THE ELEMENTS OF THE ORBIT OF A SATELLITE. 
791 
By writing y —\tt for y, we see that a term A sin (at-j-y) in the differential equation 
would generate c, a( Jp ~ cos ^ ie solution. 
From this theorem it follows that the solution of the equation 
D%=F 1 y 1 + Foy ;2 
is 
2/e 1 ('/e 1 3 — Kp) 
and the solution of 
is 
2k 1 (k 1 2 — k^) 
-f- the same with 2 and 1 interchanged 
DA=F 1 t ? 1 +F 3 7 ?2 
the same with 2 and 1 interchanged 
Also (writing the two alternatives by means of an easily intelligible notation) the 
solutions of 
are 
A 
y= 
— the same with 2 and 1 interchanged 
2*i(*i 8 —*a 2 ) 
The similar equations for D 4 £, D l y may be treated in the same way. The general 
rule is that 
y and y in the differential equations generate in the solution tz and t 'C respectively; 
and z and £ generate —ty and —ty respectively; and the terms are to he divided by 
2k 1 (k 1 3 — K.d) or 2 k 2 (k 2 — /q 2 ) as the case may he. 
Next suppose that A = 0 in the equation (199), and assume as the solution 
x =Ct 2 sin ( at-\-y)-\-Dt cos ( at-\-y ) 
Then 
~jp— C{ —aH 3 sin (at-\-y) J r kat cos (at-\- 77 )-f -2 sin (at-by)} 
+D{ —ciH cos (at A-y) — 2a sin (at-\-y)} 
cftx 
■^“= C{aH 3 sin (at-\-y) — 8aH cos (at-\-y) — 12a 3 sin (at-\-y)} 
+ D{ciH cos (at-\-i 7 ) +4a 3 sin (at-\~y)} 
5 1 2 
