1550078481-Ordinary_Differential_Equations__Roberts_

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202 Ordinary Differential Equations


The Maclaurin expansion for sine is
e3 e5

sine = 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

dt

which 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

dt

and d2B = A(A - l)tA-2.

dt^2

Substituting 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^2

Substituting 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

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