利用拉氏变换解常微分方程的初值问题{y#39;-3y#39;#39;+2y=e-ty(0)=0,y#39;(0)=1}-t为上标

问题：利用拉氏变换解常微分方程的初值问题{y#39;-3y#39;#39;+2y=e-ty(0)=0,y#39;(0)=1}-t为上标

答案：↓↓↓

董玉涛的回答：

网友采纳　　记Y(s)=L[y(t)]则L[y'(t)]=sY(s)-y(0)=sY(s)L[y''(t)]=s^2*Y(s)-sy(0)-y'(0)=s^2*Y(s)-1L[e-t]=1/(s+1)所以有sY-3(s^2*Y-1)+2Y=1/(s+1)得：Y(s)=1/(s^2-1)所以Y(t)=sinh(t)...