202 Ordinary Differential Equations
The Maclaurin expansion for sine is
e3 e5sine = e - - + - - ....
3! 5!Fore small, sine is approximately equal toe (which is written mathematically
as sine ~ B). Replacing sin e in equation ( 3) by e, we obtain the following
linearized differential equation
(4)d^2 e
L - 2 +ge = 0
dtwhich approximates the motion of the simple pendulum for e small.
We could try to solve the DE (4) by trial-and-error. That is, we could try
to guess the form of the solution of ( 4) which contains one unknown constant
A, differentiate the guessed solution twice, substitute into (4), and see if it is
possible to determine A. For instance, if we guess the solution of (4) has the
form e = tA. Then differentiating twice, we get
de= AtA-1
dtand d2B = A(A - l)tA-2.
dt^2Substituting into (4), we see that A must satisfy
LA(A - l)tA-^2 + gtA = 0.
There is no constant value A which satisfies this equation for all values oft.
Consequently, there is no solution of the DE ( 4) of the form e = tA. Guessing
e =A sin tore= sin(t +A) produces no solution to the DE (4) either. (You
may want to verify this statement.) Next, we seek a solution of (4) of the
form e =sin At. Differentiation yields
dB- =A cos At
dt
and d2B - = -A2 sm. A t.
dt^2Substituting into the DE ( 4), we find A must be chosen to satisfy
-LA^2 sin At+ gsinAt = 0 or (-LA^2 + g) sin At= 0.
Hence,
-LA^2 + g = 0 or sin At= 0.
Thus,
A= ±.Ji[L or A=O.
Notice that the choice A = 0 yields the zero solution, e = sin 0 = 0, to
the DE (4). The choice A = V9fI produces the particular solution 81 =
sin V9fit. A second linearly independent solution 82 = cos V9fit is ob-
tained by guessing a solution of ( 4) of the form e = cos Et and discov-
ering B = V9fI. (Use the Wronskian to prove that the functions sin Ct