求解非齐次线性微分方程y" -7y' +6y= sinx

特征方程：λ² - 7λ + 6 = 0
λ = 1 or 6
y = C₁e^x + C₂e^(6x)
设h = Asinx + Bcosx
dh/dx = Acosx - Bsinx
d²h/dx² = - Asinx - Bcosx,代入y'' - 7y' + 6y = sinx
- Asinx - Bcosx - 7(Acosx - Bsinx) + 6(Asinx + Bcosx) = sinx
(- A + 7B + 6A)sinx + (- B - 7A + 6B) = sinx
5A + 7B = 1,5B - 7A = 0,A = 5/74,B = 7/74
即h = (5/74)sinx + (7/74)cosx
通解为y = C₁e^x + C₂e^(6x) + (5/74)sinx + (7/74)cosx

y=c1e^x+c2e^(6x)+1/6sinx