1.Use the change of variable x =1/t to determine the general solution of the
differential equation.
4 3 3/x
x y" + 2x y'- y = 8e
-1/x 1/x 3/x
ans:y = c1e + c2e + e [台大機械]
-----------------小弟的解法----------
2 1 8 3/x
原ODE寫成 y" + ── y' - ── y = ── e
x x^4 x^4
2 1
令P(x) = ── , Q(x) = - ──
x x^4
1
- ──
Q(x) x^4
再令 ─── = ──── = A 取 A = -1
(z')^2 (z')^2
1 1 1
則 z'(x) = ── , z"(x) = -2 ── , z(x) = - ──
x^2 x^3 x
1 2 1
-2 ── + ── ───
z" + Pz' x^3 x x^2
且令 ───── = ──────────── = 0
(z')^2 1
──
x^4
則原ODE可寫成 2
d y 4 -3z
─── - y = 8z e
dz^2
-z z 1/x -1/x
yh = c1e + c2e = c1e + c2 e
1 4 -3z
yp = ──── 8z e
D^2 -1
-3z 1 4
= 8e ────── z
D^2 -6D + 8
-3z 1 3 7 2 15 3 31 4 4
= 8e [── + ── D + ── D + ─── D + ─── D +...] z
8 32 128 512 2048
= ...(感覺解答的yp有錯，不然就是我算錯了ORZ)
若照原題目令x=1/t的話要怎麼做?
2.Use the transformation z = sinx to solve [成大土木]
2
d y dy 2
─── + (tanx) ── + (cos x)y = 0
dx^2 dx
ans: y = c1cos(sinx) + c2sin(sinx)
這題我無法轉成常係數類型的，那應該怎麼做?
2 3
3.Find (x+y )y''' + (6yy')y" + 3y" +2y' [雲科環工]
1 3 c1x^2
ans: ──y + xy = ─── + c2x + c3
3 2
z
這題似乎不是齊次式，沒辦法令y = e 去做吧? 那該怎麼算 麻煩高手們了@@"
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