40 
MR, HARGREAVE ON THE SOLUTION OF 
whence we have 
+ \ tan- '.r) + A’' 
T)u=-^xXBXi ^x-\-k'x-{-y’ 
Verifying the first equation by these values, we find 
k"=~ and r=i ; 
the solution, therefore, is 
*^=i((l+‘^^) tan Vr+x)— 1+^(1 +.2^2)^ 
^ ]^i Q 
Verifying the second equation, we find k"= ^ and k"’=- ■ j and the solution is 
m=|(( 1+2?2^ tan-'j;+j?)+^(I+x2)-| . 
If the general form of the above differential equation be divided by and the 
second term be eliminated, by changing the independent variable from x to t, by 
means of the assumed equation we obtain 
dhj 
-^+A(2a— 6-f-l)(a?2-|-c2)“^2“+*^M=R(a?2+c2)-^^“+*^=RQ suppose. 
In order that the factor of u may be expressed in terms of the new variable, we must 
solve ^=(:r 2 -j-c 2 )“, and then find in terms of t. 
For most values of a, the equation between x and t is transcendental ; but parti- 
cular soluble cases may be found. 
Thus, if a= — 1 , then x=ctanc^, j;‘2_j_c2— ^ (.[jg equation becomes 
whose solution, therefore, is 
^^J)-(i+i){(^2_pg2)-(« + i)D-i{(^2_|_g2)iD4((^2_|_c2)-iR^)}} . 
which becomes, when Ro=0, 
X being ctan ct. 
The original equation is not altered by writing —b for 6 -|-l ; so that a particular 
solution may be readily deduced from the simple form 
u—kJ)^{x‘^-\-c^Y. 
This equation will reappear under another form in the sequel. 
