1.     Find a real general solution of the following system.

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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

 

by Level 11 User (81.5k points)

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