※ 引述《legenthume (沒有腳毛生不如死)》之銘言：
: x^2 y"+阿法xy'+杯塔y=0
: Consider x > 0 and let x = e^t
: A(d^2y/dt^2)+B(dy/dt)+C=0
: Let r1 and r2 be the roots of Ar^2 + Br + C = 0
: show thar if r1 and r2 are real and
: equal, then y = (c1 + c2t)e^r1t = (c1 + c2 ln x)x^r1
: y'=(c1r1+c2+c2r1t)e^r1t
: y"=(c1r1^2+c2r1+c2r1+tc2r1^2)e^r1t
: c1(Ar1^2+Br1+C)e^r1t+c2(tAr1^2+tBr1+tC+2Ar1t+B)e^r1t
: =c1(Ar1^2+Br1+C)e^r1t+c2[t(Ar1^2+Br1+C)+2Ar1t+B]e^r1t
: 右邊多出2Ar1t+B
: 沒辦法證明是0
: 怎麼辦
x^2 y"+αxy'+βy= 0
let x=exp(t) t=ln x
dy dy dt 1 dy 1 d
── = ── ── = ── ── = ── Dy D= ── 為對t微分之運算子
dx dt dx x dt x dt
2 2
d y d ┌ 1 dy ┐ 1 dy 1 dy 1
─── = ── │ ── ── │ = - ── ── + ── ── = ── D(D-1)y
dx^2 dx └ x dt ┘ x^2 dt x^2 dt^2 x^2
代回原式得
D(D-1)y+αDy+βy=0
[D^2+(α-1)D+β]y=0 , (D-r1)(D-r2)y=0
∵r1=r2為重根
y_h=(c1+c2t)exp(r1t) = (c1+c2lnx)x^r1 為通解
本題R(x)=0
故特解y_p=0
y=y_h+y_p =(c1+c2t)exp(r1t) = (c1+c2lnx)x^r1
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因為要解的ODE太多了 所以問神父吧
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※ 編輯: zi6ru04zpgji 來自: 114.37.59.203 (11/20 20:57)