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Solution:
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/ j' N, ]# n1 _6 z9 [From: d{(a+bx)*C(x)}/dx =-k C(x) + s4 f- n1 O1 F: Y3 a. v, q
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s, r* ^5 Z( r: A% t! f% }
i.e.
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(a+bx) dC(x)/dx = -(k+b)C(x) +s
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& e/ x/ [: K$ Z/ {% k& @introduce a tranform: KC(x)+s =Y(x), where K=-(k+b) ! l* P) M6 C2 r# z0 [" \
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx+ L& V, e) _4 @0 e4 F9 ^
therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)% |2 k& N9 X2 U' F$ j% u
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from here, we can get:
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G' C6 I# X6 _' _/ P. H7 A4 y' IdY(x)/Y(x) = [K/(a+bx)]dx i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)6 h: q1 B( O2 t% k% u$ a
1 K* U3 ^0 r$ d2 p8 yso that: ln Y(x) =( K/b) ln(a+bx), }2 j, W( _& [7 w
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this means: Y(x) = (a+bx)^(K/b)
! y6 T+ o2 D4 y1 }3 fby using early transform, we can have:8 ]- I- t8 H6 l
& n- C* e0 H$ v% Y, d- O @! D-(k+b)C(x)+s = (a+bx)^(k/b+1)
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0 h' M( Q3 k+ \5 X0 M! `! Nfinally:6 V2 i: o; a& Y# J) V! W+ A. x
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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