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常微分方程答案習題52-wenkub

2023-07-09 15:00:59 本頁面
 

【正文】 (t)為區(qū)間a上的連續(xù)nn實矩陣,為方程x=A(t)x的基解矩陣,而x=(t)為其一解,試證:a) 對于方程y=A(t)y的任一解y=(t)必有(t) (t)=常數(shù);b)(t)為方程y=A(t)y的基解矩陣的充要條件是存在非奇異的常數(shù)矩陣C,使(t) (t)=C.解a)[ (t) (t)]= (t)+ (t)= (t)+ (t)A(t)又因為=A(t) (t),所以=(t) A(t)[ (t) (t)]= (t) (t)A(t)+ (t) A(t) (t)=0,所以對于方程y=A(t)y的任一解y=(t)必有(t) (t)=常數(shù)b) “”假設為方程y=A(t)y的基解矩陣,則[ (t) (t)]= [(t)] +(t) (t)=[ A(t) (t)]+ (t) A(t) )+ (t)[ A(t) (t)]= (t) A(t) +(t) A(t) =0,故(t) (t)=C“”若存在非奇異常數(shù)矩陣C,detc0,使(t) (t)=C,則[ (t) (t)]= (t)+ (t)=0,故(t)(t)= (t) (t)A(t) (t)= (t)
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