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

19.2 Basic definitions 539
fromwhich
1 =B(−1)
B=−1
The particular solution becomesy = −sinx.
Sometimes the conditions involve derivatives.
Example19.6 ForthedifferentialequationofExample19.5findtheparticularsolutionwhichsatisfies
the conditions
(a) whenx = 0,theny = 0,and
(b) whenx = 0,then dy
dx = 5.
Solution Application ofthe first condition tothe general solutiony =Acosx +Bsinx gives
0=Acos0+Bsin0
=A
ThereforeA = 0 and the solution becomesy = Bsinx. To apply the second condition
wemustdifferentiatey:
dy
dx =Bcosx
Then applying the second condition weget
5=Bcos0
=B
so thatB = 5.Finallythe required particular solution isy = 5sinx.
In this example both conditions have been specified atx = 0, and are often referred
toas initialconditions.
EXERCISES19.2
1 Verifythaty = 3sin2xisasolutionof d2 y
dx 2 +4y=0.
2 Verifythat 3e x ,Axe x ,Axe x +Be x ,whereA,Bare
constants,allsatisfythe differentialequation
d 2 y
dx 2 −2dy dx +y=0
3 Verifythatx =t 2 +Alnt +Bis asolution of
t d2 x
dt 2 + dx
dt = 4t
4 Verify thaty =Acosx +Bsinx satisfiesthe
differentialequation
d 2 y
dx 2 +y=0
Verify alsothaty =Acosx andy =Bsinx each
individually satisfythe equation.
5 Ify =Ae 2x is the general solution of dy = 2y, find
dx
the particular solution satisfyingy(0) = 3.What is
the particular solution satisfying dy
dx =2whenx=0?