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this answer is the good one.2 S% g J1 q4 r7 |* S# K
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procedure:. t3 k) ^ X5 B+ |5 n T/ Y2 e' a( d
% q; E1 m. E, C4 uFrom: d{(a+bx)*C(x)}/dx =-k C(x) + s
- z) b) F# e8 Y+ S4 zso:# F- s0 ]' o# o; `. i0 X! b
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ A9 Y1 T9 G; s- U3 E3 i( \- ]i.e.
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(a+bx) dC(x)/dx = -(k+b)C(x) +s: v) C* S6 ~ b. u% L e
, H) \# [& X- R+ c+ a" d5 d8 K$ \9 n
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b) w1 V0 z+ ?; H/ ^
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx) s# G2 z/ T( f
therefore:
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& V( Q4 P: K5 g' @$ P8 o2 r{(a+bx)/K} dY(x)/dx=Y(x)5 B% V" l; R% b. l/ S9 S- D, B
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from here, we can get:3 O8 A6 y- ?$ v
: d% [0 a- l8 |* T, h$ `" HdY(x)/Y(x) = [K/(a+bx)]dx i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)/ j+ m) D& l4 I7 y; P
8 y z7 p7 m& v4 `8 Cso that: ln Y(x) =( K/b) ln(a+bx)
+ g5 k$ v9 e, A
% e3 A" _! k1 |: s: y$ T. Sthis means: Y(x) = (a+bx)^(K/b)- L* `; I1 E% W, o3 k* h5 \
by using early transform, we can have:$ _0 _( {7 W' d7 I5 \* ~
! c, ~; W) c" |9 ?( K-(k+b)C(x)+s = (a+bx)^(k/b+1)
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, M# i, L' y. X- I' B* ofinally:( w5 }( S2 |6 j
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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