solution of a nonhomogeneous equation
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8/19/2019 Solution of a Nonhomogeneous Equation
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Solution of a Nonhomogeneous Equation
Before proceeding to the theoretical basis and the actual working technique of the useful
method of undetermined coefficients, let us examine the underlying concepts as applied to a
simple numerical example.
Consider the equation
!" # $%y & ' e x ( sin x "$%
)he complementary function may be determined at once from the roots
m & *, *, $ "!%
of the auxiliary equation. )he complementary function is
yc & c$ ( c!x ( c'e x "'%
Since the general solution of "$% is
y & yc( y p
where yc is as gi+en in "'% and y p is any particular solution of "$% all that remains for us to do is
to find a particular solution of "$%.
)he righthand member of "$%,
-"x% & ' e x ( sin x "%
is a particular solution of a homogeneous linear differential equation whose auxiliary equation
has roots
m/ & $±i
"0%
)herefore, the function - is a particular solution of the equation
" # $% "! ( $%- & * "1%
2e wish to con+ert "$% into a homogeneous linear differential equation with constant
coefficients, because we know how to sol+e any such equation. But by "1%, the operator
" # $%"! ( $% will annihilate the right member of "$%. )herefore, we apply the operator to both
sides of the equation "$% and get
" # $% "! ( $% !" # $%y & * "3%
4ny solution of "$% must be particular solution of "3%. )he general solution of "3% can be written
at once from the roots of its auxiliary equation, those roots being the +alues m & *, *, $ from "!%
and the +alues m/ & $ ±i from "0%. )hus the general solution of "3% is
y & c$ ( c!x ( c'e x ( cx
e x ( c0 cos x ( c1 sin x "5%
But the desired general solution of "$% is
y & yc( y p "6%
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where yc & c$ ( c!x ( c'e x
the c$, c!, c' being arbitrary constant as in "5%. )hus there must exist a particular solution of "$%
containing at most the remaining terms in "5%. 7sing different letters as coefficients to
emphasi8e that they are not arbitrary, we conclude that "$% has a particular solution
y p & 4x e x
( B cos x ( C sin x "$*%
2e now ha+e only to determine the numerical coefficients 4, B, C by direct use of the original
equation
!" # $%y & ' e x ( sin x "$%
9rom "$*% it follows that
y p & 4"x e x ( e x% # B sin x ( C cos x,
!y p & 4"x e x ( ! e x% # B cos x # C sin x,
'y p & 4"x e x ( ' e x% ( B sin x # C cos x,
Substitution of y p into "$%then yields
4 e x ( "B(C%sin x ( "B # C%cos x & ' e x ( sin x "$$%
Because "$$% is to be an identity and because e x , sin x and cos x are linearly independent the
corresponding coefficients in the two members of "$$% must be equal: that is
4 & '
B ( C & $
B # C & *
)herefore, 4 & ', B & ; , C & ; . -eturning to "$*%, we find that a particular solution of
equation "$% is
y p & 4x e x ( B cos x ( C sin x
y p & 'x e x ( ; cos x ( ; sin x
)he general solution of the original equation
!" # $%y & ' e x ( sin x "$%
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4 careful analysis of the ideas behind the process used shows that to arri+e at the solution "$!%,
we need to perform only the following steps=
"a% 9rom "$% find the +alues of m and m/ exhibited in "!% and "0%
"b% 9rom the +alues m and m/write yc and y p as in "'% and "$*%
"c% Substitute y p into "$%, equate corresponding coefficients, and obtain the numerical +alues
of the coefficients of y p.
(d) 2rite the general solution of "$%.
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