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這是一個不顯含y的二階微分方程.令y'=p,則y''=dp/dx=dp/dy×dy/dx=p*dp/dy,代入原方程,得p*dp/dy=p*e^y.因為初始條件中y'(0)=1,即p=1,所以p≠0.兩邊消去p得dp=e^ydy,兩邊積分p=e^y+C1,所以y'=p=e^y+C1.代入初始條件...
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