1550078481-Ordinary_Differential_Equations__Roberts_

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208 Ordinary Differential Equations


Each of these series converges for all x. Substituting ix for x in the expansion
for ex, using the fact that i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = -1, etc., and
rearranging, we find


eix = f-. (ixr

L,; n!
n=O

. i2x2 i3x3 i4x4 isxs i6x6 i7 x7
= l+ix+ 2!+ 31+ 41+ 5!+6! + 7! +···
x2 x4 x6 x3 xs x 7
= (1 - 2f + 4! -6T + · · ·) + i(x - 3T + 5f - 7! + · · ·)
= cos x + i sin x.


The identity eix = cos x + i sin x is known as Euler's formula. If a + i/3
is a root of multiplicity one of the auxiliary equation associated with some
linear differential equation with constant (real or complex) coefficients, then
a solution of the differential equation corresponding to the root a + i/3 is the
complex function


y = e<a+i,B)x = eo.xei,Bx = eo.x(cosf3x + isin/3x).

Suppose the complex number a+i/3 where t -/=- 0 and its complex conjugate
a -i/3 are both roots of the auxiliary equation associated with a homogeneous
linear differential equation. Two linearly independent, complex solutions cor-
responding to these two roots are


y 1 = e< o.+i,B)x = eo.x ei,Bx = eo.x (cos /3x + i sin /3x)
y 2 = e<o.-i,B)x = eo.xe-i,Bx = eo.x(cos(-f3x) +isin(-/3x))
= eo.x(cosf3x - isin/3x).

Hence, the general solution of the differential equation will include the linear
combination


C1Y1 + c2y2 = c1eo.x(cosf3x + isin/3x) + c2eo.x(cosf3x - isin/3x)
= eo.x[(c 1 + c2) cos/3x + i(c1 - c2) sin/3x]

where c 1 and c2 are arbitrary complex constants. Choosing c 1 = c2 = 1/2,
we see that y 3 = eo.x cos f3x is a real solution to the differential equation
(since any particular linear combination of solutions is also a solution). And
choosing c 1 =-c 2 =1/(2i), we see that y 4 = eo.xsin/3x is also a real solution
of the differential equation. So two real linearly independent solutions on
( -oo, oo) corresponding to the complex conjugate pair of roots a + i/3 and
a - i/3 are y3 = eo.x cos /3x and y4 = eo.x sin /3x. (Can you prove y3 and y4 are
linearly independent? HINT: Calculate their Wronskian.) Therefore, in the
general solution the linear combination of complex solutions c 1 y 1 + c 2 y 2 can
be replaced by the linear combination of real solutions k 1 y3 + k 2 y 4 where k 1
and k 2 are arbitrary real constants.

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