Laplace transformation?

L { y'' - y' - 2y }
= s²Y - sy(0) - y'(0) - [ sY - y(0) ] - 2Y , 注意: 此處大寫的 Y 其實就是 Y(s) 的簡寫
= s²Y - sY - 2Y
= ( s² - s - 2 )Y
= ( s - 2 )( s + 1 )Y

f( t ) = t [ 1 - u( t - 1) ]

L { 1 - u( t - 1) }
= 1/s - (1/s)e^(-s) , 利用 L { f(t-a) u(t-a) } = e^(-as) * F(s)
= (1/s)[ 1 - e^(-s) ]

L { f( t ) }
= L { t [ 1 - u( t - 1) ] }
= (-1) d/ds (1/s)[ 1 - e^(-s) ] , 利用 L { t^n f(t) } = (-1)^n * [ F(s) 的 n 次微分 ]
= d/ds (1/s)[ e^(-s) - 1 ]
= - (1/s²)[ e^(-s) - 1 ] + (1/s)[ - e^(-s) ]
= [ 1 - e^(-s) - se^(-s) ] / s²

L { y'' - y' - 2y } = L { f( t ) }
( s - 2 )( s + 1 )Y = [ 1 - e^(-s) - se^(-s) ] / s²

Y(s)
= [ 1 - e^(-s) - se^(-s) ] / [ s²( s - 2 )( s + 1 ) ]
= 1/[ s²( s - 2 )( s + 1 ) ] - e^(-s)*( 1 + s )/[ s²( s - 2 )( s + 1 ) ]
= 1/[ s²( s - 2 )( s + 1 ) ] - e^(-s)*1/[ s²( s - 2 ) ]

令 1/[ s²( s - 2 )( s + 1 ) ] = A/s + B/s² + C/(s-2) + D/(s+1)
1 = As(s-2)(s+1) + B(s-2)(s+1) + Cs²(s+1) + Ds²(s-2)
代入 s = 0 得: - 2B = 1 , 故 B = - 1/2
代入 s = 2 得: 12C = 1 , 故 C = 1/12
代入 s = -1 得: - 3D = 1 , 故 D = - 1/3
s³ 項係數 = 0 = A + C + D = A + 1/12 - 1/3 = A - 1/4 , 故 A = 1/4
因此 1/[ s²( s - 2 )( s + 1 ) ] = (1/4)(1/s) - (1/2)(1/s²) + (1/12)[ 1/(s-2) ] - (1/3)[ 1/(s+1) ]

令 1/[ s²( s - 2 ) ] = E/s + F/s² + G/(s-2)
仿照以上方法可解得: E = - 1/4 , F = - 1/2 , G = 1/4
因此 1/[ s²( s - 2 ) ] = - (1/4)(1/s) - (1/2)(1/s²) + (1/4)[ 1/(s-2) ]
- e^(-s) * 1/[ s²( s - 2 ) ] = (1/4)(1/s)e^(-s) + (1/2)(1/s²)e^(-s) - (1/4)[ 1/(s-2) ]e^(-s)

L^(-1) { 1/[ s²( s - 2 )( s + 1 ) ] }
= L^(-1) { (1/4)(1/s) - (1/2)(1/s²) + (1/12)[ 1/(s-2) ] - (1/3)[ 1/(s+1) ] }
= 1/4 - (1/2)t + (1/12)e^(2t) - (1/3)e^(-t)

L^(-1) { - e^(-s)*1/[ s²( s - 2 ) ] }
= L^(-1) { (1/4)(1/s)e^(-s) + (1/2)(1/s²)e^(-s) - (1/4)[ 1/(s-2) ]e^(-s) }
= (1/4)u(t-1) + (1/2)(t-1)*u(t-1) - (1/4)e^(2t-2)*u(t-1) , 利用 L { f(t-a) u(t-a) } = e^(-as) * F(s)
= u(t-1)*[ 1/4 + (t-1)/2 - (1/4)e^(2t-2) ]

y
= L^(-1) { Y(s) }
= L^(-1) { 1/[ s²( s - 2 )( s + 1 ) ] - e^(-s)*1/[ s²( s - 2 ) ] }
= 1/4 - (1/2)t + (1/12)e^(2t) - (1/3)e^(-t) + u(t-1)*[ 1/4 + (t-1)/2 - (1/4)e^(2t-2) ]

Ans: y(t) = 1/4 - (1/2)t + (1/12)e^(2t) - (1/3)e^(-t) + u(t-1)*[ 1/4 + (t-1)/2 - (1/4)e^(2t-2) ]

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驗算 :
y = 1/4 - (1/2)t + (1/12)e^(2t) - (1/3)e^(-t) + u(t-1)*[ 1/4 + (t-1)/2 - (1/4)e^(2t-2) ]
y' = - 1/2 + (1/6)e^(2t) + (1/3)e^(-t) + u(t-1)*[ 1/2 - (1/2)e^(2t-2) ]

y'' = (1/3)e^(2t) - (1/3)e^(-t) - u(t-1)*e^(2t-2)
- y' = 1/2 - (1/6)e^(2t) - (1/3)e^(-t) + u(t-1)*[ - 1/2 + (1/2)e^(2t-2) ]
- 2y = - 1/2 + t - (1/6)e^(2t) + (2/3)e^(-t) + u(t-1)*[ - 1/2 - t + 1 + (1/2)e^(2t-2) ]

y'' - y' - 2y
= t + u(t-1)*(-t)
= t [ 1 - u( t - 1) ]
= f( t )
故驗算無誤