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此方程為一階常係數非齊次方程, y'-2y=0,對應的特徵方程為:r-2=0,解得r=2. 所以方程通解為:y=ce^(2x) 再用常數變數法,求對應的非齊次方程的通解: 設y=c(x)*e^(2x)為所求方程的通解,代入原方程,可解出c(x)=-e^(-x)+c 故方程的通解為: y=Ce^(2x)-e^x
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