Method of undetermined coefficients for so lnde with constant coefficients
Let be given the second order linear nonhomogeneous differential equation with constant coefficients
( 23 )
If the second term has a specific form, it’s possible to find a particular solution of the equation with the help of the method of undetermined coefficients, without integration.
1. If
,
( 24 )
we find a particular solution in the form
,
( 25 )
where
are undetermined coefficients and
( 26 )
If in particular
,
( 27 )
then we find a particular solution in the form
,
( 28 )
( 29 )
2. If
,
( 30 )
we find a particular solution of the form
,
( 31 )
where M, N are undetermined coefficients and
( 32 )
If in particular
,
( 33 )
then we find a particular solution of the form
,
( 34 )
( 35 )
Ex. 14.
.
1. The corresponding homogeneous equation (see Ex. 7). Roots of its characteristic equation are – 2, - 3, general solution is
.
2. We find a particular solution of the given
equation in correspondence with the formulas (24), (25), (26) (
,
is a simple root of the characteristic equation,
)
.
Finding
the derivatives of
,
,
,
we substitute the values of in the given equation, and after reducing similar terms we get
.
Equating
coefficients of the same powers of x
leads to a system of linear equations
in
,
namely
Therefore
,
and the general solution of the given differential equation is
.
Ex. 15.
.
1. The corresponding homogeneous equation
(see Ex. 8). Its characteristic equation
has real equal roots
,
and
.
2. By the same formulas (24), (25), (26) (
is a double root of the characteristic equation,
)
we seek a particular solution of the given equation in the form
,
and so
.
Substitution of in the given equation gives
and so
,
.
Ex. 16.
.
1. Corresponding homogeneous equation
was studied in the Ex. 9. Its characteristic equation has complex
roots
,
and the general solution
.
2. To find a particular solution of the given
equation we use the formulas (33), (34), (35) (
because
isn’t a root of the characteristic equation). We find
in the form
.
Since
,
substitution of values of in the given equation gives
.
Equating
coefficients of
we
obtain a system of equations in M,
N
.
Ex. 17. Solve Cauchy problem
.
1. For corresponding homogeneous equation
we have the characteristic equation
with complex roots
,
and so
.
2. To get a particular solution of the given
equation we take into account the formulas (30), (31), (32) (
is the root of the characteristic equation, and so
),
and we put
with
undetermined coefficients
.
Finding
,
and
substituting the values of
in the given equation we obtain the equality
.
Equating coefficients of
we get the system of equations in M,
N
and therefore
.
3. Taking into account the initial conditions.
.
Answer. The solution of Cauchy problem for the given equation is
.
Remark (superposition principle). We often meet with situations when a se-cond term of a differential equation (23) is a sum of several different summands of a specific form. Let for example
.
( 36 )
A
particular solution
of the equation (36) equals the sum of particular solutions
of the next equations
.
■Let
.
Then
,
and therefore
.
It
means that the sum
is a particular solution of the equation (36), that is
.■
In practice one can find with the help of one procedure.
Ex. 18. Find the general solution of a differential equation
.
1. The general solution of the associated homogeneous equation
is
because
of the characteristic equation
has two equal real roots
.
2. The second term of the given differential equation is the sum of three summands
.
On the base of the superposition principle and the formulas (27) - (28), (25) - (26), (33) - (34) we can sequentially seek three corresponding particular solutions
and then take their sum. But it’s better to find at once the particular solution of the given equation as the sum of such solutions
.
Since
the substitution of the values of in the given equation gives
.
After convenient grouping of the summands
we get
.
Therefore
,
and the general solution of the differential equation is
.