Help solving differential equation using Laplace Transforms

[abc617]

In this engineering course I'm taking I'm learning how to solve Laplace transforms. Now my ODE class only barely touched on this so I'm not sure if I'm doing this correct so I'd like some help pointing out where I went wrong, if I went wrong anywhere.


$$x''(t) + 3 x'(t) + 2x(t) = 0, x(0) = 2, x'(0) = -2$$

My Steps:
-Convert everything into its Laplace Transform
$$(s^2X - sx(0) - x'(0)) + 3(s'X - x(0)) + /2X) = 0$$

-Replace with initial values
$$(s^2X - 2s - (-2) ) + 3(s'X - 2) + 2X) = 0$$

-Distribute
$$s^2X - 2s + 2 + 3s'X - 6 + 2X = 0$$

-Add like terms
$$s^2X - 2s + 3s'X +2X - 4 = 0$$

-Group the s terms, then simplify
$$X(s^2 - 3s' + 2) - 2s - 4 = 0$$

-Isolate $$X$$
$$X = \frac{2s + 4}{s^2 - 3s' + 2}$$


Now I'm not sure if I'm done since I missed one lecture. I'd appreciate it if someone could check over and point over/highlight any errors I made. Please and thank you!

 

[Mark44]

There shouldn't be any s' terms.

Also, there is an error in the next to last line. You should have X(s² + 3s + 2)

With this correction, your last equation can be simplified:
$$X = \frac{2s + 4}{s^2 + 3s + 2} = \frac{2(s + 2)}{(s + 2)(s + 1)}$$

...
Finally, take the inverse Laplace transform to find x(t).

 

[abc617]

Thanks, apparently I put in $$s^1$$ and mistook it for $$s'$$

Alright so I rewrote
$$X = \frac{2s + 4}{s^2 + 3s + 2} = \frac{2(s + 2)}{(s + 2)(s + 1)}$$

as

$$X = \frac{2}{(s+2)(s+1)} * \frac{s+2}{(s+2)(s+1)}$$

-I was then able to cancel $$s+2$$ in the 2nd term.

-Then by using a Laplace transform table, I got:
$$x(t) = 2(e^{-t}- e^{-2t}) (e^{-t})$$

 

[Mark44]

Thanks, apparently I put in $$s^1$$ and mistook it for $$s'$$

Alright so I rewrote
$$X = \frac{2s + 4}{s^2 + 3s + 2} = \frac{2(s + 2)}{(s + 2)(s + 1)}$$

Why not just cancel the s + 2 factors? Wouldn't that be simpler? You'll get X(s) = 2/(s + 1). Finding the inverse Laplace will be pretty easy from that.

as

$$X = \frac{2}{(s+2)(s+1)} * \frac{s+2}{(s+2)(s+1)}$$

-I was then able to cancel $$s+2$$ in the 2nd term.

-Then by using a Laplace transform table, I got:
$$x(t) = 2(e^{-t}- e^{-2t}) (e^{-t})$$

I don't think this is right, but I didn't take the time to check. You can check by verifying that this function satisfies the initial conditions and the differential equation.

 

[abc617]

Alright so factoring out and canceling:

$$X = \frac{2s + 4}{s^2 + 3s + 2} = \frac{2(s + 2)}{(s + 2)(s + 1)} = 2\frac{1}{s+1}$$

So then I would take the inverse Laplace transform of
$$2\frac{1}{s+1}$$

To get:
$$x(t) = e^{-t}$$

 

[Mark44]

It's easy enough to check whether your solution is correct. As I mentioned before, your solution has to satisfy the initial conditions (it doesn't) and the differential equation (it doesn't).

$$\mathcal{L}^{-1}(\frac{1}{s + 1}) = e^{-t}$$

but you have 2/(s + 1), so ...?

 

[abc617]

Oh, I forgot the 2, so it should've been

$$x(t) = 2e^{-t}$$?

 

[Mark44]

Why the question mark? Are you asking me whether this is the solution?

Check it for yourself.