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this answer is the good one.
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4 V' A( g2 M; t) S" ^procedure:
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) l7 Z4 P1 r: x% K2 ]. YFrom: d{(a+bx)*C(x)}/dx =-k C(x) + s
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
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6 B* \; u: X& G, X. S' @5 ]- l- ?(a+bx) dC(x)/dx = -(k+b)C(x) +s
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- r2 R2 f" F! F9 Ointroduce a tranform: KC(x)+s =Y(x), where K=-(k+b) 9 _4 u* a- w" x, {; O
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
0 T$ O- E% |+ R4 Ytherefore:
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' e( `# z4 S6 @9 D{(a+bx)/K} dY(x)/dx=Y(x)
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1 A, u7 }; @$ ?. }5 Zfrom here, we can get:2 s' D: J- L( E
/ w) D/ x" l' a; k9 P2 ydY(x)/Y(x) = [K/(a+bx)]dx i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)% e$ q$ |6 `& `; _& H4 ^5 M
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so that: ln Y(x) =( K/b) ln(a+bx)
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, F2 f( @: X: f R5 Mthis means: Y(x) = (a+bx)^(K/b)" z' \; P/ i- v; S- H
by using early transform, we can have:9 @" w( w5 P& d; D( A% y
5 g# ~, T0 Y6 {2 y' x R-(k+b)C(x)+s = (a+bx)^(k/b+1)
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5 I8 T! j; D+ S9 d. u: O. N! afinally:5 |. ]4 ^/ [9 S+ @! A Q
5 n) u: Z6 N u7 w) I! U8 gC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s) |
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