工程式數學解ODE

法一: 令 u(x)=x-y(x), so, y'=1-u'
1-u'=(u+6)/(3u+4), then u'=(2u-2)/(3u+4)
∫(3u+4)/(2u-2) du = ∫dx
∫[(3/2)+7/(2u-2)] du= x+c
(3/2)u+(7/2)ln| 2u-2 | = x+c
thus 3(x-y)+ 7ln| 2x-2y-2 | = 2x+C

法二:令 u=x-y+6, so, u'=1-y'
1-u'= u/(3u-14)
u'=(2u-14)/(3u-14)
∫(3/2)+ 7/(2u-14) du= x+c
(3/2)u+(7/2) ln| 2u-14| = x+c
3(x-y+6)+7 ln| 2x-2y-2| = 2x+ C

法三: integral factor , exact
(x-y+6)dx+(3y-3x-4)dy=0
Let integral factor u(t), t=x-y, then we hope
u(t)+(t+6)u'(t)=3u(t)-u'(t)(-3t-4)
or (2t-2)u'(t) = -2u(t), then u'(t)/u(t)=1/(1-t)
u(t)= - ln(1-t), thus u=1/(1-x+y) is an integral factor
(x-y+6)/(1-x+y) dx + (3y-3x-4)/(1-x+y) dy=0
∫(x-y+6)/(1-x+y) dx=∫[-1 + 7/(1-x+y)] dx= -x - 7 ln| 1-x+y|+C1(y)
∫(3y-3x-4)/(1-x+y) dy=∫[3- 7/(1-x+y)]dy= 3y - 7 ln| 1-x+y|+C2(x)
so, answer is -x+3y-7 ln| 1-x+y | = C

我講方法
你算算~~

另x-y=u
對x微分

則1+y'=u' 帶入原式

題目變成
u'-1=(u+6)/(3u+4)

接著就是基本的積分了(可變數分離)

還有問題再問吧~~