(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'd like to solve a DE using Laplace transform.

[itex]\ddot y (t)+\omega ^2 y(t)=f(t)[/itex] for all t>0.

Initial conditions: [itex]y(0)=\dot y (0)=0[/itex]. The dot denotes the derivative with respect to t.

2. Relevant equations

[itex]\mathbb{L}( \dot y )=s\mathbb{L}( y )-y(0)[/itex].

Convolution: if [itex]h=f*g[/itex] then [itex]H(s)=G(s)F(s)[/itex].

3. The attempt at a solution

I apply the LT on the DE: [itex]s^2 Y(s)-sy(0)-\dot y(0)+\omega ^2 Y(s)=F(s)[/itex]. Therefore [itex]Y(s)=\frac{F(s)}{s^2+\omega ^2 }[/itex]. Now I can use the convolution property: [itex]y(t)=f(t)*\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )[/itex].

Unfortunately I do not find the inverse Laplace transform of [itex]\left ( \frac{1}{s^2+\omega ^2} \right )[/itex] in any table.

So it means I must perform the integral [itex]\frac{1}{2\pi i }\int _{\sigma -i \infty}^{\sigma + i \infty }e^{st}\frac{1}{s^2+\omega ^2 }ds[/itex]. Any help for this is appreciated.

Edit: Let [itex]f(z)=\frac{e^{zt}}{z^2+\omega ^2}[/itex]. I need to employ the residue theorem. The denominator can be rewritten into [itex](z-i \omega )(z+i \omega)[/itex]. So the residue at [itex]z=i\omega[/itex] is worth [itex]\lim _{z \to i \omega } \frac{e^{zt}}{z+i \omega }=\frac{e^{i \omega t }}{2i \omega}[/itex]. While the residue at [itex]z=-i \omega[/itex] is worth [itex]-\frac{e^{-i \omega t}}{2i \omega }[/itex].

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# Laplace transform on a DE

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