Complexification for finding a particular solution

AI Thread Summary
The discussion revolves around solving the differential equation y'' + 4y = 4sec(2t) using complexification versus variation of parameters. The user attempts to find a particular solution by complexifying the equation, leading to a complex exponential form. However, the resulting real part does not match the expected answer, which includes a logarithmic term, indicating a need for integration. The conversation concludes that complexification is only effective for certain functions, specifically those involving sines, cosines, and exponentials, while variation of parameters is deemed more suitable for this problem. Ultimately, the effectiveness of complexification depends on the nature of the forcing term in the equation.
marmot
Messages
55
Reaction score
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)]
 
Last edited:
Mathematics news on Phys.org
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.
 
thanks a lot! so it means that complexfication is only efficient if i have sines and cosines?
 
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.
 
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top