Re: [理工] [工數]-ode ode ^^"

[ytyty (該換個版潛水了™ )]

※ 引述《topee (eason)》之銘言：
: :      (-2/3) ln │1-3y^(1/2)│ = x + c
: 除y^1/2
: y^-(1/2)dy + 3y^1/2 dx  = dx
: 令v= y^1-(1/2) =  y^1/2
: dv = (1-1/2)y^-1/2 dy
: 2dv + 3v dx = dx
: (D+3/2)v = 1
             ^^這裡錯了= =
             應為1/2
: v    = c1 e^-3/2x
:   h
: v   =   2/3
:   p

  v   =   1/3
    p
:                         2
: y^1/2 = c1 e^-(3/2)x + ---
:                         3

                          1
  y^1/2 = c1 e^-(3/2)x + ---
                          3
把我的答案轉換一下~

(-2/3) ln │1-3y^(1/2)│ = x + c


                        (-3/2)x     (-3/2)c
          1-3y^(1/2) = e         * e

                (-3/2)x           (-3/2)c
     y^(1/2) = e         *(-1/3) e        + 1/3

          (-3/2)c
令(-1/3) e        為另一常數c1


所以               (-3/2)x
     y^(1/2) = c1e         + 1/3

答案是一樣的^^

: 請問這答案有一樣嗎???
: 或 ln(3y^1/2 -1)^2 + x = c

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推 topee:感謝了= = 我忘了令一邊也要除2..