172 
PROF. H. F. BAKER ON CERTAIN EINEAR 
Wherefore 
This gives 
£2 -1 (u) = /cos jft, — sin ^t\. 
v sin cos \t J 
12 -1 (u) .v.Q (u) = i / cos \t, — sin \t \ / — cos t, sin t\ (u) 
v sm cos \t / \ sin t, cos t, 
= \ / — cos \t, sin ^t\ Q ( u) 
sin | -t, cos \t / 
- 1 ’ °y 
0 , 1 / 
Denoting this by ^o-, we find a 2 — 1, and hence 
_ i 
- 2 
12 [12 1 ( u ) vQ (w)] = 
i +^+— (i^) 2 + 3 j 
cll^t + ashwt, 
0 
e 
it 
,-it 
2 -u, o.xx 2 (/ \ / e *% 0 \ (x°, y°), 
0, e 3 
Thus the solution is 
{x, y) = Q{u+v ) (as 0 , y°) = 
\ — sin ■g'h cos j%t/ \ 0, & 
x — x (> e~ iJ cos j^t + y 0 ^ sin \t, 
y — —x°e~ it sin \t + y°e- t cos \t 
cos bt, sin brt 
if if / \ n At 
namely, 
The period of the coefficients in the original equation is 2tt. The functions cos D, 
sin i \t have only the period 4x. To bring the result into the form given by the 
general theory we may write 
x = x e 
, 0 -,— 5 ( 1 +i)t l l At 
(e"+1 )+y»e Ul+i)t . (l - e""), 
2% 
y = —x°e~* {1+i)t . —.(e il -l)+y°e w+i)t . |(l+e- < ‘), 
the so-called characteristic exponents being 
±Hi+0- 
