Therefore, when 
r=0 : R(O)+a;R(1)+aR(2)=02 (5.82) 
r=1 : R(1)+a,R(0)+ a2R(1)=0 (5.83) 
r=2 : R(2)+a;R(1)+a2R(0) =0 (5.84) 
r2=2_ generally R(r)+a,R(r—1) + a2R(r —2) = 0. (5.85) 
From Eq. 5.83 
R(1)(1 + a2) = — a4R(0) = —ayo2, 
RO) = 03, 
and therefore 
R(1) -( ml a. (5.86) 
1+a2 
Inserting this value into Eq. 5.84 gives 
ROS ee (5.87) 
l+a2 
Substituting R(1), R(2) into Eq. 5.82, gives 
a (1+ ay)02 
eee (5.88) 
(1 —az2)(1 —a) + a2)(1 + ay + a2) 
This equation connects o? witha. From Eq. 5.86 
I Ge) a ea (5.89) 
(1 —a2)(1 — a, + a2)(1 +a; +. a2) 
From Eq. 5.87 
2 
R(2) = + aj —a2(1 + a2) £2 
1+a2 
+a} —as—ay 
ee Ue a ee ee 
5 Ce Catern Carn (5.90) 
Equation 5.85 is a homogeneous equation with the same coefficient as the homogeneous 
Eq. 5.60. Its solution is in the shape of 
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