作者doom8199 (～口卡口卡 修～)
看板Grad-ProbAsk
标题Re: [理工] [工数]-ODE
时间Thu Oct 29 17:10:53 2009
※ 引述《a1133333 (阿杰)》之铭言：
: y''-4y'+4y=x^3e^2x + xe^2x
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(D-2)(D-2)y = (x^3 + x)e^(2x) ____(1)
令 u = (D-2)y
所以 (1)式 → ┌ u' - 2u = (x^3 + x)e^(2x) ____(2)
└ y' - 2y = u ____(3)
from (2) → [u*e^(-2x)]' = x^3 + x
→ u*e^(-2x) = ∫ (x^3 + x) dx
= x^4/4 + x^2/2 + C1
→ u = (x^4/4 + x^2/2 + C1)*e^(2x) 带回(3)式
from (3) → y' - 2y = (x^4/4 + x^2/2 + C1)*e^(2x)
→ [y*e^(-2x)]' = (x^4/4 + x^2/2 + C1)
→ y*e^(-2x) = ∫ x^4/4 + x^2/2 + C1 dx
= x^5/20 + x^3/6 + C1*x + C2
or y = (x^5/20 + x^3/6 + C1*x + C2)*e^(2x)
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其它正规方法: (找 yp)
known yc = (C1 + C2*x)e^(2x) by solving y''-4y'+4y=0
<1> Method of Undetermined Coefficients:
set yp = x^2*(ax^3 + bx^2 + cx + d)e^(2x)
determine a、b、c、d
<2> Variation of Parameters:
set yp = u(x)*y1 + v(x)*y2
then solve : ┌ u'(x)*y1 + v'(x)*y2 = 0 for u'(x)、v'(x)
└ u'(x)*y1' + v'(x)*y2' = f(x)
with ┌ y1 = e^(2x)
│ y2 = xe^(2x)
└ f(x) = (x^3 + x)e^(2x)
<1> <2> are incomplete
for exercise (老早想学某人说这句话XD)
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◆ From: 140.113.141.151
※ 编辑: doom8199 来自: 140.113.141.151 (10/29 17:14)
1F：推 a1133333: 请问为什麽是多乘上x^2不是乘上x 10/30 23:24
2F：→ doom8199:因为会和 yc 的根重复到 10/31 03:32