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9.1: Introduction to Linear Higher Order Equations

( \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(n1)++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(n1)++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(n1)++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, …, kn1 be arbitrary real numbers. Then the initial value problem

Ly=F,y(x0)=k0,y(x0)=k1,,y(n1)(x0)=kn1

has a unique solution on (a,b).

Homogeneous Equations

Equation ??? is said to be homogeneous if F0 and nonhomogeneous otherwise. Since y0 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

x3yx2y2xy+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+3c2x2c3x2=02c1+6c2x+2c3x3=0.

If Equation ??? holds for all x in (0,), then it certainly holds at x=1; therefore,

2c1+3c2+2c3=02c1+3c22c3=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

|111231262|=|100213240|=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

2c13c22c3=02c1+3c22c3=02c16c22c3=0.

Since the determinant of this system is

|111231262|=|100213240|=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)+y7yy+6y=0

is normal and has the solutions y1=ex, y2=ex, y3=e2x, and y4=e3x 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+c2ex+c3e2x+c4e3x=0

for all x. We must show that c1=c2=c3=c4=0. Differentiating Equation ??? three times yields the system

c1ex+c2ex+2c3e2x+27c4e3x=0c1exc2ex+2c3e2x23c4e3x=0c1ex+c2ex+4c3e2x+29c4e3x=0c1exc2ex+8c3e2x27c4e3x=0.

If Equation ??? holds for all x, then it certainly holds for x=0. Therefore

c1+c2+2c3+27c4=0c1c2+2c323c4=0c1+c2+4c3+29c4=0c1c2+8c327c4=0.

The determinant of this system is

|11111123114911827|=|11110214003802728|=|2140382728|[5pt]=|2140380624|=2|38624|=240,

so the system has only the trivial solution c1=c2=c3=c4=0. Now Theorem 9.1.2 implies that

y=c1ex+c2ex+c3e2x+c4e3x

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 n1 times leads to the n×n system of equations

c1y1(x)+c2y2(x)++cnyn(x)=0c1y1(x)+c2y2(x)++cnyn(x)=0c1y(n)1(x)+c2y(n1)2(x)+cny(n1)(x)n=qc1y(n1)1(x)+c2y(n1)2(x)++cny(n1)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)y1(x)y2(x)yn(x)y(n1)1(x)y(n1)2(x)y(n1)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)yn1++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 W0 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:

  1. The general solution of Ly=0 on (a,b) is y=c1y1+c2y2++cnyn.
  2. {y1,y2,,yn} is a fundamental set of solutions of Ly=0 on (a,b).
  3. {y1,y2,,yn} is linearly independent on (a,b).
  4. The Wronskian of {y1,y2,,yn} is nonzero at some point in (a,b).
  5. The Wronskian of {y1,y2,,yn} is nonzero at all points in (a,b).
Example 9.1.3

In Example 9.1.1 we saw that the solutions y1=x2, y2=x3, and y3=1/x of

x3yx2y2xy+6y=0

are linearly independent on (,0) and (0,). Calculate the Wronskian of {y1,y2,y3}.

Solution

If x0, then

W(x)=|x2x31x2x3x21x226x2x3|=2x3|1x1x323x1x313x1x3|

where we factored x2, 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)=2x3|34x023x1x336x0|=2x3(1x3)|34x36x|=12x

Therefore W(x)0 on (,0) and (0,).

Example 9.1.4

In Example 9.1.2 we saw that the solutions y1=ex, y2=ex, y3=e2x, and y4=e3x of

y(4)+y7yy+6y=0

are linearly independent on every open interval. Calculate the Wronskian of {y1,y2,y3,y4}.

Solution

For all x,

W(x)=|exexe2xe3xexex2e2x3e3xexex4e2x9e3xexex8e2x27e3x|.

Factoring the exponential common factor from each row yields

W(x)=ex|11111123114911827|=240ex,

from Equation ???.

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 α1, α2, …, are real numbers, not all zero. Under the assumptions of Theorem 10.3.3, suppose y1 and y2 are solutions of Equation 5.1.35 such that

αyi(c)+yi(c)++y(n1)i(c)=0,1in.

Then {y1,y2,yn} isn’t linearly independent on (a,b).

Proof

Since α1, α2, …, αn are not all zero, Equation ??? implies that

|y1(c)y1(c)y(n1)1(c)y2(c)y2(c)y(n1)2(c)yn(c)yn(c)y(n1)n(c)|=0,

so

|y1(c)y2(c)yn(c)y1(c)y2(c)yn(c)y(n1)1(c)y(n1)2(c)(c)y(n1)n(c)(c)|=0

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 yp be a particular solution of Ly=F on (a,b), and let {y1,y2,,yn} 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=yp+c1y1+c2y2++cnyn,

where c1,c2,,cn 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 ypi is a particular solution of Ly=Fi on (a,b). Then

yp=yp1+yp2++ypr

is a particular solution of

Ly=F1(x)+F2(x)++Fr(x)

on (a,b).

We’ll apply Theorems 9.1.6 and 9.1.7 throughout the rest of this chapter.


This page titled 9.1: Introduction to Linear Higher Order Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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