Question

微分方程：用待定系数法解：x''-2x'+x=24t^2e^t

Answer 1

求齐次方程通特征方程r^2-2r+1=0,r1=r2=+1,则通解为x=C1*e^t+C2*t

设非齐次方程特解x*=t^2*(At^2+Bt+C)*e^t=(At^4+Bt^3+Ct^2)*e^t,

x*’=[At^4+(4A+B)t^3+(3B+C)t^2+2Ct]e^t,x*’’=[At^4+(8A+B)t^3+(12A+6B+C)t^2+(6B+4C)t+2C]e^t,

代入原方程,[12At^2+6Bt+2C]e^t=24t^2e^t,

比较系数得：A=2,B=C=0,

则原方程通解为x= C1*e^t+C2*t+2t^4* e^t=(C1+2t^4) e^t+C2*t