Fourier transform of translated Dirac Delta.

Question

Solve: $$ y'' + 6y' +5y = \delta(t-3)$$ using Fourier's Transform.

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$$\mathcal{F}\{y'' + 6y' +5y\} = \mathcal{F}\{\delta(t-3)\}$$

$$ \mathcal{F}\{y\}\left[(iw)^2 + 6iw + 5\right]= \mathcal{F}\{\delta(t-3)\}$$

$$\mathcal{F}\{y\} = \frac{\mathcal{F}\{\delta(t-3)\}}{(iw + 5)(iw + 1)}$$

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I know that ${F}\{\delta(t)\}=1$, what happens with ${F}\{\delta(t-a)\}$?

Answer 1

$$ \frac{d^2y(t)}{dt^2} + 6\frac{dy(t)}{dt} +5y(t) = \delta(t-3)$$ Of course, one can compute the $\mathcal{F}\{\delta(t-3)\}$ from the definition of the Fourier transform.

But, knowing $\mathcal{F}\{\delta(t)\}$ the solution is straightforward with the change of variable : $$T=t-3\qquad\to\qquad \frac{d^2y(T)}{dT^2} + 6\frac{dy(T)}{dT} +5y(T) = \delta(T)$$