# Complexification for finding a particular solution

1. Feb 28, 2009

### marmot

So I have this:

y'' + 4 y = 4 sec(2 t).

which translates to

p(D)y=4sec(2t)

where

p(D)=D^2+4

where D is a differential operatior

I know i have two choices for this, which is either looking for the particular solution through variable parameters which involved the winkonsian and some integrals, or just complexifying

if i complexify

i get p(D)y=4exp(-2ti) because cos(2t) is the real part of this exponential

because the equation is linear I can do this

y_p=4exp(-2ti)/(p(-2i))

where y_p is the particular solution.

after a lot of algebra, i find that the real part is

y_p=(-16cos(-2t)-32sin(-2t))/80

which doesnt look like all like the correct answer, which has a logarithm which means there is probably some integration involved. why does not this work?

y=4 * [2^(-2)cos(2*t)ln( abs(cos(2*t))) + t*2^(-1)*sin(2 t)]

Last edited: Feb 28, 2009
2. Feb 28, 2009

### lurflurf

if I complexify

I get p(D)y=4/[exp(2ti)+exp(-2ti)]

You appear assume that
Re[1/f]=1/Re[f]
which is untrue
infact
Re[1/exp(2i t)]=cos(2 t)
1/Re[exp(2i t)]=sec(2 t)

Variation of parameters is better for this problem.

3. Feb 28, 2009

### marmot

thanks a lot! so it means that complexfication is only efficient if i have sines and cosines?

4. Mar 2, 2009

### lurflurf

It depends what you mean by complexification. You were using it to try to solve the equation by undetermined coefficients, that can only work when the forcing term can be annihilated by a differentiation operator with constant coefficients. Such functions are sums and products of sin cos and exp.