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this answer is the good one.- h6 h0 V U' [% j. K% g
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+ t3 Y8 d2 J) i B4 H) xFrom: d{(a+bx)*C(x)}/dx =-k C(x) + s! {' T& S# ^) R. o/ P6 a& e
so:, M' p5 z! u2 E3 v$ `) j$ Q" b
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
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) o6 V3 T* m* {: O+ h2 B(a+bx) dC(x)/dx = -(k+b)C(x) +s1 B# S* n$ Y9 {& W$ K
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introduce a tranform: KC(x)+s =Y(x), where K=-(k+b) 0 C9 e; i5 d/ c+ P% D9 L* w
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx2 P6 P; ~ R; ?7 ^& O
therefore:
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: Y) t$ Q$ B% T; F1 ~{(a+bx)/K} dY(x)/dx=Y(x)9 }2 b2 _5 y; o
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from here, we can get:# S: o) F, }7 S" o
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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)9 y3 j5 t; i* W# ^1 _
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so that: ln Y(x) =( K/b) ln(a+bx)6 v) s3 } h% Y# U7 ~! @7 a
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this means: Y(x) = (a+bx)^(K/b), ?2 Y! [& I0 C
by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:4 e+ o/ A! b% j7 Z
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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