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題目應該是y“+3y'+2y=e^x吧?特徵方程為r^2+3r+2=0,得r=-1,-2即齊次方程的通解y1=C1e^(-x)+C2e^(-2x)設特解y*=ae^x,代入方程得:ae^x+3ae^x+2ae^x=e^x即6ae^x=e^x得6a=1a=1/6故原方程通解y=y1+y*=C1e^(-x)+C2e^(-2x)+…
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