请教大家一道微分题，多谢！(原题贴错了，现纠正）

醉酒当歌

请教大家一道微分题，多谢！(原题贴错了，现纠正）

5 D9 P$ E0 k8 V; E6 {d [(a+bx)c ] /dx = -kc +s
8 b6 o5 t; K1 N% ?  l/ cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
' Q7 r  W* f5 r, D9 A, \
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
3 s- A, c  C. L) f0 F( k9 f
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
& k9 y9 c+ z6 _7 w( y: }(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 v1 E9 a" ]) I, S- T7 f. Z+ cso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: ?+ `# E8 o4 g) f7 `6 li.e.
$ u- X6 u. h$ S8 J7 Z2 E( @
/ V) K8 G$ K$ e' l- g1 `" d(a+bx) dC(x)/dx  = -(k+b)C(x) +s
2 K6 z+ Y& X4 o$ |) x* s
1 ?  s1 y* L1 L
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
1 s% r! T9 Z; @0 F  ]: O/ z: ]
8 }' t: `# w2 \# J{(a+bx)/K} dY(x)/dx=Y(x)

6 B; M! G: H" B* l0 I2 Ffrom here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
) I0 O3 ^3 j8 O/ E# a% j
( X$ e" ?) u) I) |9 zso that:   ln Y(x) =( K/b) ln(a+bx)
1 a! H9 v) f% w! N) Z( g7 G0 z
" d. K6 A+ N" wthis means:  Y(x) = (a+bx)^(K/b)
) o7 e. ^1 Y7 @7 ?5 C; S; rby using early transform, we can have:
5 [0 W- M" X- K2 @/ R
-(k+b)C(x)+s = (a+bx)^(k/b+1)
% m7 H# Y2 e+ n/ \4 J6 v
finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)