Re: [理工] [工數]-變係數ode
看板Grad-ProbAsk作者ggyy940 (★GY大★ 研究所目標四大)時間13年前 (2012/07/11 18:15)推噓1(1推 0噓 5→)留言6則, 2人參與討論串12/12 (看更多)
※ 引述《kevinchu26 (吹風機達人)》之銘言：
: 題目如下
: (2x^2 + 3x + 1)y" + 2xy'- 2y = 0
[(2x^2+3x+1)y']' + [(-2x-3)y]' = 0
(2x^2+3x+1)y' + (-2x-3)y = C1
(-2x-3)
dy + ------------ ydx = C1 /(x+1)(2x+1) dx
(x+1)(2x+1)
I=exp[∫(-2x-3)/(x+1)(2x+1) dx]
=exp[∫ 1/x+1 - 4/2x+1 dx]
=(x+1)/(2x+1)^2
兩邊同成積分因子I
dy[(x+1)/(2x+1)^2] =C1 /(2x+1)^3 dx
y[(x+1)/(2x+1)^2] =k1 /(2x+1)^2 +k2
y = k1 /(x+1) + k2 2(x+1)^2/(x+1) 為通解
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