Why is there a sin term in the particular solution for cosine equation?

[quietrain]

Homework Statement

Q'' + 100Q' +50000Q = 4000cos(100t)
i found the general solution to be e^-50t[Acos(50sqrt19)t +Bsin(50sqrt19)t]

but i have a problem with the particular solution

i tried Ce^i100t

did i try the wrong expression? because when i compared coefficients, i found C=4000/40000 = 0.1
but the answer given by my lecturer is 16/170 cos100t + 4/170sin100t

idon't understand where the sin term comes from? when i try Ce^i100t, i should be getting just the cosine terms right? since my target is 4000cos100t. so how come there's a sin term?

help appreciated!

 

[Mark44]

Homework Statement

Q'' + 100Q' +50000Q = 4000cos(100t)
i found the general solution to be e^-50t[Acos(50sqrt19)t +Bsin(50sqrt19)t]

No, this isn't the general solution. It looks more like the solution to the homogeneous problem, namely Q'' + 100Q' + 50000Q = 0. I didn't check your solution, but you can by making sure that it satisfies Q'' + 100Q' + 50000Q = 0.

BTW, the t factor should be inside the parentheses: e.g., Acos(50sqrt(19)t)

but i have a problem with the particular solution

i tried Ce^i100t

For reasons too long to go into here, I would recommend that you try a particular solution of y_p = Ccos(100t) + Dsin(100t).

did i try the wrong expression? because when i compared coefficients, i found C=4000/40000 = 0.1
but the answer given by my lecturer is 16/170 cos100t + 4/170sin100t

idon't understand where the sin term comes from? when i try Ce^i100t, i should be getting just the cosine terms right? since my target is 4000cos100t. so how come there's a sin term?

help appreciated!

 

[quietrain]

oh. i always thought that was called the general solution? where you let the characteristic equation = 0

oh my mistake, t should be in the brackets yes

thats the problem. i don't understand. in another question, y'' +4y = cos2x. i could try Z = Ae^2ix, so that the real part would just be cos2x and i discard the imaginary part

so why in this case i have to use Acosx + Bsinx ? also, how do i know when i have to use this instead of Z = Ae^2ix?

many thanks!

 

[Mark44]

The general solution consists of two parts: the solution to the homogeneous problem + the particular solution.

For your second question, you're making two mistakes.
1. In the homogeneous equation (y'' + 4y = 0), the characteristic equation has two solutions - r = 2i and r = -2i. Rather than have a homogeneous solution of y = Ae^2ix + Be^-2ix, it's much easier to use y = Acos(2x) + Bsin(2x).
2. Since the nonhomogeneous equation is y'' + 4y = cos(2x), a particular solution involving e^2ix or e^-2ix isn't going to work. When a function is a solution of the homogeneous problem, it can't also be a solution of the corresponding nonhomogeneous problem.

For your particular solution, try y = Cxcos(2x) + Dxsin(2x).

 

[quietrain]

ah i see... so as long as i see cosx or sinx, i should use Ccosx+Dsinx? when i see tan i should use method of variation of parameters?

i went to look through my lecturer's notes again, and i realize what he did was to introduce a imaginary part in the exact same format as the real part

i.e , if it was y" + y' + 4y = cosx
he introduced a iv" + iv' +4iv = isinx

then he combine to get z = y + iv. so he says y = Re(Z).
and he trys e^{ix or something like this}

so his method would work also?

but if i use Ccosx+Dsinx, i don't have to worry about imaginary or real parts right?

 

[Mark44]

Yes, both methods work, but if you use the method I described, you don't have to be concerned with the imaginary parts.

 

[quietrain]

ah i see! thanks