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Solution:7 o7 _ _9 ~: B9 G- i$ s6 g
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From: d{(a+bx)*C(x)}/dx =-k C(x) + s
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s3 i! u7 G; e- s5 X
i.e.
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/ f# { Z& o% ~/ ~! _5 ?0 T(a+bx) dC(x)/dx = -(k+b)C(x) +s% o2 Z n2 V* M" O
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& G5 B# v; R2 P, ^7 {) M5 Jintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
/ v! K' d& L: }0 o1 `: Vwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx! W. G6 v& \) ?0 J. q3 L
therefore:" v* l" v1 R+ j2 k: k$ D5 U5 Z
/ Q5 V j' ?/ C5 F8 _8 q{(a+bx)/K} dY(x)/dx=Y(x)
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4 K# Y0 I( F5 U4 E: P& ?' u) Mfrom here, we can get:; r A- ?. w& x: t5 J- a c
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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)
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4 {3 Z9 P1 G K! nso that: ln Y(x) =( K/b) ln(a+bx)* G* U6 G0 @+ |# [* ~6 Y2 N
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this means: Y(x) = (a+bx)^(K/b). ^' }$ }$ c7 u
by using early transform, we can have:. k% e' D, w) r+ ~8 j
5 Y: f% r9 W) Q* q- I-(k+b)C(x)+s = (a+bx)^(k/b+1). k8 _" u4 h* X
% h# o# P+ n0 [* Vfinally:- m' Y" C; R$ G1 ? { I/ K
4 l# U( M0 p6 q/ C/ AC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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