The name is absent

⇒ λ(t)a(t) = -(n1 + n2)

⇒   λ(t) = —        ■■

^{v j}            a(t)

From (8)

From (10):

Combined with (15) gives:

and because ai (t) = aj (t), from equation (11),

(ɪi(t)                   , , ,                         ,           1

= = = 2x(t)(nιcι + n2C2)--₇-

ai(t)aj(t)                                   ai(t)

Multiplying by (n1 + n2)ai(t)aj (t), one obtains:

(nɪai(t) + n2aj (t)) = 2ai(t)(ai(t)nι + aj (t)^)(nɪeɪ + n2C2)x(t)

Using results (11) and (13), the above is equivalent to:

t⅛(t) = 2a(t)a_i(t)(n₁ c₁ + n₂c₂)x(t) — a(t) b + ρ

And noting that a_i(t) = ^aa^n from equations (14) and (10) gives:

With constant loading da/dt = 0, and so:

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