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y'''+8y=0 的特徵方程為: λ^3+8=(λ+2)(λ^2 -2λ+4)=0 有根:λ1=-2 ,λ2=1+i√3 ,λ3=1-i√3 故方程有 y1=e^-2x y2=e^x*cos√3x y3=e^x*sin√3x ∴微分方程y'''+8y=0的一般解: y=C1e^(-2x)+C2(e^x*cos√3x)+C3(e^x*sin√3x)
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