微分方程 y"+2y'+y=xe^x通解,

特征方程：r^2+2r+1=0,r=-1所以y1=(C1x+C2)e^(-x)设y2=(Ax+B)e^x则y2'=(Ax+A+B)e^xy2''=(Ax+2A+B)e^x则Ax+2A+B+2Ax+2(A+B)+Ax+B=xA=1/4,B=-1/4所以y=y1+y2=(C1x+C2)e^(-x)+(x-1)e^x/4