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.