2nd order ODE's odd and even functions

[ppy]

Hi

I have been looking at some lecture notes inc the following example. Solve :
y'' + ω^2y = (some even function)
The particular integral is then found using Fourier series. As the function on the RHS is even this only includes cosine terms.
The complementary function is found from the homogeneous equation and is
y= Acos(ωt) + Bsin(ωt)

The general solution is the P.I. + C.F. My question is ; as the original function is even does that mean the constant B must equal zero in all cases ?

In general for a 2nd order inhomogeneous linear ODE if the function on the RHS is odd or even does that imply anything about the general solution ?

 

[tiny-tim]

hi ppy!

The general solution is the P.I. + C.F. My question is ; as the original function is even does that mean the constant B must equal zero in all cases ?

no

if y is a solution of y'' + ω²y = f(t),

then y + Bsinωt will be also, since:

(y + Bsinωt)'' + ω²(y + Bsinωt)

= y'' + ω²y + (Bsinωt)'' + ω²Bsinωt

= f(t) + 0

 

[ppy]

thanks for that.
So when working out the particular integral the odd or even-ness of the function affects the Fourier series ? But for the complementary function it has no effect ?

 

[tiny-tim]

that's right

you can add any complementary solution to your particular solution, and still have a solution …

(that's fairly easy to prove quite generally, by the same method i used above)

 

[blue_raver22]

Thank you for your question. In regards to the example you provided, it is correct that the particular integral will only include cosine terms since the function on the right-hand side is even. As for the complementary function, it is determined by the homogeneous equation and can take on different forms depending on the initial conditions. Therefore, the constant B may or may not equal zero in all cases. It ultimately depends on the specific problem being solved.

In general, if the function on the right-hand side is odd, it will only contain sine terms in the particular integral. If it is even, it will only contain cosine terms. However, this does not necessarily imply anything about the general solution. The general solution will still depend on the particular solution and the complementary function, which can have different forms depending on the initial conditions. It is important to consider all aspects of the problem when determining the general solution. I hope this helps clarify your question.