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this answer is the good one./ b& y2 ~% |9 A9 u8 a) T, ^
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7 N2 P; a( s# m! j0 t$ Yprocedure:
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* ?% D: Q. j9 L( A" j( ?$ j7 JFrom: d{(a+bx)*C(x)}/dx =-k C(x) + s
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s' C$ t" B- e6 t
i.e.
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(a+bx) dC(x)/dx = -(k+b)C(x) +s; e5 U" X/ h8 Z/ [: r. G5 C) \3 F
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4 _- q& @+ F$ ?$ A' Sintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
" e1 }# F8 N2 d' Ewhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
+ a" ]0 X7 f& d' l. {& D& X F8 K5 ptherefore:3 W5 @/ |9 `8 m3 P) M& n' E4 ?/ d$ f
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{(a+bx)/K} dY(x)/dx=Y(x)' \! ^/ q8 e" J3 C h# a, A
' u+ h- P' b/ y$ `6 d6 Bfrom here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)2 [+ i1 z$ W& Y- X% n" e# t8 D1 p
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so that: ln Y(x) =( K/b) ln(a+bx)% U F* ]9 f2 L, n5 W
1 D3 c7 ]0 y* y1 j0 @. c, a/ dthis means: Y(x) = (a+bx)^(K/b)
7 t8 m1 n9 D i5 ]; F+ u* aby using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)( O! `: j5 a! h- V
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finally:4 j, n4 K" p3 Q& N* B* I
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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