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Solution:4 Y! e* d+ E) N2 u* t W9 A
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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) +s2 G X) _" R F% a
i.e.
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(a+bx) dC(x)/dx = -(k+b)C(x) +s
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q b) R4 K. C" }( Y/ eintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b) 9 h' {& Z* i6 i U
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx" w+ [$ C! R" F3 R r
therefore:& T% G% U k( g, P& \
" P9 o0 I: S4 d& M2 P; [{(a+bx)/K} dY(x)/dx=Y(x)
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_' Q" e c- R, t4 G) K# cfrom here, we can get:8 R* F6 N0 B5 Y& ?
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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)+ |$ G6 D5 y; |
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so that: ln Y(x) =( K/b) ln(a+bx)( ?# k! [ C% @8 t8 B
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this means: Y(x) = (a+bx)^(K/b)1 g( Y8 R0 {& x( M# a$ X- j4 y( x
by using early transform, we can have:
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) i1 t! O* b5 P6 B3 H-(k+b)C(x)+s = (a+bx)^(k/b+1)
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& s) c7 `+ ^; g" a4 |! Bfinally:
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* ]6 K& Y f; i4 _; FC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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