利用拉氏变换解常微分方程的初值问题{y'-3y''+2y=e-t y(0)=0,y'(0)=1} -t为上标
问题描述：
利用拉氏变换解常微分方程的初值问题{y'-3y''+2y=e-t y(0)=0,y'(0)=1} -t为上标
答
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)...