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Solution:
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X. G$ k5 P, U+ A t9 f6 VFrom: d{(a+bx)*C(x)}/dx =-k C(x) + s5 k! |- I- h2 j
so:
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1 o" U/ X+ |' u" e0 b5 e2 n0 mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s7 ?0 M* e, s' W( `" F3 ~3 @
i.e.
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! O! M1 Q4 Y# {1 a/ i" X6 Y. b(a+bx) dC(x)/dx = -(k+b)C(x) +s5 f- a5 {( l, q1 y$ W
+ |: g. ~+ Y; c& p* N) b' e8 r3 S' ^/ x
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b) 3 A, U, A- c N- I
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx) X% o' k. V5 O) i/ {; U' w6 A# G
therefore:) y0 x0 w4 s- C; W' J
& U+ T1 K4 w: O, l( u{(a+bx)/K} dY(x)/dx=Y(x)4 g' f6 U: m( q
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from here, we can get:8 h! z+ L) a2 ~$ Y1 o, K
9 m: W9 {) H" U" w7 O5 @dY(x)/Y(x) = [K/(a+bx)]dx i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
: G1 q6 T& O9 {+ I, ?& d$ r4 \! T* n
# ]# G+ m5 q" W3 f% A5 L0 P, m9 Oso that: ln Y(x) =( K/b) ln(a+bx)
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this means: Y(x) = (a+bx)^(K/b)9 r8 f% z5 i# z- ]
by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)+ f q& p7 l8 V4 Y4 U: k# a
" O* H$ i2 d/ _, P+ r/ x
finally:
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5 A. [( d" {& p0 J# ~4 d# J8 hC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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