Using Fourier transform to solve ODE

[progrocklover]

Homework Statement

Homework Equations

The Attempt at a Solution

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 haven't use the boundary conditions yet.
Please gives me some idea on this.

 

[gomunkul51]

Are you sure you need to solve the ODE by Fourier transform and not by Laplace transform?

 

[klondike]

However, for this question I am not sure whether should I stop here as I noticed that I haven't use the boundary conditions yet.
Please gives me some idea on this.

Your are taking convolution of h(t) and f(t) to find the particular integral of the ODE. The impulse response h(t) -- you can think of a Green's function of the system. In the process of finding h(t) from H(j\omega), you're using Fourier and inverse Fourier transformations. It requires the boundary value of the GF vanishes at infinity such that the FT converges.

 

[progrocklover]

Your are taking convolution of h(t) and f(t) to find the particular integral of the ODE. The impulse response h(t) -- you can think of a Green's function of the system. In the process of finding h(t) from H(j\omega), you're using Fourier and inverse Fourier transformations. It requires the boundary value of the GF vanishes at infinity such that the FT converges.

I don't quite understand what you mean. I know the green's function is
2je^(-2(t-k)) - 3je^(-3(t-k)), but I have no idea of how to utilize the boundary condition.

 

[hunt_mat]

Just take the Fourier transform to obtain:

<br /> (-k^{2}+5ik+6)\hat{y}(\xi})=\hat{f}(\xi)<br />

Then re-arrange and take the inger Fourier transform to obtain:

<br /> y(x)=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{\hat{f}(\xi )e^{ix\xi}}{-k^{2}+5ik+6}d\xi<br />

This will be your particular solution, the complete solution will be given by:

<br /> y(x)=Ae^{-2x}+Be^{-3x}+\frac{1}{2\pi}\int_{\mathbb{R}}\frac{\hat{f}(\xi )e^{ix\xi}}{-k^{2}+5ik+6}d\xi<br />

You may now use the boundary conditions to obtain A and B.