Calculus: Analytic Geometry and Calculus, with Vectors

446 Trigonometric functions

for which (2) holds, then there must be constants I and B for which (1) holds. The
latter and more difficult result is often needed by students who have not yet
made reliable contacts with differential equations. Our solution of the more
difficult problem will be postponed until we have solved the easy problem by a
new method which involves only reversible steps. Starting with the formulas

(3)

r y = .4 sin kt + B cos kt

jay7t =Akcoskt - Bksinkt,

we eliminate B, and then eliminate Z, from these equations to obtain

(4)

cos kt d1 + (k sin kt)y = A

d y

sin ktdt- (k cos kt)y = -Bk.

Therefore,

dt[cos ki dt + (k in kt)y] = 0

(5) d

dt[sin kt dt - (k cos kt)y] = 0

or

(6)

cos ktd2ydt2-

d^2

k sin kt

dy

+ k sin kt

dy

dt dt+ (k2 cos kt)y = 0

d d

r

sin kt t2 +

d

y y

k cos kt dt- k cos kt + (k2 sin kt)y = 0

dt

(7)

cos kt

[zdt

+ key ] = 0

sin kt [dt + key] = 0.

Since there is not for which cos kt and sin kt are both 0, we conclude that

(8) ddt+ key = 0.

We are now ready to prove the more difficult result. Suppose that y is a given

function for which (8) holds. Then we can multiply by cos kt and sin kt to

obtain (7) and hence (6) and hence (5). Since the derivatives of the quantities

in brackets are zero, these quantities must be constants which we can call A

and -Bk to obtain (4). Solving the equations (4) for y and dy/dt gives (3)

and our problem is solved.

(^14) It is very easy to show that if k, A, B are constants for which k> 0 and if
(1) y = dew + Be *%
then
(2) ''-key=0.