Using Laplace Transform to Solve Differential Equations

[fluidistic]

Homework Statement

I'd like to solve a DE using Laplace transform.
$\ddot y (t)+\omega ^2 y(t)=f(t)$ for all t>0.
Initial conditions: $y(0)=\dot y (0)=0$. The dot denotes the derivative with respect to t.

Homework Equations

$\mathbb{L}( \dot y )=s\mathbb{L}( y )-y(0)$.
Convolution: if $h=f*g$ then $H(s)=G(s)F(s)$.

The Attempt at a Solution

I apply the LT on the DE: $s^2 Y(s)-sy(0)-\dot y(0)+\omega ^2 Y(s)=F(s)$. Therefore $Y(s)=\frac{F(s)}{s^2+\omega ^2 }$. Now I can use the convolution property: $y(t)=f(t)*\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )$.
Unfortunately I do not find the inverse Laplace transform of $\left ( \frac{1}{s^2+\omega ^2} \right )$ in any table.
So it means I must perform the integral $\frac{1}{2\pi i }\int _{\sigma -i \infty}^{\sigma + i \infty }e^{st}\frac{1}{s^2+\omega ^2 }ds$. Any help for this is appreciated.

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

 

[vela]

Homework Statement

I'd like to solve a DE using Laplace transform.
$\ddot y (t)+\omega ^2 y(t)=f(t)$ for all t>0.
Initial conditions: $y(0)=\dot y (0)=0$. The dot denotes the derivative with respect to t.

Homework Equations

$\mathbb{L}( \dot y )=s\mathbb{L}( y )-y(0)$.
Convolution: if $h=f*g$ then $H(s)=G(s)F(s)$.

The Attempt at a Solution

I apply the LT on the DE: $s^2 Y(s)-sy(0)-\dot y(0)+\omega ^2 Y(s)=F(s)$. Therefore $Y(s)=\frac{F(s)}{s^2+\omega ^2 }$. Now I can use the convolution property: $y(t)=f(t)*\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )$.
Unfortunately I do not find the inverse Laplace transform of $\left ( \frac{1}{s^2+\omega ^2} \right )$ in any table.

It is in the table. Look for $\frac{\omega}{s^2+\omega^2}$. Remember ω is just a constant here.

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

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

So what do you get when you sum those and simplify?

 

[fluidistic]

It is in the table. Look for $\frac{\omega}{s^2+\omega^2}$. Remember ω is just a constant here.

Ok thanks! I'd rather look when I'm done, just for the surprise of getting a good result.

So what do you get when you sum those and simplify?

What I had done is $\mathbb{L ^{-1}} \left ( \frac{1}{s^2+\omega ^2} \right )=\frac{1}{2\pi i }\int _{\sigma -i \infty}^{\sigma + i \infty }e^{st}\frac{1}{s^2+\omega ^2 }ds$. When performing the complex integral, I had that it's worth $2\pi i \sum _i res f(z)$. So that the $2 \pi i$'s terms cancel out. So that the inverse LT of $\frac{1}{s^2+\omega^2}$ should be just what you said, the sum of the residues. This gave me $\frac{i}{2\omega } (e^{i\omega t }-e^{-i \omega t })=\frac{i \sin (\omega t)}{\omega}$.

Therefore $y(t)=f(t)*\frac{i\sin (\omega t )}{\omega}=\int _0 ^t f(t-\tau )i \frac{\sin (\omega \tau)}{\omega}d\tau$.
Does this look possible? I'm going to check out the table.

Edit: I'm wrong it seems. I should have found $\frac{\sin (\omega t )}{\omega}$?

 

[vela]

Yeah, sine has an i in the denominator, so you should have just left it down there. Other than that extra factor of i, your results look good.

 

[fluidistic]

Yeah, sine has an i in the denominator, so you should have just left it down there. Other than that extra factor of i, your results look good.

Thank you vela, I found out the mistake.