082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

570 Chapter 19 Ordinary differential equations I
sothaty P
willbeasolutionif αischosensothat −4α = 1,thatis α = − 1 4 .Thereforethe
particularintegralisy P
(x) = − e2x
4 .FromProperty2thegeneralsolutionoftheinhomogeneous
equation is found by summing this particular integral and the complementary
function
y(x) =Ae 3x +Be −2x − 1 4 e2x
Example19.27 Obtain a particularintegral ofthe equation
d 2 y
dx 2 −6dy dx +8y=x
Solution In the last example, we found that a fruitful approach was to assume a solution in the
same form as that on the r.h.s. Suppose we assume a solutiony P
(x) = αx and proceed
to determine α. This approach will actually fail, but let us see why. Ify P
(x) = αx then
dy P
y P
dx =αandd2 = 0.Substitution into the differential equation yields
dx2 0−6α+8αx=x
and α ought now to be chosen so that this expression is true for allx. If we equate the
coefficients ofxwe find 8α = 1 so that α = 1 , but with this value of α the constant
8
terms are inconsistent. Clearly a particular integral of the form αx is not possible. The
problemarisesbecausedifferentiationoftheterm αxproducesconstanttermswhichare
unbalanced on the r.h.s.So, wetry a solution of the form
y P
(x)=αx+β
with α, β constants. Proceeding as before, dy P
dx = α, d2 y P
= 0. Substitution in the
dx
differential equation now gives
2
0−6α+8(αx+β)=x
Equating coefficients ofxwefind
8α=1 (19.15)
and equating constant termswe find
−6α+8β=0 (19.16)
From Equation (19.15), α = 1 and then from Equation (19.16)
8
( 1
−6 +8β=0
8)
so that
8β = 3 4