A serious disadvantage of the method of variation of parameters is that the
integrations required are often quite difficult. In certain cases, it is usually
much simpler to guess a particular solution. In this section we will establish
a systematic method for guessing solutions of the equation

(1)

where , and are constants, and has one of several
special forms.

PART I: Consider first the differential equation

(2)

We seek a function such that the three functions
, and add
up to a given polynomial of degree .
One can show that the differential equation (2) has a solution of the form

(3)

EXAMPLE: Find a particular solution of the equation

(4)

Solution: Since , we set

and compute

So,

therefore

which gives

Hence,

is a particular solution of (4).

PART II: Consider now the differential equation

(5)

One can show that (5) has a particular solution of the form

Equivalently,

EXAMPLE: Find the general solution of the equation

(a)
(b)

Solution: We first note that the characteristic polynomial in both cases is

Thus, the characteristic equation has roots
, and therefore two linearly independent solutions
to the associated homogeneous equation are

Consequently, the general solution to the associated homogeneous equation is

(a) Since is not a root of , it follows that is
not a solution of the homogeneous equation

Thus, we set

Computing

and cancelling off the factor from both sides of the equation
gives

This implies that and

Therefore the general solution to
is

(b) Since is a root of of multiplicity 1 (that is, not a repeated
root), we set

We have

and

therefore

and cancelling off the factor from both sides of the equation
gives

This implies that and

Therefore the general solution to
is

EXAMPLE: Find the general solution of the equation

Solution: The characteristic polynomial is

Thus, is a repeated root of the characteristic equation, and therefore two linearly independent solutions
to the associated homogeneous equation are

Consequently, the general solution to the associated homogeneous equation is

We now find a particular solution of the equation .
Since is a repeated root of , we set

We have

and

therefore

and cancelling off the factor from both sides of the equation
gives

This implies that and

Therefore the general solution to
is

EXAMPLE: Find a particular solution of the equation

Solution: Since is not a root of the characteristic polynomial

we set

We have

and

therefore

Cancelling off the factor from both sides of the equation

gives

This implies that

Hence,

and

PART III: Let be a particular solution of the
equation

(6)

Since

the real part of the right-hand side of (6) is

while
the imaginary part is

One can show that

is a solution of

while

is a solution of

EXAMPLE: Find a particular solution of the equation

(7)

Solution: We first note that the right-hand side of (7) is the imaginary part of

(8)

Therefore we will find a particular solution of equation
(7) as the imaginary part of a complex-valued solution
of the equation

(9)

To this end, observe that the characteristic equation

has complex roots . So, is a root of of multiplicity
1, therefore we set

(10)

We have

and

therefore

Using this in (9) gives

hence

This implies that

Substituting this into (10) we get

Therefore

is a particular solution of (7).

EXAMPLE: Find a particular solution of the equation

(11)

Solution: From the previous Example,

is a complex-valued
solution of (9). Therefore,

is a particular solution of (11).

EXAMPLE: Find a particular solution of the equation

(12)

Solution: We first note that the right-hand side of (12) is the real part of
. Indeed, by Euler's formula we have

Therefore we will find a particular solution of equation
(12) as the real part of a complex-valued solution
of the equation

(13)

To this end, observe that is not a root of the characteristic equation

Thus, we set

(14)

We have

and

therefore

Using this in (13), we obtain

which implies

We have

which implies

Grouping together the terms containing and , we get