solution of a nonhomogeneous equation

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 "$%.

solution of a nonhomogeneous equation

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