Download Discuss on Variation of Parameters

Discuss on Variation of
Parameters
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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 y 1 and y 2 are known functions. The next step is to vary the parameters; that is, to
replace the constants c 1 and c 2 by (as yet unknown) functions v 1( x) and v 2( x) to obtain
the form of a particular solution y of the given nonhomogeneous equation:
The
goal
is
to
determine
these
functions v 1 and v 2.
Then,
since
the
functions y 1 and y 2 are already known, the expression above for y yields a particular
solution of the nonhomogeneous equation. Combining y with y hthen gives the general
solution of the non‐homogeneous differential equation, as guaranteed by Theorem B.
Since there are two unknowns to be determined, v 1 and v 2, 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 v 1 and v 2—set
This will always be the first condition in determining v 1 and v 2; 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 c 1 and c 2 to obtain
Differentialtion yields
Nest, remember the first condition to be imposed on v 1 and v 2:
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 v 1 and v 2 are
To solve these two equations for v 1′ and v 2′, first multiply the first equation by sin x;
then multiply the second equation by cos x:
Adding these equations yields
Substituting v 1′ = sin x back into equation (1) [or equation (2)] then gives
Now, integrate to find v 1 and v 2 (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 = v 1( x) y 1 + v 2( x) y 2 (where y h = c 1 y 1 +c 2 y 2 is the general solution of the
corresponding homogeneous equation), the two conditions on v 1 and v 2 will always be
So after obtaining the general solution of the corresponding homogeneous equation
( y h = c 1 y 1 + c 2 y 2) and varying the parameters by writing y = v 1 y 1 + v 2 y 2, go
directly to equations (1) and (2) above and solve for v 1′ and v 2′.
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 v 1 and v 2 from v 2′ and v 2′:
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 y 1 = x and y 2 = x 3 are solutions of its corresponding homogeneous equation:
Since the functions y 1 = x and y 2 = x 3 are linearly independent, Theorem A says that the
general solution of the corresponding homogeneous equation is
Varying the parameters c 1 and c 2 gives the form of a particular solution of the given
nonhomogeneous equation:
where the functions v 1 and v 2 are as yet undetermined. The two conditions
on v 1 and v 2 which follow from the method of variation of parameters are
which in this case ( y 1 = x, y 2 = x 3, a = x 2, d = 12 x 4) become
Solving this system for v 1′ and v 2′ yields
from which follow
Therefore, the particular solution obtained is
and
the
general
solution
of
the
given
nonhomogeneous
equation
is