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

564 Chapter 19 Ordinary differential equations I
To find the particular solution we mustnow imposethe given conditions:
y(0)=1 gives 1=A+B
y ′ (0)=1 gives 1=2A−5B
from which A = 6 7 andB = 1 . Finally, the required particular solution is
7
y = 6 7 e2x + 1 7 e−5x .
Example19.21 Findthe generalsolution of
d 2 y
dx 2 +4y=0
Solution As before, lety = e kx so that dy
dx = kekx and d2 y
dx = 2 k2 e kx . The auxiliary equation is
easily found tobe
that is
so that
k 2 +4=0
k 2 =−4
k=±2j
that is,we have complex roots.The two independent solutions ofthe equation arethus
y 1
(x) = e 2jx and y 2
(x) = e −2jx
so thatthe general solution can be written inthe form
y(x) =Ae 2jx +Be −2jx
However, in cases such as this, it is usual to rewrite the solution in the following way.
Recall from Chapter 9 thatEuler’s relations give
and
so that
e 2jx = cos2x +jsin2x
e −2jx = cos2x −jsin2x
y(x) =A(cos2x +jsin2x) +B(cos2x −jsin2x)
If we now relabel the constants such that
A+B=C and Aj−Bj=D
we can write the general solution inthe form
y(x) =Ccos2x +Dsin2x