用拉氏变换求方程特解用拉氏变换求方程y"+2y'-3y=e^-t满足初始条件y(0)=0,y'(0)=1的特解

Question

用拉氏变换求方程特解
用拉氏变换求方程y"+2y'-3y=e^-t满足初始条件y(0)=0,y'(0)=1的特解

Answer 1

s^2 Y(S)-sy(0)-y'(0)+2sY(S)-3Y(S)=1/(s+1)
s^2 Y(S)-1+2sY(S)-3Y(S)=1/(s+1)
Y(S)=(s+2)/(s^2+2s-3)
=(s+2)/[(s-3)(s+1)]
=(5/4)/(s-3) + (-1/4)/(s+1)
y(t)=(5/4)e^3t +(-1/4)e^-t