Re: [理工] [工數]-高階ODE
※ 引述《mdpming (ｎＥｗ)》之銘言：
: 1.
: 2 2
: yy'' - (y') - y y' = 0
: 答案
: c1
: y = ----------------
: -c1*x
: c2*e - 1
: 我算到
: 1 1 1
: ----(--- - ------)dy = dx
: c1 y y+c1
: y c1*x
: ----- = c2*e
: y+c1
: 怎麼移都移不出來 還是我算錯了@@
: 2.
: 2 2 2 2
: yy'' - (y') = y lny - x y
: 答案
: x -x 2
: y = exp(c1*e + c2*e + x + 2)
: 感謝 QQ
------
1.
yy'' - (y')^2 - (y^2)y' = 0
→ (y^2)(y'/y)' - (y^2)y' = 0
→ (y^2)(y'/y - y)' = 0
<1> if y^2 = 0 → y = 0 (special solution)
<2> if (y'/y - y)' = 0
→ y' - y^2 = c1*y
1
→ ─── dy = dx
y(y+c1)
1 y
→ ── ln｜───｜ = x + c2
c1 y + c1
y c1*x c1*c2
or ─── = c3*e , c3 = e
y + c1
若想整理成 y = f(x) 型態
我都是這樣想:
令 A = c3*e^(c1*x)
所以 y/(y+c1) = A → y = Ay + c1*A
→ (1-A)y = c1*A
→ y = c1*A/(1-A)
------
2.
yy'' - (y')^2 = (y^2)lny - (x^2)(y^2)
→ (y^2)(y'/y)' = (y^2)lny - (x^2)(y^2)
→ (y^2)[ (y'/y)' - lny + x^2 ] = 0
<1> if y^2 = 0 → y = 0 (special solution)
<2> if (y'/y)' - lny + x^2 = 0
→ (lny)'' - lny = -x^2 ____ 視 lny 為 x 的二階線性 O.D.E.
yc = c1*e^x + c2*e^(-x)
yp = x^2 + 2
(過程省略)
因此 lny = c1*e^x + c2*e^(-x) + x^2 + 2
--
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