推 ntust661:推! 03/31 12:17
推 peacrackers:嘩...感謝不吝指教! 04/01 00:50
※ 引述《peacrackers (歡樂可樂果)》之銘言：
: 各位板眾大家好，小弟有2題ODE求解
: 雖然已大略知曉方法，卻仍不得其精要之處而苦思
: 在這裡請教大家，請不吝賜教。感謝大家。
: 1. dy y-xy^2-x^3
: ---- = ------------
: dx x+x^2y+y^3
: 2. dy y+xy^3(1+lnx)
: ---- = ---------------
: dx x
: 兩題我都用合併法，但餘項不知如何處理
1.
xdy-ydx+y(x^2+y^2)dy+x(x^2+y^2)dx=0
=>
xdy-ydx
----------- + ydy + xdx =0
x^2+y^2
=>
-1
tan (y/x) + 0.5y^2 + 0.5x^2 + c = 0
2.
xdy = ydx + xy^3(1+lnx)dx
=>
xdy - ydx
----------- = xy(1+lnx)dx
y^2
=>
-d(x/y) = xy(1+lnx)dx
=>
-x/y d(x/y) = x^2 dx + x^2*lnx dx
=>
-0.5(x/y)^2 = (1/3)x^3 + F(x) + c1 ... (1)
又
F(x) = ∫x^2*lnx dx
= 0.5∫ x*lnx d(x^2)
= 0.5 [x^3*lnx - ∫x^2 d(x*lnx) ]
= 0.5 [x^3*lnx - ∫x^2*(lnx + 1) dx]
= 0.5 [x^3*lnx -(1/3)x^3 - F(x) + c2]
=>
3*F(x) = x^3*lnx - (1/3)x^3 + c2
=>
F(x) = (1/3)x^3*lnx - (1/9)x^3 + c3
帶回(1)式
=>
-0.5(x/y)^2 = (1/3)x^3 + (1/3)x^3*lnx - (1/9)x^3 + c4
=>
2 1
y = ------------------------------
-(4/9)x - (2/3)x*lnx + c/x^2
1
y = ±------------------------------------
√[ -(4/9)x - (2/3)x*lnx + c/x^2]
--
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※ 編輯: jack0711 來自: 140.113.123.237 (03/31 02:32)