laplace transform?

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 - 1 - sY + 2Y
= ( s² - s + 2 )Y - 1

L { 6t² + 8t + 7 }
= 6*2/s³ + 8/s² + 7/s
= ( 7s² + 8s + 12 ) / s³
≡ F(s)

利用公式: L { e^(at) * f( t ) } = F(s-a)
令 a = 1 , f( t ) = 6t² + 8t + 7 得 :

L { ( 6t² + 8t + 7 )e^t }
= F(s-1)
= [ 7(s-1)² + 8(s-1) + 12 ] / (s-1)³
= [ 7( s² - 2s + 1 ) + 8s - 8 + 12 ] / (s-1)³
= ( 7s² - 6s + 11 ) / (s-1)³

L { y'' - y' + 2y } = L { ( 6t² + 8t + 7 )e^t }
( s² - s + 2 )Y - 1 = ( 7s² - 6s + 11 ) / (s-1)³
( s² - s + 2 )Y = 1 + ( 7s² - 6s + 11 )/(s-1)³
( s² - s + 2 )Y = ( s³ - 3s² + 3s - 1 + 7s² - 6s + 11 )/(s-1)³
( s² - s + 2 )Y = ( s³ + 4s² - 3s + 10 )/(s-1)³ = ( s² - s + 2 )( s + 5 )/(s-1)³
Y = ( s + 5 )/(s-1)³

Y(s)
= ( s + 5 )/(s-1)³
= (s-1)/(s-1)³ + 6/(s-1)³
= 1/(s-1)² + 6/(s-1)³

y
= L^(-1) { Y(s) }
= L^(-1) { 1/(s-1)² + 6/(s-1)³ }
= L^(-1) { 1/(s-1)² } + L^(-1) { 6/(s-1)³ } , 因為 L 與 L^(-1) 皆屬 linear transform
= te^t + 3t²e^t , 請參考底下註解
= e^t ( 3t² + t )

Ans: y(t) = e^t ( 3t² + t )

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註解.
由乘 t 性質: L { t^n * f( t ) } = (-1)^n * [ F(s) 的 n 次微分 ] 得:
L { t * f( t ) } = (-1) * F'
L { t² * f( t ) } = F''

令 f( t ) = e^t
F(s) = L { e^t } = 1/(s-1)
L { t * e^t } = (-1) * F' = (-1) * (-1)/(s-1)² = 1/(s-1)²
L { t² * e^t } = F'' = ( F' )' = [ (-1)/(s-1)² ]' = 2/(s-1)³

L^(-1) { 1/(s-1)² } = te^t
L^(-1) { 6/(s-1)³ } = 3 * L^(-1) { 2/(s-1)³ } = 3t²e^t

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驗算.
y = e^t ( 3t² + t )
y' = e^t( 3t² + t ) + e^t( 6t + 1 ) = e^t ( 3t² + 7t + 1 )
y'' = e^t( 3t² + 7t + 1 ) + e^t( 6t + 7 ) = e^t ( 3t² + 13t + 8 )

y'' - y' + 2y
= e^t [ 3t² + 13t + 8 - ( 3t² + 7t + 1 ) + 2( 3t² + t ) ]
= e^t ( 6t² + 8t + 7 )
故驗算無誤