y''-8y'+16y=x+e^4x
1个回答

齐次方程的通解用特征多项式

kk - 8k + 16 = (k-4)^2 = 0

k=4(二重)

所以线性无关基本解为:e^4t ,te^4t

特解用算子方法:

(D-4)^2 y = x+e^4x

=>

y = 1/((D-4)^2) * (x+e^4x) = (1/16 + D/32 + 0(D))x + 1/(D-4)^2 e^4x

= x/16 + 1/32 + e^4x * 1/DD (1)

= xx*e^4x/2 + x/16 + 1/32

通解

c1*e^4t + c2*te^4t + xx*e^4x/2 + x/16 + 1/32