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特徵方程為:λ^2-2λ=0,得:λ=0,2 故y1=c1e^(2x)+c2 設y*=ax^2+bx y*'=2ax+b y*"=2a 代入:2a-2(2ax+b)=x,比較係數得:-4a=1,2a-2b=0,得:a=b=-1/4,故y*=-(x^2+x)/4 因此通解y=y1+y*=c1e^(2x)+c2-(x^2+x)/4
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