推 zendla:看到一半學術網路一直斷= = 11/17 12:28
※ 引述《zendla (夏夜薄荷)》之銘言：
: 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@@"
xy''' + y^2 y''' + 6yy'y'' + 3y'' + 2(y')^3 = 0
(1) + (2) + (3) + (4) + (5) = 0
∫(1) => ∫ xy''' dx = xy'' - ∫y''dx
∫[(1) + (4)] => xy'' + 2 ∫y''dx = xy'' + 2y'
(6) (7)
∫(6) => ∫ xy'' dx = xy' - ∫y'dx
∫[(6) + (7)] => xy' + ∫y'dx = xy' + y
(8) (9)
∫(8) = > ∫ xy' dx = xy - ∫y'dx
∫[(8) + (9)] => xy
=> ∫∫∫ [(1) + (4)] (dx)^3 = xy
∫(2) => ∫y^2 y'''dx = y^2 y'' - ∫2yy'y''dx
∫[(2) + (3)] => y^2 y'' + 4∫yy'y''dx
(10) (11)
(11) => 令 u = yy', v'=y''
u' = (y')^2 + yy'', v=y'
∫yy'y''dx = y(y')^2 - ∫(y')^3dx - ∫yy'y''dx
=>∫yy'y''dx = [y(y')^2]/2 - [∫(y')^3dx]/2
=> ∫[(2) + (3)] = y^2 y'' + 2y(y')^2 - 2∫(y')^3dx
=> ∫[(2) + (3) + (5)] = y^2 y'' + 2y(y')^2
(12) (13)
∫(12) => ∫y^2 y''dx = y^2 y' - ∫ 2y(y')^2 dy
∫[(12) + (13)] = y^2 y'
∫y^2 y'dx = y^3 - ∫2y^2 y'dx
=> ∫y^2 y'dx = y^3/3
=> ∫∫∫[(2) + (3) + (5)](dx)^3 = y^3/3
右邊積分三次補三個積分常數 => C1 x^2/2 + C2 x + C3
=> xy + y^3/3 = C1 x^2/2 + C2 x +C3
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