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特徵方程為r²+2r=0,得r=0,-2 齊次方程的解為y1=c1+c2e^(-2x) 設特解為y*=x(ax+b) y*'=2ax+b,y*"=2a,代入原方程: 2a+4ax+2b=-x+3 比較係數:4a=-1,2a+2b=3 得a=-1/4,b=7/4 所以原方程的通解y=y1+y*=C1+C2e^(-2x)+x(-x/4+7/4)
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