equation in normal form can be thought of as a Schr¨dinger equation,

o

d2 ψ

’ 2 + (V (x) ’ E)ψ = 0, (3.42)

dx

and we can gain insight into the properties of the solution by bringing our

physics intuition and experience to bear.

3.3 Inhomogeneous Equations

A linear inhomogeneous equation is one with a source term:

p0 (x)y (n) + p1 (x)y (n’1) + · · · + pn (x)y = f (x). (3.43)

It is called “inhomogeneous” because the source term f (x) does not contain

y, and so is di¬erent from the rest. We will devote an entire chapter to

the solution of such equations by the method of Green functions. Here, we

simply review some elementary material.

3.3.1 Particular Integral and Complementary Function

One method of dealing with inhomogeneous problems, one that is especially

e¬ective when the equation has constant coe¬cients, is simply to try and

guess a solution to (3.43). If you are successful, the guessed solution yP I

is then called a particular integral . We may add any solution yCF of the

homogeneous equation

p0 (x)y (n) + p1 (x)y (n’1) + · · · + pn (x)y = 0 (3.44)

to yP I and it will still be a solution of the inhomogeneous problem. We

use this freedom to satisfy the boundary or initial conditions. The added

solution, yCF , is called the complementary function.

Example: Charging capacitor. The capacitor is initially uncharged, and the

switch is closed at t = 0

78 CHAPTER 3. LINEAR ORDINARY DIFFERENTIAL EQUATIONS

C

Q

V

R

The charge on the capacitor, Q, obeys

dQ Q

R + = V, (3.45)

dt C

where R, C, V are constants. A particular integral is given by Q(t) = CV .

The complementary-function solution of the homogeneous problem is

Q(t) = Q0 e’t/RC , (3.46)

where Q0 is constant. The solution satisfying the initial conditions is

Q(t) = CV 1 ’ e’t/RC . (3.47)

3.3.2 Variation of Parameters

We now follow Lagrange, and solve

p0 (x)y (n) + p1 (x)y (n’1) + · · · + pn (x)y = f (x) (3.48)

by writing

y = v 1 y 1 + v2 y 2 + · · · + v n y n (3.49)

where the yi are the n linearly independent solutions of the homogeneous

equation and the vi are functions of x that we have to determine. This

method is called variation of parameters.

Now, di¬erentiating gives

y = v1 y1 + v2 y2 + · · · + vn yn + {v1 y1 + v2 y2 + · · · + vn yn } . (3.50)

We will chose the v™s so as to make the terms in the braces vanish. Di¬eren-

tiate again:

y = v1 y1 + v2 y2 + · · · + vn yn + {v1 y1 + v2 y2 + · · · + vn yn } . (3.51)

3.3. INHOMOGENEOUS EQUATIONS 79

Again, we will chose the v™s to make the terms in the braces vanish. We

proceed in this way until the very last step, at which we demand

(n’1) (n’1) n’1

+ · · · + v n yn

v1 y1 + v2 y 2 = f (x)/p0 (x). (3.52)

If you substitute the resulting y into the di¬erential equation, you will see

that the equation is satis¬ed.

We have imposed the following conditions on vi :

v1 y1 + v2 y2 + · · · + vn yn = 0,

v1 y1 + v2 y2 + · · · + vn yn = 0,

.

.

.

(n’1) (n’1) n’1

+ · · · + v n yn

v1 y1 + v2 y 2 = f (x)/p0 (x). (3.53)

This system of linear equations will have a solution for v1 , . . . , vn , provided

the Wronskian of the yi is non-zero. This, however, is guaranteed by the

assumed linear independence of the yi . Having found the v1 , . . . , vn , we obtain

the v1 , . . . , vn themselves by a single integration.

Example: First-order linear equation. A simple and useful application of this

method solves

dy

+ P (x)y = f (x). (3.54)

dx

The solution to the homogeneous equation is

x

’ P (s) ds

y1 = e . (3.55)

a

We therefore set x

’ P (s) ds

y = v(x)e , (3.56)

a

and ¬nd that x

’ P (s) ds

v (x)e = f (x). (3.57)

a

We integrate once to ¬nd

x ξ

P (s) ds

v(x) = f (ξ)e dξ, (3.58)

a

b

and so x x

’ P (s) ds

y(x) = f (ξ) e dξ. (3.59)

ξ

b

We select b to satisfy the initial condition.

80 CHAPTER 3. LINEAR ORDINARY DIFFERENTIAL EQUATIONS

3.4 Singular Points

So far in this chapter, we have been assuming, either explicitly or tacitly,

that our coe¬cients pi are smooth, and that p0 never vanishes. If p0 does

become zero then bad things happen, and the location of the zero of p0 is

called a singular point of the di¬erential equation. All other points are called

ordinary points.

If, in the di¬erential equation

p0 y + p1 y + p2 y = 0, (3.60)