(B2)22(B)2=2+p2 lv=0 (B) dd(z d(z dz dz Bre2-d= d (z (Br dz (Bdz (Br)dz (Br)zr-dz Iv=0 d1() (By)2=2d(B)2=2- (B) d(2)(22+-1 dv Iv=0 (B) l=0 d(Bz. B2'Brds*l7 (Br )2v+ [1 ]v=0 d(Brr) Bz d(Bz. (Br
dz d z dz d d z d ( ) / ( ) = z dz d −1 = ) ( ) ( ( ) ) ( ) ( 1 1 2 z dz d z dz d d z d − − = z dz d z dz d 1 2 2 2 2 1 2 ( ) ( 1) ( ) − − − = − ] 0 ( ) [1 ( ) ( ) 2 2 1 2 2 2 2 1 2 + + − = − − v z dz z dv z dz d v ] 0 ( ) [1 ( ) ( ) ( 1) ) ( ) ( 2 2 2 2 1 2 2 1 2 + + − = − + − − v z dz z dv z dz dv v d z d ] 0 ( ) [1 ( ) ( ) ( 2 2 2 2 1 2 + + − = − v z dz z dv v d z d ] 0 ( ) [1 1 ) ( ) ( 2 2 1 2 + + − = − v z dz z dv z v d z d ] 0 ( ) [1 ( ) 1 ) ( ) ( 2 2 2 + + − v = d z z dv z v d z d
这是v一阶贝塞耳函数。通解为 v(=)=C1J()+C2N(B) l()=z(=) 5.求下列方程的通解 a.l"+au=0; b.z2l"-2zl4(=4-1)=0 cu'-3u+zu=0 1-2c Z"+ l+(B) =0 b=2(y-1) b u+laz+ a=0 +1a=(6y) 4z24 ? 2 B b+2 2b+2(2)=C1VEJ(2a ×)、b+2)+C 2√ab+2 b b+2
u(z) z v(z) = ( ) ( ) ( ) 1 2 v z = C J z +C N z 这是 − 阶贝塞耳函数。通解为 5. 求下列方程的通解 . '' 3 ' 0 . '' 2 ' 4( 1) 0; . '' 0; 2 4 − + = − + − = + = c zu u zu b z u zu z u a u az u b a. ' [( ) ] 0 1 2 '' 2 2 2 2 1 2 = − + + − + − u z u z z u 2 1 = ] 0 4 1 4 1 '' [ 2 2 + + − u = z z u az b b = 2( −1) 1 2 = + b 2 a = () 2 2 + = b a 2 1 = 2 1 + = b ) 2 2 ) ( 2 2 ( ) ( 2 1 2 2 1 1 + + + + + = b b z b a z C zN b a u z C z J