Re: [理工] 工數 微分方程

[endlesschaos (佐佐木信二)]

※ 引述《jody0113 (peter123)》之銘言：
: 1.
: y"-2y'+2y=e^(x)tan(x)

  2
(D  - 2D + 2)y_h = 0

            x               x
y_h = C1 * e cos(x) + C2 * e sin(x)

                x               x
Let y_p = v1 * e cos(x) + v2 * e sin(x)

                x                          x
    y_p'= v1 * e [cos(x) - sin(x)] + v2 * e [cos(x) + sin(x)]

                 x                x
        + v1' * e cos(x) + v2' * e sin(x)
          ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 0

                 x                           x
    y_p"= v1' * e [cos(x) - sin(x)] + v2' * e [cos(x) + sin(x)]

                 x                x
        - 2v1 * e sin(x) + 2v2 * e cos(x)


Substitute them into the ODE

   {
   { v1' * [cos(x) - sin(x)] + v2' * [cos(x) + sin(x)] = tan(x)
=> {
   { v1' * cos(x) + v2' * sin(x) = 0
   {

   {
   { v1 = sin(x) - ㏑[sec(x) + tan(x)]
=> {
   { v2 = -cos(x)
   {


=> y = y_h + y_p

             x               x          x
     = c1 * e cos(x) + c2 * e sin(x) - e cos(x) * ㏑[sec(x) + tan(x)]


: 2.
: y''-2y'+2y=e^(-x)ln(x)
: 麻煩版上的高手解題
: 我算到最後的積分都積不出來...

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