Question

拉式变换求微分方程的解

y'''(t)-3y''(t)+3y'(t)-y(t)=t^2*e^t。y(0)=1,y(0)'=0,y''(0)=-2

Answer 1

方程两边作拉普拉斯变换
L[y'''(t)-3y''(t)+3y'(t)-y(t)]=L(t^2*e^t)
[s^3*F(s)-s^2*y(0)-s*y'(0)-y''(0)]-3[s^2*F(s)-s*y(0)-y'(0)]+3[s*F(s)-y(0)]-F(s)=2/(s-1)^3
因为y(0)=1，y'(0)=0，y''(0)=-2，所以
s^3*F(s)-s^2+2-3[s^2*F(s)-s]+3[s*F(s)-1]-F(s)=2/(s-1)^3
(s^3-3s^2+3s-1)*F(s)-s^2+3s-1=2/(s-1)^3
(s-1)^3*F(s)=2/(s-1)^3+s^2-3s+1
F(s)=1/(s-1)^6+[(s^2-2s+1)-(s-1)-1)/(s-1)^3
F(s)=1/(s-1)^6-1/(s-1)^3-1/(s-1)^2+1/(s-1)
原函数y(t)=L^(-1)[F(s)]

=L^(-1)[1/(s-1)^6-1/(s-1)^3-1/(s-1)^2+1/(s-1)]
=[(1/120)*t^5-(1/2)*t^2-t+1]*e^t