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Math Help - ODE part ii)

  1. #1
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    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)

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  2. #2
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    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 i(x) = e^{Ax} does it. In this case you should use Eulers formula for the sin.

    Edit: I guess, i(x) = e^{\i \omega x} will totally work.
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  3. #3
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    Re: ODE part ii)

    Sorry for sounding like a complete novice but im clueless on your method.
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  4. #4
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    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 " Acos(\omega t)+ B sin(\omega t)" rather tham Ae^{i\omega t} because your equation uses "sin(t)" rather than a complex exponential.
    (Of course, e^{i\omega t}= cos(\omega t)+ sin(\omega t).)

    For part iii, that won't work because, with \omega^2= 400, " sin(\omega t)" and " cos(\omega t)" are already solutions to the associated homogeneous equation. Do you remember what to do in that case?
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  5. #5
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    Re: ODE part ii)

    Quote Originally Posted by HallsofIvy View Post
    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 " Acos(\omega t)+ B sin(\omega t)" rather tham Ae^{i\omega t} because your equation uses "sin(t)" rather than a complex exponential.
    (Of course, e^{i\omega t}= cos(\omega t)+ sin(\omega t).)

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

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