工程數學 Laplace Transform問題 求解

y"+3y'+2.25y=9t^3+64, y(0)=1, y'(0)=31.5=> 4y"+12y'+9y=36t^3+256L(4y")=4s^2*Y(s)-4s-126L(12y')=12sY(s)-12L(9y)=9Y(s)L(9t^3+256)=6/s^4+256/s(4s^2+12s+9)Y(s)=6/s^4+256/s+4s+148Y(s)=(4s^5+148s^4+256s^3+6)/[s^4*(2s+3)^2]=a/s+b/s^2+c/s^3+d/s^4+e/(2s+3)+f/(2s+3)^2(4s^5+148s^4+256s^3+6)=as^3(2s+3)^2+bs^2(2s+3)^2+cs(2s+3)^2+d(2s+3)^2+es^4(2s+3)+fs^4=(4a+2e)s^5+(12a+4b+3e)s^4+(9a+12b+4c+f)s^3+(9b+12c+4d)s^2+(9c+12d)s+9d獲得5ㄍ方程式:3d=2 => d=3/23c+4d=0 => c=-4d/3=-29b+12c+4d=0 => b=-4(3c+d)/9=29a+12b+4c+f=256 => 9a+f=2462a+e=212a+4b+3e=148 => 12a+3e=142=> a=68/3, e=-130/3, f=42代入原式裡面:Y(s)=68/3s+2/s^2-2/s^3+3/2s^4-130/3(2s+3)+42/(2s+3)^2=68/3s+2/s^2-2/s^3+3/2s^4-65/3(s+3/2)+21/2(s+3/2)^2y(t)=68/3+2t^2-t^2+t^3/2-65e^(-3t/2)/3+21t*e^(-3t/2)/2

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