- #1
marmot
- 55
- 1
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 doesn't look like all like the correct answer, which has a logarithm which means there is probably some integration involved. why does not this work?
this is the answer btw:
y=4 * [2^(-2)cos(2*t)ln( abs(cos(2*t))) + t*2^(-1)*sin(2 t)]
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 doesn't look like all like the correct answer, which has a logarithm which means there is probably some integration involved. why does not this work?
this is the answer btw:
y=4 * [2^(-2)cos(2*t)ln( abs(cos(2*t))) + t*2^(-1)*sin(2 t)]
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