Research Space for Linear Algebra & Discrete Mathematics
y1'=-2y1-te^(-2t)先解齊次解y1'=-2y1得y1(h) = c1e^(-2t)令特解y1(p)= dt^2e^(-2t)代入原式解d則y1 = y1(h)+y1(p)y2'=5y2-7te^(-2t)先解齊次解y2'=5y2得y2(h)=c2e^(5t)令特解y2(p)=(d0+d1t)e^(-2t)代入原式即可解得d0, d1, 則y2 = y2(h)+y2(p)這個可以參考一些工數的書
張貼留言
1 則留言:
y1'=-2y1-te^(-2t)
先解齊次解y1'=-2y1
得y1(h) = c1e^(-2t)
令特解y1(p)= dt^2e^(-2t)代入原式解d
則y1 = y1(h)+y1(p)
y2'=5y2-7te^(-2t)
先解齊次解y2'=5y2
得y2(h)=c2e^(5t)
令特解y2(p)=(d0+d1t)e^(-2t)代入原式即可解得d0, d1, 則y2 = y2(h)+y2(p)
這個可以參考一些工數的書
張貼留言