Question

Дифференциальная уравнения 4.11
y''-3y'+2y=-e^(-2x), y(0)=1, y'(0)=0

Answer 1

$y''-3y'+2y = -e^{-2x},y(0)=1, y'(0)=0.\\\\y''-3y'+2y=0\\y=Ae^{2x}+Be^{x}\\\\Guess\,\, that\,\, ke^{-2x} is \,\, the\,\, particular\,\, solution.\, We\,\, have$

$4ke^{-2x}+6ke^{-2x}+2ke^{-2x}=-e^{-2x}\\\Rightarrow k =-1/12\\y = Ae^{2x}+Be^{x}-1/12e^{-2x}\\\\\left \{ {A+B-1/12=1} \atop {2A+B+2/12=0}} \right. \Leftrightarrow A=-15/12, B=28/12\\\\y=-15/12e^{2x}+28/12e^{x}-1/12e^{-2x}$