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Solution:" W" ?2 x' Z0 V1 k P- X' `
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From: d{(a+bx)*C(x)}/dx =-k C(x) + s
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0 k1 q: S5 w# [5 E% |9 tbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
h$ i8 @1 }- u. |i.e.
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(a+bx) dC(x)/dx = -(k+b)C(x) +s3 P; t, A2 l i+ n, e
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* I1 Q2 E* [2 X" s+ f2 Z1 Y: f Ointroduce a tranform: KC(x)+s =Y(x), where K=-(k+b) 8 x1 y' C. G' X
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx2 d) C5 {: t) t
therefore:
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2 [ [: v& U, q{(a+bx)/K} dY(x)/dx=Y(x)
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from here, we can get:
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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). ~& r! |' u I, H2 j- d( ~
0 ^' ?0 U/ } m- W) Y5 n. oso that: ln Y(x) =( K/b) ln(a+bx)" N4 x/ V2 y/ F: h! j3 a G# ?
2 W9 \% p0 O6 b4 `this means: Y(x) = (a+bx)^(K/b)4 c9 n0 Q: z- ~, {
by using early transform, we can have:
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) l( P8 Q1 Y7 ?" L/ F' r$ |-(k+b)C(x)+s = (a+bx)^(k/b+1)3 _( ~2 e; D9 R5 o& G" q$ D
/ i; C. N# D9 e0 b) B. hfinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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