DIFFERENTIAL EQUATIONS OF THE SECOND ORDER. 71 



43. To see how this solution is included in the general solution of 41, it is neces- 

 sary to determine a function w such that 



2i -- -- ypw = v (T 4 y *"7 F (s, s') = 3> (s, s'), 



say. Now, for the derivatives with regard to y, s is parametric, being xp -J- u ; and 

 s' = s 2iy. We thus find on integration 



< being arbitrary, and s parametric in the integral. Evaluating the integral through 

 integration by parts, and dividing by the exponential factor on the left-hand side, we 

 find 



w = e *' ypy d>(x. u) --<<& (s, s 



r \ i 





say, so that u^ is expressed in terms of x, y, 11, and w. 2 in terms of s, s, x. 



Denoting derivatives of w 2 with regard to s, s', x, by 8/Ss, S/S.s-', S/8x, respectively, 



we have 



fl i S_ . B_\ . $_ . /_S_ _, 3_\ _ _ _4 &;_ 



I cj y I I ^ /^ I ., I o / I "" cv 1 cj y ) 



this last change arising from the fact that every term in w. 2 satisfies the transformed 

 equation because 3? satisfies it ; 



Consequently 



OS iy O-S ( 



. ^.- a 

 ^ ?./ ) 



^ _i_ i . 



j?3 " " 7 S* 7 ^- 



Further, 



dw, 



y^ = 



so that 



