Fox 發問時間: 科學數學 · 2 0 年前

[Laplace] 麻煩數學高手幫我解題

initial Value problem

1. y' 3y=10sint y(0)=0

2. y'-5y=1.5exp(- 4t) y(0)=1

可以幫我寫出詳細步驟嗎?

實在是想不出來怎麼解........

謝謝各位大大啊~~~~

已更新項目:

1. y' +3y=10sint y(0)=0 少了一個+ 抱歉

1 個解答

評分
  • 龍昊
    Lv 7
    2 0 年前
    最佳解答

      基本觀念:£{ ƒ'(t) }= sF(s) - ƒ(0)£{ e+ωt }= 1/( s - ω ) , £{ e-ωt }= 1/( s + ω )£{ sin ωt }= ω/( s2 + ω2 )£{ cos ωt }= s/( s2 + ω2 )  有了以上的觀念我們就可以來解題。1. y' + 3y = 10sin t,y(0) = 0sol:  £{ y' + 3y }= £{ 10sin t }  → sY(s) + 3Y(s) = 10/( s2 + 1 )  → Y(s) = 10/[ ( s2 + 1 )( s + 3 ) ]      = U[ 1/( s2 + 1 ) ] + V[ s/( s2 + 1 ) ] + A/( s + 3 )  U + jV = [ 10/( s + 3 ) ]s = j     = 10( 3 - j )/10 = 3 - j  → U = 3 , V = - 1  A = [ 10/( s2 + 1 ) ]s = -3   = 10/10 = 1  → Y(s) = 3[ 1/( s2 + 1 ) ] - [ s/( s2 + 1 ) ] + 1/( s + 3 )  y(t) = £-1{ Y(s) }  → y(t) = 3sin t - cos t + e-3t #2. y' - 5y = 1.5e-4t,y(0) = 1sol:  £{ y' - 5y }= £{ 1.5e-4t }  → sY(s) - 1 - 5Y(s) = 1.5/( s + 4 )  → ( s - 5 )Y(s) = 1.5/( s + 4 ) + 1  → Y(s) = 1.5/[ ( s +  4 )( s - 5 ) ] + 1/( s - 5 )      = A/( s + 4 ) + B/( s - 5 ) + 1/( s - 5 )  A = [ 1.5/( s - 5 ) ]s = - 4   = 1.5/( - 9 ) = - 1/6  B = [ 1.5/( s + 4 ) ]s = 5   = 1.5/9 = 1/6  → Y(s) = ( - 1/6 )[ 1/( s + 4 ) ] + ( 1/6 )[ 1/( s - 5 ) ] + 1/( s - 5 )      = ( 7/6 )[ 1/( s - 5 ) ] - ( 1/6 )[ 1/( s + 4 ) ]  y(t) = £-1{ Y(s) }  → y(t) = ( 7/6 )e5t - ( 1/6 )e-4t #  希望以上解答能幫助您!

    參考資料: 自己
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