微積分二課后題答案,復(fù)旦大學(xué)出版社第十章

1、1第十章第十章習(xí)題習(xí)題10?11指出下列各微分方程的階數(shù)：(1)x(y′)2?2yy′?x?0(2)(y″)3?5(y′)4?y5?x6?0(3)?2y″?x2y?0(4)(x2?y2)dx?(x2?y2)dy?0yx???解:(1)因?yàn)榉匠讨形粗瘮?shù)y的最高階導(dǎo)數(shù)的階數(shù)為1故該方程為一階微分方程.(2)二階.(3)三階.(4)一階.2驗(yàn)證下列給定函數(shù)是其對(duì)應(yīng)微分方程的解：(1)y?(x?C)e?xy′?y?e?x(2)xy?C1ex?

2、C2e?xxy″?2y′?xy?0(3)x?cos2t?C1cos3t?C2sin3t，x″?9x?5cos2t(4)?1，xyy″?x(y′)2?yy′?02212CyCx?解:(1)()()()()eeeeeeexxxxxxxyxcyyxcxcyxc????????????????????????是微分方程的解.exyy????(2)在方程兩邊對(duì)x求導(dǎo)有上方程兩邊對(duì)x求12eexxxycc???12eexxyxycc?????導(dǎo)有即

3、即122eexxyxycc???????2yxyxy?????20xyyxy??????所以所確定的函數(shù)是方程的解.12eexxxycc???()yyx?20xyyxy??????(3)121212122sin23sin33cos34cos29cos39sin394cos29cos39sin39cos29cos39sin35cos2xtctctxtctctxxtctcttctctt????????????????????????所以是微

4、分方程的解.12cos2cos3sin3xtctct???95cos2xxt????3即1lnln11cyx?????1ln(1)(1)cxy???即1(1)(1)ecxy???1(1)(1)ecxy????記有而當(dāng)即時(shí)顯然是方程的解上1ecc??(1)(1)(0)xycc????10y??1y??式取時(shí)包含了故方程的解為(c為任意常數(shù))0c?1y??(1)(1)xyc???(2)分離變量得:兩邊積分得221001ddxxyxyyx??

5、????可知即211lnxcy???211eecxy???211cxyee?????又顯然是方程的解.0y?方程的通解為(c為任意常數(shù)).?21exyc??(3)分離變量得兩邊積分得即222211ddyxyxyx???221ln(1)ln1ycx????從而記有.2121ln1ycx???1221(1)ecyx????1ecc??22(1)1ycx???(4)分離變量得兩邊積分得即22sincoscosddyxxyx??1tancosy

6、cx???.tansecyxc??(5)原方程可化為:兩邊積分得(1)(1)ddyyyxxx???23232323yyxxc????由得所以原方程滿足初始條件的特解為01xy??115236c???即.2323523236yyxx????33222()3()5xyxy????(6)分離變量得兩邊積分得eddyyyxx???22eeyyxyc?????由得故原方程滿足初始條件的特解為(1)0y?12c?.21(1)(1)2eyyx????