How to Solve a Differential Equation Using Laplace Transforms?

[Saladsamurai]

Homework Statement

Given:

dx/dt + 3*x = exp(-3*t) and all initial conditions are zero.

Homework Equations

Laplace

The Attempt at a Solution

L[dx/dt + 3*x = exp(-3*t)]

s*X(s) + 3*X(s) = 1 / (s + 3)

X(s) = 1 / (s + 3)^2

So here is where I get mixed up. For some reason, I thought I was supposed to use a partial fraction expansion here. But those of you who know better will probably get a good chuckle out of hearing about how I did do the partial fraction expansion only to re-discover that

X(s) = 1 / (s + 3)^2

So...2 questions:

1) Do you know why I thought I needed a PFE?

2) I have a list of Laplaces and their inverses. 1 / (s + a)^n is NOT one of them.

1 / (s + a) IS one of them.

I presume I am supposed to use this rule in conjunction with some other rule to find the inverse Laplace of 1 / (s + a)^n .
Can I get a hint on how to do this?

 

[rock.freak667]

If you had

X(s)=1/s², you can easily get x(t)

So you'd need to apply the shift property.

You probably thought 1/(s+3)(s+3) would have given you something simpler, but it was already in its simplest form. Don't worry, I've done that exacts same thing.

 

[Saladsamurai]

I will have to Google the "shift theory." This is for a Controls class, so the treatment of the mathematics is really brief and incomplete. So...I'll be back in a moment.

EDIT:So the shift theory is just

L^-1[F(s − a)] = exp(at)*f(t)

I am not too sure how to use this. This is going to take a while Okay. I am looking at one of my old texts. I have that:

$$L[t*u(t)] = \frac{1}{s^2}$$

where u(t) is the unit step function. What does that really mean? Is that just a fancy name for '1' ?

I also have the Frequency Shift Theorem :

$$L[e^{-at}f(t)] = F(s + a)$$

so...

 

[rock.freak667]

So the shift theory is just

L^-1[F(s − a)] = exp(at)*f(t)

I am not too sure how to use this. This is going to take a while

Say you need to get

$$L^{-1} \left( \frac{3}{(s-1)^2+9} \right)$$

You easily know that


$$L^{-1}\left( \frac{3}{s^2+9} \right)=sin3t$$

So to get the require inverse transform, you need to shift the 's' to 's-1', and to do this you multiply the usual transform by e^1t.

 

[Saladsamurai]

Say you need to get

$$L^{-1} \left( \frac{3}{(s-1)^2+9} \right)$$

You easily know that$$L^{-1}\left( \frac{3}{s^2+9} \right)=sin3t$$

So to get the require inverse transform, you need to shift the 's' to 's-1', and to do this you multiply the usual transform by e^1t.

Hmmm...so it's simply

$$L^{-1} \left( \frac{3}{(s-1)^2+9} \right) = e^t*\sin(3t)$$

?

 

[Saladsamurai]

So for my particular case:

F(s) = 1 / s^2 --> f(t) = t

and

F(s+a) = 1 / (s+3)^2 --> f(t) = exp(-3*t)*t

?

 

[rock.freak667]

Hmmm...so it's simply

$$L^{-1} \left( \frac{3}{(s-1)^2+9} \right) = e^t*\sin(3t)$$

?

Yep, the F(s-a) is your 3/((s-1)²+9), your f(t)=sin(3t) and your shift is -1, so it is e^t