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freedombless, 這個題很簡單,y'=e^x+y,變為y'-y=e^x,方程兩端同乘以e^(-x),就變為e^(-x)y'-ye^(-x)=1,而此等式左端凑微分為[y*e^(-x)]',兩邊同時積分得ye^(-x)=x+c,這個求通解的過程叫積分因數法. 上式為通解,當初始條件y(0)=0時,交x=0,y=0,代入上式得c=0,故原微分方程的特解為y=xe^x.
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