5.方程 y”-(0y)2=0 的特解为y=() yl-0=0.l-0=-1 1 A. 1 B. -1 (x+1)2 (x+1)2 C.Inx+1 D.-Inx+1 解 令y=p,y=p=中，故 dp dx dx =p2→ dx =x+c 因plx-0=ylx-o=-1 代入上式c=1 p =p=- dx x+1 ylg ylo=0代入上式c2=0 故y=-lnx+l, 故选D7 5.方程 2 0 0 ( ) 0 | 0, | 1 x x y y y y = =    − =  = = −   的特解为y =( ) A． 2 1 1 ( 1) x + + B． 2 1 1 ( 1) x − + C．ln | 1| x + D．− + ln | 1| x 解 令y p  = , dp y p dx   = = ，故 dp 2 p dx = 2 1 dp dx p  = 1 1 x c p − = + 因 0 0 p y | | 1 x x = = = = −  代入上式c1 =1 故 1 1 dy p dx x = = − + 1 1 dy dx x = − + 2 y x c = − + + ln | 1| 0 | 0 x y = = 代入上式c2 = 0 故y x = − + ln | 1|， 故选 D