Transformation of certain Differential Equations. 495 



Let ,(g +i ^) + 2 ^ = , 



where p is a positive integer. Make 



2/=D 2 «+F«=(D 2 + ^)«. 



Then, reducing by (a.), we have 



(Pv 

 x- r \=x B 4 u + k^xDhi=B 4 xu-4 ! D 3 u + ^D 2 xu-2^Du 

 ax 1 



= (D 2 + P)(D 2 .™-4Dm) + 2PDz<. 



In like manner 



2p a% = 2p D3 " + 2p * 9 D M= ( D2 + ^) 2 P D w * 

 Also 



^j/=FxD 2 « + F^m=A 2 D 2 j;w-2FDm + A 4 ^k 



Substituting these values in the given equation, it becomes 



(D 2 + F) {D^«+(2jo-4)Da + F«M}=0, 

 or D 2 .rM-f- (2^-4)D7< + & 2 ^ = 0, 



which is equivalent to 



(d 2 u JO \ . J. du 



an equation similar to the given one, p being changed into 

 p — 1. Hence, by continuing to repeat the same operations, 

 the last term may be taken away. If therefore y = {D 2 + k' 2 )pu f 



d*u 

 we shall have -j— z 2 + k* u = 0, the integral of which is known. 



Again, let j^ + kx-^ -pky = 0. 



Here make y=T)u + k xu= (D + kx) u. As before, re- 

 ducing by (a.), we have 



f|=D 3 « + itD^M=D 3 M + ^D 2 a + 2^D M =(D + h')D 2 w 

 dx l v ' 



+ 2 k D u. 



And 



d xi t-v 



kx -,— = &# D 2 w + F.rD xu = #D 2 #w—2/f Dm + ^a'D.z?/ 

 a.r 



= (D + k x) A D x u— 2 ft D u. 



Substituting these values in the given equation, it becomes 



(D + k x) (D* u + k D x u — pku)=0. 



Or ~D 2 u + k Dxu — p ku — 0, 



which is equivalent to 



