The three simultaneous eqns are
y1’ = y2 ------------------- (1)
y2’ = -y1 + y3 ------------ (2)
y3’ = -y2 ------------------ (3)
By simple observation, we see that y1’ = - y3’, i.e. y1(x) = -y3(x) + const
Differentiating (2) and substituting for y3’ from (3),
y2’’ = -y1’ – y2
Substituting for y1’ from (1),
y2’’ = -y2 – y2
y2’’ + 2y2 = 0
Auxiliary eqn
m^2 + 2 = 0
m = +/- i.rt(2)
y2 = A.cos(rt(2)x) + B.sin(rt(2)x)
from (1),
y1’ = y2 = A.cos(rt(2)x) + B.sin(rt(2)x)
Integrating,
y1 = (A/rt(2)).sin(rt(2)x) – (B/rt(2)).cos(rt(2)x) + C
And, y3’ = -y2
y3 = -y1 + const = -(A/rt(2)).sin(rt(2)x) + (B/rt(2)).cos(rt(2)x) + D
The final solutions are:
y1(x) = (A/rt(2)).sin(rt(2)x) – (B/rt(2)).cos(rt(2)x) + C
y2(x) = A.cos(rt(2)x) + B.sin(rt(2)x)
y3(x) = -(A/rt(2)).sin(rt(2)x) + (B/rt(2)).cos(rt(2)x) + D