x2+c2x3+c3x=02c1x+3c2x2−c3x2=02c1+6c2x+2c3x3=0.
If Equation ??? holds for all x in (0,∞), then it certainly holds at x=1; therefore,
2c1+3c2+2c3=02c1+3c2−2c3=02c1+6c2+2c3=0.
By solving this system directly, you can verify that it has only the trivial solution c1=c2=c3=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 ??? is
|11123−1262|=|10021−3240|=12,
it follows that Equation ??? has only the trivial solution, so {y1,y2,y3} is linearly independent on (0,∞). Now Theorem 9.1.2 implies that
y=c1x2+c2x3+c3x
is the general solution of Equation ??? on (0,∞). To see that this is also true on (−∞,0), assume that Equation ??? holds on (−∞,0). Setting x=−1 in Equation ??? yields
−2c1−3c2−2c3=0−2c1+3c2−2c3=0−2c1−6c2−2c3=0.
Since the determinant of this system is
|1−1−1−23−12−6−2|=|100−21−32−40|=−12,
it follows that c1=c2=c3=0; that is, {y1,y2,y3} is linearly independent on (−∞,0).
Example 9.1.2
The equation
y(4)+y‴−7y″−y′+6y=0
is normal and has the solutions y1=ex, y2=e−x, y3=e2x, and y4=e−3x on (−∞,∞). (Verify.) Show that {y1,y2,y3,y4} is linearly independent on (−∞,∞). Then find the general solution of Equation ???.
Solution
Suppose c1, c2, c3, and c4 are constants such that
c1ex+c2e−x+c3e2x+c4e−3x=0
for all x. We must show that c1=c2=c3=c4=0. Differentiating Equation ??? three times yields the system
c1ex+c2e−x+2c3e2x+27c4e−3x=0c1ex−c2e−x+2c3e2x−23c4e−3x=0c1ex+c2e−x+4c3e2x+29c4e−3x=0c1ex−c2e−x+8c3e2x−27c4e−3x=0.
If Equation ??? holds for all x, then it certainly holds for x=0. Therefore
c1+c2+2c3+27c4=0c1−c2+2c3−23c4=0c1+c2+4c3+29c4=0c1−c2+8c3−27c4=0.
The determinant of this system is
|11111−12−311491−18−27|=|11110−21−400380−27−28|=|−21−4038−27−28|[5pt]=|−21−403806−24|=−2|386−24|=240,
so the system has only the trivial solution c1=c2=c3=c4=0. Now Theorem 9.1.2 implies that
y=c1ex+c2e−x+c3e2x+c4e−3x
is the general solution of Equation ???.
The Wronskian
We can use the method used in Examples 9.1.1 and 9.1.2 to test n solutions {y1,y2,…,yn} of any nth order equation Ly=0 for linear independence on an interval (a,b) on which the equation is normal. Thus, if c1, c2,…, cn are constants such that
c1y1+c2y2+⋯+cnyn=0,a<b,
then differentiating n−1 times leads to the n×n system of equations
c1y1(x)+c2y2(x)+⋯+cnyn(x)=0c1y′1(x)+c2y′2(x)+⋯+cny′n(x)=0c1y(n)1(x)+c2y(n−1)2(x)⋮⋯+cny(n−1)(x)n=qc1y(n−1)1(x)+c2y(n−1)2(x)+⋯+cny(n−1)n(x)=0
for c1, c2, …, cn. For a fixed x, the determinant of this system is
W(x)=|y1(x)y2(x)⋯yn(x)y′1(x)y′2(x)⋯y′n(x)⋮⋮⋱⋮y(n−1)1(x)y(n−1)2(x)⋯y(n−1)n(x)|.
We call this determinant the Wronskian of {y1,y2,…,yn}. If W(x)≠0 for some x in (a,b) then the system Equation ??? has only the trivial solution c1=c2=⋯=cn=0, and Theorem 9.1.2 implies that
y=c1y1+c2y2+⋯+cnyn
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 nth order equation
P0(x)y(n)+P1(x)yn−1+⋯+Pn(x)y=0
is normal on (a,b), let y1, y2, …, yn be solutions of Equation ??? on (a,b), and let x0 be in (a,b). Then the Wronskian of {y1,y2,…,yn} is given by
W(x)=W(x0)exp{−∫xx0P1(t)P0(t)dt},a<b.
Therefore, either W has no zeros in (a,b) or W≡0 on (a,b).
Formula Equation ??? 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 y1, y2, …, yn 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}} \\[4pt] {2 x} & {3 x^{2}} & {-\frac{1}{x^{2}}} \\[4pt] {2} & {6 x} & {\frac{2}{x^{3}}}\end{array}\right|=2 x^{3}\left|\begin{array}{ccc}{1} & {x} & {\frac{1}{x^{3}}} \\[4pt] {2} & {3 x} & {-\frac{1}{x^{3}}} \\[4pt] {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} \\[4pt] {2} & {3 x} & {-\frac{1}{x^{3}}} \\[4pt] {3} & {6 x} & {0}\end{array}\right|=2 x^{3}\left(\frac{1}{x^{3}}\right)\left|\begin{array}{cc}{3} & {4 x} \\[4pt] {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}\\[4pt] e^x&-e^{-x}&2e^{2x}&-3e^{-3x}\\[4pt] e^x&e^{-x}&4e^{2x}&9e^{-3x}\\[4pt] 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\\[4pt]1&-1&2&-3\\[4pt]1&1&4&9\\[4pt] 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)\\[4pt] y_{2}(c)&y_{2}'(c)&\cdots&y_{2}^{(n-1)}(c)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_{n}(c)&y_{n}'(c)&\cdots&y_{n}^{(n-1)}(c)\\[4pt] \end{array}\right|=0,\nonumber
so
\left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)&\cdots& y_{n}(c)\\[4pt] y_{1}'(c)&y_{2}'(c)&\cdots& y_{n}'(c)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_{1}^{(n-1)}(c)&y_{2}^{(n-1)}(c)(c)&\cdots& y_{n}^{(n-1)}(c)(c)\\[4pt] \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.