﻿16 
  Lord 
  Kelvin 
  : 
  The 
  Problem 
  of 
  

  

  And 
  the 
  coefficients 
  a 
  u 
  a 
  2 
  , 
  a 
  s 
  , 
  etc. 
  can 
  be 
  determined 
  from 
  

   the 
  following 
  equations 
  : 
  — 
  

  

  2.3.a 
  1= 
  =i; 
  4.5.a 
  2 
  = 
  (/3.1 
  2 
  -2.3)a 
  1 
  2 
  ; 
  6.7.a 
  3 
  = 
  (2/3.1.2. 
  -2.3. 
  -4.5) 
  ai 
  a 
  3 
  ; 
  

  

  i-2 
  

  

  2i(2i 
  + 
  l) 
  ai 
  = 
  S 
  {2f5.r(i-r)-2r(2r+l)~(2i-2r)(2i-2r+l)}a 
  r 
  ai-r 
  

  

  + 
  

  

  \P\2j 
  -K 
  i 
  + 
  1 
  )\^i( 
  ie 
  ^ 
  n 
  ); 
  

  

  >(14 
  

  

  t-1 
  

  

  2 
  

  

  2i(2i+l) 
  a< 
  = 
  S 
  {2j3.r(i-r)-2r(2r 
  + 
  l)-(2i-2r)(2{-2r 
  + 
  l)}o 
  r 
  oj- 
  f 
  <t 
  odd); 
  

  

  r 
  =i 
  

  

  where 
  4dt4-n 
  

  

  Similarly, 
  if 
  we 
  choose 
  n 
  so 
  that 
  (n 
  + 
  1) 
  = 
  ten, 
  and 
  then 
  choose 
  

   ol 
  x 
  so 
  that 
  2.3.«i.n 
  = 
  l, 
  the 
  first 
  term 
  of 
  the 
  second 
  form 
  

   of 
  %" 
  K 
  {pc) 
  becomes 
  — 
  [® 
  K 
  (x)] 
  K 
  /x 
  4 
  ; 
  and 
  the 
  second 
  term, 
  

   equated 
  to 
  zero, 
  gives 
  us 
  the 
  following 
  equations 
  to 
  determine 
  

   «2i 
  a 
  3 
  j 
  e 
  tc 
  : 
  — 
  

  

  4.5.a 
  2 
  =/3a 
  1 
  2 
  ; 
  6.7.a 
  3 
  = 
  (2/3.1.2-4.5)a 
  1 
  a 
  2 
  ; 
  8.9.a, 
  = 
  (2/3.1.3. 
  - 
  6.7)a 
  ia3 
  + 
  (/3.2 
  2 
  -4.5)a 
  2 
  2 
  ; 
  

  

  i-2 
  

  

  2i(2* 
  + 
  l)a 
  i 
  ={2/3.1.(t-l)-(2t-2)(2i-l)}fl 
  1 
  a._ 
  1 
  + 
  S 
  {2&.r(i-r)-2r(2r 
  + 
  l) 
  

  

  r=2 
  

  

  -(2i-2rX2i-2r 
  + 
  l)\a 
  r 
  ai- 
  r 
  + 
  j 
  fS^)" 
  -i(i 
  + 
  l) 
  } 
  a- 
  (i 
  even) 
  

  

  5 
  (I 
  

  

  »— 
  l 
  

  

  r 
  = 
  2 
  

  

  2t(2i 
  + 
  l)a 
  < 
  ={2/3.1.(i-l)-(2z-2)(2i-l)}a 
  l 
  a 
  i 
  _ 
  1 
  + 
  S 
  {2/3.<i-r)-2r(2r+l) 
  

  

  r 
  = 
  2 
  

  

  4* 
  

  

  (2i-2r)(2i-2r 
  + 
  l)}a 
  r 
  ai_r(i 
  odd); 
  where 
  /3 
  = 
  4(n 
  + 
  l)= 
  

  

  («-!)■ 
  

  

  § 
  12. 
  In 
  the 
  particular 
  case 
  of 
  equation 
  (1) 
  for 
  which 
  

   /e 
  = 
  c© 
  (& 
  = 
  1), 
  and 
  which 
  corresponds 
  to 
  an 
  ideal 
  gas 
  which 
  

   obeys 
  Boyle's 
  Law 
  for 
  all 
  pressures, 
  the 
  differential 
  equation 
  

   becomes 
  

  

  t3P-5 
  a* 
  

  

  Assume 
  as 
  the 
  solution 
  of 
  this 
  equation 
  which 
  gives 
  p 
  = 
  l, 
  

   when 
  a?=oc, 
  

  

  *-*(«)- 
  

  

  /-, 
  a, 
  a„ 
  a, 
  \» 
  . 
  (17) 
  

  

  