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Math Help - using Fourier transform to solve ODE

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    using Fourier transform to solve ODE

    using Fourier transform to solve ODE-0odeft.jpg
    For part i) I got the answer 1/((jw)^2 + 5jw +6)

    For part ii)
    I first consider input to be a unit impulse

    Thus, Y(w)=H(w)F(w) and F(w)=1
    yI(t)=-1/2pi integrate from -infinity to infinity (e^jwt)/(w^2 - 5jw - 6) dw
    where yI(t) is the output when f(t) is a unit impulse
    using complex contour integration,
    I got yI(t) = 0 for t<0
    and yI(t) = 2je^(-2t) - 3je^(-3t) for t>0
    Then using y(t)=yI(t)*f(t)
    I got y(t) = integrate from -infinity to infinity ( 2je^(-2(t-k)) - 3je^(-3(t-k)) )f(k)dk

    For a similar example in my notes, it just stops at this step.

    However, for this question I am not sure whether should I stop here as I noticed that I havent use the boundary conditions yet.
    Please gives me some idea on this.
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  2. #2
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    Quote Originally Posted by progrocklover View Post
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    For part i) I got the answer 1/((jw)^2 + 5jw +6)

    For part ii)
    I first consider input to be a unit impulse

    Thus, Y(w)=H(w)F(w) and F(w)=1
    yI(t)=-1/2pi integrate from -infinity to infinity (e^jwt)/(w^2 - 5jw - 6) dw
    where yI(t) is the output when f(t) is a unit impulse
    using complex contour integration,
    I got yI(t) = 0 for t<0
    and yI(t) = 2je^(-2t) - 3je^(-3t) for t>0
    Then using y(t)=yI(t)*f(t)
    I got y(t) = integrate from -infinity to infinity ( 2je^(-2(t-k)) - 3je^(-3(t-k)) )f(k)dk

    For a similar example in my notes, it just stops at this step.

    However, for this question I am not sure whether should I stop here as I noticed that I havent use the boundary conditions yet.
    Please gives me some idea on this.
    I agree with your Fourier transform this gives that

    \displaystyle [(jw)^2+5(jw)+6]\hat{y}=\hat{f}(w) \iff (jw+2)(jw+3)\hat{y}=\hat{f}(w)


    Now solving for \hat{y} and using partial fractions I get

    \displaystyle \hat{y}=\frac{\hat{f}(w)}{jw+2}-\frac{\hat{f}(w)}{jw+3}

    Since we don't know what f is we will have to make some assumptions later.

    Since we have the product of Fourier transforms we will use the convolution theorem

    Note that

    \displaystyle \mathcal{F}^{-1}\left( \frac{1}{a+jw}\right)=e^{-ax}u(x)

    Where u(x)=\begin{cases}1, \text{ if } x> 0 \\ 0, \text{ if } x< 0\end{cases}

    The Heaviside step function.

    Now using the convolution theorem we get that

    \displaystyle y=\int_{-\infty}^{\infty}f(\tau)e^{-2(x-\tau)}u(x-\tau)d\tau-\int_{-\infty}^{\infty}f(\tau)e^{-3(x-\tau)}u(x-\tau)d\tau

    So we will need to make assumptions on f so the above integrals converge.
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