1. ODE part ii)

Hi.
Im currently studying for a maths exam and im really stuck with part ii) of this question.
I have solved the homogenous equation i(x)=5cosx + 0sinx
would greatly appreciate someone walking me through part ii)

2. Re: ODE part ii)

Hello,

you need any solution for the inhomogeneous problem, because you have already solve the homogeneous ODE. For this reason an "ansatz" like $\displaystyle i(x) = e^{Ax}$ does it. In this case you should use Eulers formula for the sin.

Edit: I guess, $\displaystyle i(x) = e^{\i \omega x}$ will totally work.

3. Re: ODE part ii)

Sorry for sounding like a complete novice but im clueless on your method.

4. Re: ODE part ii)

He is talking about the "method of undetermined coefficients". If you are expected to be able to solve a problem like this, you must have been taught that. It might be the complex exponential that is confusing you. I would recommend something like "$\displaystyle Acos(\omega t)+ B sin(\omega t)$" rather tham $\displaystyle Ae^{i\omega t}$ because your equation uses "sin(t)" rather than a complex exponential.
(Of course, $\displaystyle e^{i\omega t}= cos(\omega t)+ sin(\omega t)$.)

For part iii, that won't work because, with $\displaystyle \omega^2= 400$, "$\displaystyle sin(\omega t)$" and "$\displaystyle cos(\omega t)$" are already solutions to the associated homogeneous equation. Do you remember what to do in that case?

5. Re: ODE part ii)

Originally Posted by HallsofIvy
He is talking about the "method of undetermined coefficients". If you are expected to be able to solve a problem like this, you must have been taught that. It might be the complex exponential that is confusing you. I would recommend something like "$\displaystyle Acos(\omega t)+ B sin(\omega t)$" rather tham $\displaystyle Ae^{i\omega t}$ because your equation uses "sin(t)" rather than a complex exponential.
(Of course, $\displaystyle e^{i\omega t}= cos(\omega t)+ sin(\omega t)$.)

For part iii, that won't work because, with $\displaystyle \omega^2= 400$, "$\displaystyle sin(\omega t)$" and "$\displaystyle cos(\omega t)$" are already solutions to the associated homogeneous equation. Do you remember what to do in that case?
Small correction:

$\displaystyle e^{i\omega t}=\cos(\omega t)+i\sin(\omega t).$