Re: [理工] [工數]-一階ODE
※ 引述《bbdirty5566 (張苞)》之銘言:
: 1
: y'+ -y = y^2
: 2
: 1 1 1
: ans:y= - ----------- or y= - ,但y=1/2 隱含於通解中
: 2 1-(ce^1/2x) 2
: 我算出來還是有點不一樣
: 感覺怪怪的~"~
此為Bernoulli Eq
y=1/2為 trivial solution
解法
2
兩邊同除y
1 1 1
原式等於 -----y'+ ---*---=1
y^2 2 y
令u=1/y
u'=(-y^-2)y'
代回可得 -u'+u/2=1
整理可得 u'-u/2=-1
此為一階線性ODE
∫(-1/2)dx -x/2
I=e =e
-x/2 -x/2 -x/2
Iu=∫IQ(x)=∫e *-1dx=2∫e d(-x/2)=2e +C
x/2 -1
u=2+C*e =y
1 1 1
y= ---------= ---*------------
x/2 2 1+Ce^(x/1)
2+Ce
跟你不一樣的負號可以併入常數
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