推 honestonly:m(_ _)m 01/06 23:37
推 kagato: m(_ _)m 01/07 01:13
推 ntust661: m(_ _)m 01/07 01:23
推 QQkimi: m(_ _)m 01/07 19:33
推 asdf322505:感謝大大的幫忙 01/07 22:35
※ 引述《asdf322505 ()》之銘言：
: 2x 2
: y''— ------y'+ ----- y =12(x^2+1)
: x^2+1 x^2+1
: ans:y=c1x+c2(x^2-1)+2x^4+6x^2
: 這題
: 想好久
: 誰能幫我解答@@
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1
令 m = ────
x^2 + 1
所以原 ode
→ y'' - 2xmy' + 2my = 12/m
→ (xy'-y)'/x - 2m(xy'-y) = 12/m
→ m(xy'-y)' - 2x(m^2)(xy'-y) = 12x
→ m(xy'-y)' + m'(xy'-y) = 12*x since m' = -2x(m^2)
→ [m(xy'-y)]' = 12x
→ [m(x^2)(y/x)']' = 12x
→ m(x^2)(y/x)' = 6x^2 + c1
→ (y/x)' = 6(x^2 + 1) + c1*[1 + 1/(x^2)]
→ y = (2x^4 + 6x^2) + c1*(x^2 - 1) + c2*x
ps:
也能用 [(x-i)D - 1][(x+i)D - 2]y = 12(x^2+1)^2
(x+i)^3 y
→ { ────*[────]' }' = 12(x+i)^2
(x-i) (x+i)^2
(x+i)^3 y
→ ────*[────]' = 4(x+i)^3 + c1
(x-i) (x+i)^2
y 1 -2i
→ [────]' = 4(x-i) + c1*[ ──── + ──── ]
(x+i)^2 (x+i)^2 (x+i)^3
y -1 i
→ ──── = 2(x-i)^2 + c1*[ ─── + ──── ] + c2
(x+i)^2 (x+i) (x+i)^2
→ y = 2(x^2 + 1)^2 + c1*(-x) + c2*(x+i)^2
= 2(x^2 + 1)^2 + c1'*x + c2'*(x^2 - 1)
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