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THE METHOD OF UNDETERMINED COEFFICIENTS, Lecture notes of Algebra

The method of undetermined coefficients requires that the operator L have constant coefficients and that the function g be of a particular class. It works for ...

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THE METHOD OF UNDETERMINED COEFFICIENTS
In standard mathematical terminology, “the” method of undetermined coefficients is a method
used to solve relatively simple nonhomogeneous linear differential equations with constant coef-
ficients. However, the method of undetermined coefficients for differential equations is just one
example of a family of methods used when you know the structure of a solution to a problem but
not the specific solution. In partial fraction decomposition, we write a general form of terms in
a sum, using an as yet undetermined coefficient for each. Then we do some work to reduce the
original problem to a system of simple equations for the undetermined coefficients. In differential
equations, the method of undetermined coefficients involves identifying the structure of a particular
solution to a differential equation and then using that structure to reduce the original problem to
a simple algebra problem.
The method of undetermined coefficients can be described using a large number of examples, or
it can be described in general terms using an abstract formulation. The advantage of the abstract
formulation is that once you understand it, there is very little to remember. A smaller number of
examples will suffice to illustrate the method.
Linear Differential Operators
Any linear differential equation of order nhas the general form
an(x)dny
dxn+an1(x)dn1y
dxn1+· · · +a0(x)y=g(x).
All such equations can be written in the abstract form
Ly =g,
where the notation Ly simply refers to the left side of the differential equation.
The symbol Ltaken by itself represents a linear differential operator. These are like functions,
except that the input and output of functions are numbers or vectors, while the input and output
of linear differential operators are functions of a common variable. The notation f(x) = xcos x
says that the function fconsists of instructions to multiply the argument by its cosine. Similarly,
the notation Ly =y00 +ysays that the operator Lconsists of instructions to add the argument
function to its second derivative. The name f is often used to refer to an unspecified function
and only takes a particular form in examples. We’ll use the name L the same way. It will be a
generic function in theoretical statements and have particular forms in examples.
Operator notation has two important advantages. As noted above, it allows all linear equations
to be expressed in the same abstract notation Ly =g. The other advantage is that solving an
equation y00 +y=ex(for example) is equivalent to identifying all the functions ythat yield exas
the output of the operator L, defined in this context as instructions to add the function and its
second derivative.
The method of undetermined coefficients requires that the operator Lhave constant coefficients
and that the function gbe of a particular class. It works for y00 +y= 1, but not for y00 +xy = 1.
From here on, we’ll assume that Lis a linear differential operator of order nwith constant
coefficients.
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THE METHOD OF UNDETERMINED COEFFICIENTS

In standard mathematical terminology, “the” method of undetermined coefficients is a method

used to solve relatively simple nonhomogeneous linear differential equations with constant coef-

ficients. However, the method of undetermined coefficients for differential equations is just one

example of a family of methods used when you know the structure of a solution to a problem but

not the specific solution. In partial fraction decomposition, we write a general form of terms in

a sum, using an as yet undetermined coefficient for each. Then we do some work to reduce the

original problem to a system of simple equations for the undetermined coefficients. In differential

equations, the method of undetermined coefficients involves identifying the structure of a particular

solution to a differential equation and then using that structure to reduce the original problem to

a simple algebra problem.

The method of undetermined coefficients can be described using a large number of examples, or

it can be described in general terms using an abstract formulation. The advantage of the abstract

formulation is that once you understand it, there is very little to remember. A smaller number of

examples will suffice to illustrate the method.

Linear Differential Operators

ˆ Any linear differential equation of order n has the general form

an(x)

dny

dxn^

+ an− 1 (x)

dn−^1 y

dxn−^1

+ · · · + a 0 (x)y = g(x).

All such equations can be written in the abstract form

Ly = g,

where the notation “Ly” simply refers to the left side of the differential equation.

The symbol L taken by itself represents a linear differential operator. These are like functions,

except that the input and output of functions are numbers or vectors, while the input and output

of linear differential operators are functions of a common variable. The notation f (x) = x cos x

says that the function f consists of instructions to multiply the argument by its cosine. Similarly,

the notation Ly = y′′^ + y says that the operator L consists of instructions to add the argument

function to its second derivative. The name “f ” is often used to refer to an unspecified function

and only takes a particular form in examples. We’ll use the name “L” the same way. It will be a

generic function in theoretical statements and have particular forms in examples.

Operator notation has two important advantages. As noted above, it allows all linear equations

to be expressed in the same abstract notation Ly = g. The other advantage is that solving an

equation y′′^ + y = ex^ (for example) is equivalent to identifying all the functions y that yield ex^ as

the output of the operator L, defined in this context as instructions to add the function and its

second derivative.

The method of undetermined coefficients requires that the operator L have constant coefficients

and that the function g be of a particular class. It works for y′′^ + y = 1, but not for y′′^ + xy = 1.

ˆ From here on, we’ll assume that L is a linear differential operator of order n with constant

coefficients.

Some Preliminary Examples

Before we develop the general method of undetermined coefficients, it is helpful to start with some

simple examples.

ˆ Suppose yp is any one particular solution of an equation Ly = g and that the general solution

of the associated homogeneous equation is an n-parameter family yh. Then the general

solution of Ly = g is y = yp + yh.

This result follows immediately from the property that a linear operator can be distributed among

sums. Thus,

L[yp + yh] = Lyp + Lyh = g + 0 = g.

All of the examples below will use the operator L defined by Ly = y′^ + y. This means that each

example will have the same homogeneous solution: yh = c 1 e−x.

Example 1

Suppose we want to find a particular solution for y′^ + y = 8e^3 x. Since L simply adds the derivative and the function, it makes sense that the function we want should be a multiple of e^3 x. Since we don’t know which

one, we can use the generic form y = Ae^3 x. Then y′^ = 3Ae^3 x^ and Ly = 4Ae^3 x. We want Ly to be 8e^3 x,

which is accomplished by taking A = 2. The particular solution is yp = 2e^3 x^ and the general solution is y = 2e^3 x^ + c 1 e−x. 

Example 2 Suppose we want to find a particular solution for y′^ + y = x. This time it makes sense that we should try a

first degree polynomial, but we don’t know which one. Using the generic form y = A + Bx, we get y′^ = B and Ly = A + B + Bx. We want Ly to be x, which requires A + B = 0 and B = 1. Thus, A = −1,

yp = −1 + x, and the general solution is y = −1 + x + c 1 e−x. 

Example 3

To find a particular solution for y′^ + y = e−x, we can try y = Ae−x, as we did in Example 1. However, we then get Ly = 0 and we can’t choose A to get Ly = e−x. The correct trial solution turns out to be

Axe−x. Then y′^ = Ae−x^ − Axe−x, so Ly = Ae−x. This works, because A = 1 gives us the desired g. So the particular solution is yp = xe−x^ and the general solution is y = xe−x^ + c 1 e−x^ = (c 1 + x)e−x. 

Example 4

To find a particular solution for y′^ + y = 2 cos x, our first thought is that we could use y = A cos x. But then y′^ = −A sin x and Ly = A cos x − A sin x. We can get the right cosine coefficient by choosing

A = 2, but this does not work because Ly = 2 cos x − 2 sin x and g = 2 cos x are not the same function. Instead, the correct trial solution turns out to be A cos x + B sin x. Then y′^ = −A sin x + B cos x and

Ly = (A + B) cos x + (B − A) sin x. We can get cos x from this family by choosing A + B = 2 and B − A = 0. The result is the particular solution yp = cos x + sin x and the general solution is y = cos x + sin x + c 1 e−x.



From these examples, we have the following observations:

1. In many cases, the correct trial solution is merely a version of the function g with specific

coefficients replaced by as yet undetermined coefficients.

These results are enough to identify a general form for generalized exponential functions.

1. The family

y(x) = Pd(x)erx, (1)

where Pd is an arbitrary polynomial of degree d with nonzero xd^ coefficient, includes all of

the generalized exponential functions having characteristic value r and degree d.

2. The family

y(x) = Pd(x)eαx^ cos βx + Qd(x)eαx^ sin βx, (2)

where Pd and Qd are arbitrary polynomials of degree d (but with different coefficients),

includes all of the generalized exponential functions having characteristic value r and degree

d. Only one of Pd and Qd has to have a nonzero xd^ coefficient.

Example 6 The family of generalized exponential functions with characteristic value pair 1 ± 2 i and degree 1 can be written as y 1 ± 2 i, 1 = (Ax + B)ex^ cos 2x + (Cx + D)ex^ sin 2x,

where A and C are not both zero, in which case the degree would be 0 instead of 1. 

Solving Ly = g

Now that we know about generalized exponential functions, it is easy to describe the method

of undetermined coefficients. We assume that L has constant coefficients and g is a generalized

exponential function.

1. Identify the (possibly complex) characteristic value(s) c and degree d for g(x).

2. Find the family of functions yc,d using the appropriate formula (1) or (2).

3. Identify whether c is a characteristic value (or pair) for the equation Ly = 0. If so, let s be

the multiplicity. If not, just take s = 0.

4. The particular solution is one of the members of the trial solution y = xsyc,d(x).

5. Determine the coefficients by substituting the trial solution into the left side of the differential

equation and choosing the coefficients so that the result of Ly is the function g.

Example 7 Let Ly = y′^ + y. We now apply our method to Examples 1–4.

  1. The function g(x) = 8e^3 x^ is a generalized exponential function with characteristic value c = 3 and degree d = 0. This means that yc,d = Ae^3 x. The characteristic value of g is not a characteristic value of L, so s = 0. The trial solution is y = Ae^3 x.
  2. The function g(x) = x is a generalized exponential function with characteristic value c = 0 and degree d = 1. This means that yc,d = (A + Bx)e^0 x. The characteristic value of g is not a characteristic value of L, so s = 0. The trial solution is y = A + Bx.
  3. The function g(x) = e−x^ is a generalized exponential function with characteristic value c = −1 and degree d = 0. This means that yc,d = Ae−x. The characteristic value of g is a characteristic value of L, with multiplicity s = 1. The trial solution is y = x^1 Ae−x^ = Axe−x.
  1. The function g(x) = cos x is a generalized exponential function with characteristic value pair c = 0 ± i and degree d = 0. This means that yc,d = A cos x + B sin x. The characteristic value of g is not a characteristic value of L, so s = 0. The trial solution is y = A cos x + B sin x.



Example 8 To solve y′′^ − 4 y′^ + 13y = xe^2 x^ cos 3x + 3e^2 x^ sin x, we first identify g as a generalized exponential function

with c = 2 ± 3 i and d = 1. The family of such functions is yc,d = (A + Bx)e^2 x^ cos 3x + (C + Dx)e^2 x^ sin 3x.

The operator L has one complex pair of characteristic values, which happens to be the same as c. Thus, s = 1 and the trial solution is y = (Ax + Bx^2 )e^2 x^ cos 3x + (Cx + Dx^2 )e^2 x^ sin 3x.