Lecture Notes in Differential Equations - Bruce E. Shapiro

376 LESSON 34. CRITICAL POINTS
The initial conditions for (34.11) give
hence the solution is
The trajectory is the unit circle because
0 = A cos 0 + B sin 0 = A (34.18)
1 = −A sin 0 + B cos 0 = B (34.19)
y = sin t (34.20)
x = cos t (34.21)
x 2 + y 2 = 1 (34.22)
for all t, and the solution is the set of all points starting at an angle of 0
from the x axis.
The initial conditions for (34.13), on the other hand, give
adding the two equations
√ √
2 2
0 = A
2 + B √
2
√
2 2
1 = −A
2 + B 2
(34.23)
(34.24)
1 = 2B
√
2
2 = B√ 2 =⇒ B = 1 √
2
=
√
2
2
(34.25)
hence
which gives a solution of
√
2
A = −B = −
2
(34.26)
√ √
2 2
y = −
2 cos t + sin t (34.27)
√ √
2
2 2
x =
2 sin t + cos t (34.28)
2