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

19.6 Constant coefficient equations 561
A linear time-invariant system has components whose properties do not vary with
time and as such can be modelled by a linear constant coefficient differential
equation.
19.6.1 Findingthecomplementaryfunction
AsstatedinProperty2(Section19.5.2),findingthegeneralsolutionofay ′′ +by ′ +cy = f
is a two-stage process. The first task is to determine the complementary function. This
is the general solution of the corresponding homogeneous equation, that isay ′′ +by ′ +
cy = 0.We now focus attention on the solution of such equations.
Example19.17 Verify that y 1
= e 4x and y 2
= e 2x both satisfy the constant coefficient homogeneous
equation
d 2 y
dx −6dy +8y=0 (19.12)
2 dx
Writedown the general solution of thisequation.
Solution Ify 1
= e 4x , differentiation yields
dy 1
dx =4e4x
and similarly,
d 2 y 1
dx 2 = 16e4x
Substitution into Equation (19.12) gives
16e 4x −6(4e 4x )+8e 4x =0
so that y 1
= e 4x is indeed a solution. Similarly if y 2
= e 2x , then dy 2
dx = 2e2x and
d 2 y 2
dx 2 = 4e2x . Substitution into Equation (19.12) gives
4e 2x −6(2e 2x )+8e 2x =0
so that y 2
= e 2x is also a solution of Equation (19.12). Now e 2x and e 4x are linearly
independent functions. So,from Property 1 we have
y H
(x)=Ae 4x +Be 2x
as the general solution of Equation (19.12).
Example19.18 Findvalues ofkso thaty = e kx isasolution of
d 2 y
dx − dy
2 dx −6y=0
Hence statethe general solution.