Re: [理工] [工數]-ODE
※ 引述《abcxyz123 (阿里八八與四十大盜)》之銘言：
: xy''-y'=(3+x)(x^2)e^x
: 答案:Y=c1+c2x^2+(x^2)e^x
: 重點在於特解= =
: 我算了幾次都算不出來
: 卡在後面有幾項積分消不掉 不知錯在哪
: 高手幫解 感恩~
令p = y' 代回ode中
1
p'- ──p = (3x+x^2)e^x
x
∫ - 1/x dx
I(x)= e = 1/x
Ip = ∫ (1/x)(3x+x^2)e^x dx
= (2+x)e^x + c1
y' = p = (2x+x^2)e^x + c1x
y = ∫ (2x+x^2)e^x + c1x dx
= c1x^2 + c2 +(x^2)e^x
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11/15 12:06, , 1F
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (11/15 13:13)
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