Question

解微分方程：y''-2y'+5y=(e^x)(cos2x-sin2x)的通解！！

Answer 1

y''-2y'+5y = (e^x)(cos2x-sin2x)
特征方程 r^2 - 2r + 5 = 0, r = 1±2i,
故设特解 y = x(e^x)(Acos2x+Bsin2x)
则 y' = (e^x)(Acos2x+Bsin2x) + x(e^x)(Acos2x+Bsin2x) + x(e^x)(-2Asin2x+2Bcos2x)
= (e^x)[Acos2x + Bsin2x + (A+2B)xcos2x + (B-2A)xsin2x]
y'' = (e^x)[Acos2x + Bsin2x + (A+2B)xcos2x + (B-2A)xsin2x]
+(e^x)[-2Asin2x + 2Bcos2x + (A+2B)cos2x + (B-2A)sin2x
- 2(A+2B)xsin2x + 2(B-2A)xcos2x]
= (e^x)[2(A+2B)cos2x + 2(B-2A)sin2x + (4B-3A)xcos2x - (3B+4A)xsin2x]
代入微分方程得
[2(A+2B)cos2x + 2(B-2A)sin2x + (4B-3A)xcos2x - (3B+4A)xsin2x]
-2[Acos2x + Bsin2x + (A+2B)xcos2x + (B-2A)xsin2x]
+5(Axcos2x+Bxsin2x) = cos2x - sin2x
即 4Bcosx - 4Asin2x = cos2x - sin2x
A = B = 1/4
特解 y = (1/4)x(e^x)(cos2x+sin2x)
通解 y = (e^x)(C1cos2x+C2sin2x) + (1/4)x(e^x)(cos2x+sin2x)