For the differential equation
the method of undetermined coefficients works only when the coefficients
a,
b, and
c are constants and the right-hand term
d(
x) is of a special form. If these restrictions do not apply to a given nonhomogeneous linear differential equation, then a more powerful method of determining a particular solution is needed: the method known as
variation of parameters.
The first step is to obtain the general solution of the corresponding homogeneous equation, which will have the form
where
y1 and
y2 are known functions. The next step is to
vary the parameters; that is, to replace the constants
c1 and
c2 by (as yet unknown) functions
v1(
x) and
v2(
x) to obtain the form of a particular solution
y of the given nonhomogeneous equation:
The goal is to determine these functions
v1 and
v2. Then, since the functions
y1 and
y2 are already known, the expression above for
y yields a particular solution of the nonhomogeneous equation. Combining
y with
y
h
then gives the general solution of the non-homogeneous differential equation, as guaranteed by Theorem B.
Since there are two unknowns to be determined,
v1 and
v2, two equations or conditions are required to obtain a solution. One of these conditions will naturally be satisfying the given differential equation. But another condition will be imposed first. Since
y will be substituted into equation (*), its derivatives must be evaluated. The first derivative of
y is
Now, to simplify the rest of the process—and to produce the first condition on
v1 and
v2—set
This will always be the first condition in determining
v1 and
v2;
the second condition will be the satisfaction of the given differential equation (*).
Example 1: Give the general solution of the differential equation
y″ +
y = tan
x.
Since the nonhomogeneous right-hand term,
d = tan
x, is not of the special form the method of undetermined coefficients can handle, variation of parameters is required. The first step is to obtain the general solution of the corresponding homogeneous equation,
y″ +
y = 0. The auxiliary polynomial equation is
whose roots are the distinct conjugate complex numbers
m = ±
i = 0 ± 1
i. The general solution of the homogeneous equation is therefore
Now, vary the parameters
c1 and
c2 to obtain
Differentialtion yields
Nest, remember the first condition to be imposed on
v1 and
v2:
that is,
This reduces the expression for
y′ to
so, then,
Substitution into the given nonhomogeneous equation
y″ +
y = tan
x yields
Therefore, the two conditions on
v1 and
v2 are
To solve these two equations for
v1′ and
v2′, first multiply the first equation by sin
x; then multiply the second equation by cos
x:
Adding these equations yields
Substituting
v1′ = sin
x back into equation (1) [or equation (2)] then gives
Now, integrate to find
v1 and
v2 (and ignore the constant of integration in each case):
and
Therefore, a particular solution of the given nonhomogeneous differential equation is
Combining this with the general solution of the corresponding homogeneous equation gives the general solution of the nonhomogeneous equation:
In general, when the method of variation of parameters is applied to the second-order nonhomogeneous linear differential equation
with
y =
v1(
x)
y1 +
v2(
x)
y2 (where
y
h
=
c1
y1 +
c2
y2 is the general solution of the corresponding homogeneous equation), the two conditions on
v1 and
v2 will always be
So after obtaining the general solution of the corresponding homogeneous equation (
y
h
=
c1
y1 +
c2
y2) and varying the parameters by writing
y =
v1
y1 +
v2
y2, go directly to equations (1) and (2) above and solve for
v1′ and
v2′.
Example 2: Give the general solution of the differential equation
Because of the In
x term, the right-hand side is not one of the special forms that the method of undetermined coefficients can handle; variation of parameters is required. The first step requires obtaining the general solution of the corresponding homogeneous equation,
y″ – 2
y′ +
y = 0:
Varying the parameters gives the particular solution
and the system of equations (1) and (2) becomes
Cancel out the common factor of
e
x
in both equations; then subtract the resulting equations to obtain
Substituting this back into either equation (1) or (2) determines
Now, integrate (by parts, in both these cases) to obtain
v1 and
v2 from
v2′ and
v2′:
Therefore, a particular solution is
Consequently, the general solution of the given nonhomogeneous equation is
Example 3: Give the general solution of the following differential equation, given that
y1 =
x and
y2 =
x3 are solutions of its corresponding homogeneous equation:
Since the functions
y1 =
x and
y2 =
x3 are linearly independent, Theorem A says that the general solution of the corresponding homogeneous equation is
Varying the parameters
c1 and
c2 gives the form of a particular solution of the given nonhomogeneous equation:
where the functions
v1 and
v2 are as yet undetermined. The two conditions on
v1 and
v2 which follow from the method of variation of parameters are
which in this case (
y1 =
x, y2 =
x3,
a =
x2,
d = 12
x4) become
Solving this system for
v1′ and
v2′ yields
from which follow
Therefore, the particular solution obtained is
and the general solution of the given nonhomogeneous equation is
Review and Introduction
Second-Order Equations
