laplace transform-－rlz87tm45t的部落格

標題:

laplace transform?

發問:

solve y''-y'+2y=(6t^2+8t+7)e^t, y(0)=0, y'(0)=1 by laplace transform

最佳解答:

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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 ) ------------------------------------------------------------------------------------ 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 ------------------------------------------------------------------------------------ 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

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