Form of the particular solution
First derivative of partic. soln
Substitute in the diff eqn
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and
Equate coefficients of similar terms
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Using Mathcad to solve for A and B
So, we can solve for A and B. But what about C?.
The coefficient C can not be determined because the term appears in both the nonhomogenus term and the complementary solution.
Example
Solve the diff eqn
The terms and derivatives of the nonhomog functions are

, , , , , ,

Non appear in the complementary solution. Therefore the form of the particular solution is
CASE 2 (The terms of f(x) or its derivatives appear in the complementary solution)
Example (failing)

Solve
Solution:
Associated homog. eqn
Roots of the charac. poly
Complementary solution
1 , x and
Nonhomgenous terms and their derivatives
So the general solution is given by
Check
Example

If the roots of the characteristic poly of the diff eqn
are 4, 4, 4 , 2i, -2i. Find the form of the complementary solution of the diff eqn
Solution
Particular solution before modification
Since appears in the complementary solution with multiplicity 3, so we multiply by . Similarly the cos 2x and sin 2x appear in the comp solution with multiplicity 1. so we multiply the terms of the sin and cos by x.
Thus, the modified particular solution is
Solution
Assoc homog eqn
Roots of the char poly
complementary solution
If we ignore the fact that the term appears in the complementary solution, then the particular solution will have the form
To avoid duplicating the terms that appear in the complementary solution we multiply the corresponding terms in yp(x) by , where s is the multiplicity of the root that produced the duplicated
terms (in this case 2) (See the justification page 330 in text)
Thus, we modify by multiplying it by
Substitute the derivatives in the diff eqn gives
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The above eqn says that every other solution of the nonhomogenous eqn is the sum of the associated solution of homogenous eqn and a particular solution of the nonhomogenous eqn.
Thus, to find the general solution of the nonhomogenous eqn we first solve the associated homogenous eqn and try to find one particular solution of the nonhomogenous eqn.
Method of Undetermined Coefficients
This method will work of nonhomogenous linear diff. eqns with constant coefficients with the nonhomogenous term f(x) that are linear combinations of products of the functions

1. polynomials in x
2. erx
3. cos(bx) and sin(bx)
Examples of f(x)
The method of solution depends on wither the terms of the function f(x) or any of their derivatives appear in the solution of the associated homogenous eqn.
CASE 1 (non of the terms of f(x) or their derivative appear in the solution of the homog. eqn)
Example

Solve the diff eqn
y'' + 4y = 3x - 2 e-x
Solution:
y'' + 4y = 0
The associated homog. eqn.
5.5 Undetermined Coefficients and Variation of Parameters
The general nonhomog linear diff. eqn has the form
(1)
with the associated homogenous eqn
(2)
Let
(3)
Then eqn (1) can by written as
The operator L defined in (3) is a linear operator. That is
Let be a particular solution of the nonhomogenous eqn (1), i.e., . If y is any other solution of eqn (1), i.e., then
This means that the difference of the two solutions is a solution of the associated homogenous eqn. (2)
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and
Therefore, when substituting in the form of the particular solution , we get
And the general solution is given by
Example

Find the form of the particular soln of the diff eqn.
Solution
Associated homog. eqn
Roots of the charac. poly.
Complementary solution
Roots of the characteristic eqn.
The complementary solution
The terms and their derivatives of the nonhomogenous function f(x) =3x - 2 e-x are
1, x and e-x
Note that non of them appear in the complementary solution. Therefore, let us try a particular solution of the form
where A and B are coefficient to be determined by substituting in the diff Eqn.
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Substitute in the diff. eqn.
Equating the coefficients of similar terms gives