Case 2 If α^2 −β^2 = 0, the characteristic equation has the repeated roots λ=−α. The
first root gives the first member of the fundamental set as e−αt and using the rule
for repeated roots, the second member of the fundamental set of solutions is te−αt.
The general solution to equation (9.128) can then be expressed in the form
y =c 1 e−αt +c 2 te −αt
Case 3 If α^2 −β^2 = −ω^2 < 0 , then the fundamental set of solutions is
{e−αtcos ωt, e−αt sin ωt}and the general solution to equation (9.128) is given by
y=c 1 e−αt cos ωt +c 2 e−αtsin ωt
The solution to the homogeneous vector equation is then given by
y c=c 1 y 1 (t) + c 2 y 2 (t)
where {y 1 (t), y 2 (t)}are the functions from one of the cases previously examined.
To find a particular solution which gives the right-hand side sin 3 tˆe 3 examine
this function and its first couple of derivatives 3 cos 3 tˆe 3 , −9 sin3tˆe 3. The basic terms
in the set containing the function and its derivatives are the terms sin 3 tand cos 3t
multiplied by some constant. One can then assume a particular solution has the
form
yp=y p(t) = Asin3tˆe 3 +Bcos 3tˆe 3 (9 .132)
where A, B are undetermined coefficients. Substitute the equation (9.132) into the
nonhomogeneous equation (9.133) and show there results after simplification
[6 αA + (β^2 −9)B] cos3t+ [(β^2 −9)A− 6 αB ] sin 3 t= sin3t (9 .133)
Compare like terms in equation (9.133) and show Aand Bmust be selected to satisfy
the simultaneous equations
6 αA + (β^2 −9)B=0
(β^2 −9)A− 6 αB =1
One finds
A=
β^2 − 9
(β^2 −9)^2 + 36 α^2 B=
− 6 α
(β^2 −9)^2 + 36α^2
and so the particular solution is given by
y p=yp(t) =
[
(β^2 −9)
(β^2 −9)^2 + 36 α^2
sin 3 t−^6 α
(β^2 −9)^2 + 36α^2
cos 3t
]
ˆe 3
The general solution can then be represented
y =y(t) = yc+y p
