278 Integration of Differential Equations. 
We will J : d?z dz : Qu 
e will now reduce the equation w dak? Hi —2a'uz—F7, (m), 
d'z dz 
to the form of (2); (m) is easily changed to dar — 2a’ 
(- 2 — FHSS and if we put y=z+ = we have se a 
2a'y= 0, (nv). If we put 2g—-2=0 or q=1, a?=1, b?=2a’, 
c=1 or 2gp—q+1=2p=1, we get p=3, and (2) wil become 
the same as(n). Hence if in (e) we put g==1,a?=1, b=V —2a’, 
we have y=a/fda(1 2?)-*eos.urV — HBB faa (1- 22) -2 
log.[u(1—z?)]cos.ueV —2a’, (0), for the complete value of y, 
which is the integral of (n). If weuse e for the hyperbolic base, 
= urd 2a) ual Bal 
we get by known fonts cos.ur V patents 5} . 
and if we put r=cos.p we get dz (1—2?)-? = — dy, 1l-7?= 
¥ rani 2008 4 u/2a'.cos.p 
sin.?9, cos.uxV = 2a! ; by substituting 
21 caekaeat a 
these values in(o) we have y= — ‘dy ( ov tal cos. 4 icicae 
B’ 
fad log.usin.*9), the abi beipg taken from cos.g=0 to 
B/ 
cos.y=1; or if we pm Ed 5 and = >, by A and B, we have by 
interchanging the limits of Me eal and changing its sign, 
(which makes no alteration in its value,) y= ef dy(e “Se aid 
tLe oan p (a +B log.usin.? *); the integral being taken from 
cos. = 1 to cos. = 0, or if we put 7 = 3.14159, &c. (= the 
semi-circumference of a circle whose radius =1,) we get fe 
fia (¢ ~ ww Ba.208.p pe ee) [A+B log.usin. 9), 0 and 5 
denoting the limits. of g; or (since cos. becomes negative in the 
second quadrant) we shall obtain the same value of y by putting 
y=f- dye S91 4 4 Blog. usin. 9), (m’), the integral being 
taken’ Hom'>=0; topes, (mt) will be found on trial to satisfy 
(nm), and since iti jent arbitrary constant 5; 
y is the complete i integral of (n) as required. 
