Review of the neutron-capture process in fission reactors

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N X xm (iy)n−m . (n − m)! n=0 m=0 59 (10) In the double summation above, for a fixed n which runs from 0 to ∞, we sum over m from 0 to n. 1. 1 Let us now interchange the order of the double summation, that is, we fix m first and sum over n. 1, we find that we have to let m run from 0 to ∞, and for a fixed m, we sum over n from m to ∞. (n − m)! k! m=0 k=0 ∞ ∞ m X X m=0 x m! (iy)k k! k=0 x+iy ⇒e = ex · eiy . 60 (11) With (10) and (11), ex+iy can be evaluated in terms of real exponential, cosine and sine functions, that is, ex+iy = ex · eiy = ex (cos(y) + i sin(y)).

We use y = eλx , y 0 = λeλx and y 00 = λ2 eλx . Substituting into the ODE, we find that λ2 − 4λ + 13 = 0 ⇒ λ = 2 + 3i, λ = 2 − 3i. Thus, the general solution is given by y = Ae(2+3i)x + Be(2−3i)x where A and B are arbitrary constants. It is useful to rewrite the general solution as y = Ae(2+3i)x + Be(2−3i)x = e2x (Aei(3x) + Be−i(3x) ) = e2x (A [cos(3x) + i sin(3x)] + B [cos(3x) − i sin(3x)]) = e2x ([A + B] cos(3x) + i [A − B] sin(3x)) . Thus, we obtain y = e2x [C cos(3x) + D sin(3x)] where C = A + B and D = i[A − B] are arbitrary constants.

5. To convert the 2nd order ODE to a 1st order one, let s0 (x) = p(x). Thus, dp p = 1 + p2 . dx The ODE above is separable and can be solved as follows. Z Z dp p = dx. 1 + p2 The integral on the left hand side can be evaluated as explained on page 30 by letting p = sinh θ. This leads to θ = x + C. From p = sinh θ and s0 (x) = p(x), it follows that p(x) = sinh(x + C) ⇒ s0 (x) = sinh(x + C) ⇒ s(x) = cosh(x + C) + F. From the endpoints (0, 0) and (1, 0), we know that s(0) = 0 and s(1) = 0. This gives eC + e−C + 2F C −1 −C e·e +e e + 2F = 0 = 0.