y''-4y' +13y=0的特征方程s^2 -4s +13 =0s= 2 +3i, 2-3i所以y''-4y'+13y = 0的解为y=e^(-2x)(acos3x + b sin(3x)特解考虑多项式情况,设y'' -4y' +13y =26x的特解是cx +d带入得到-4c + 13(cx+d) = 9cx +13d = 26x, c= 26/9, d=0所以通解就是y=e^(-2x)(acos3x + b sin(3x) -26x/9