Use Laplace to solve this differential equation

You're confusing the function x with the variable t. L[x] = L[x(t)] = X(s) while L[t] = 1/s. X(s), the Laplace transform of x(t), is what you're solving for.

 

Yup, looks right.

 

After you solve for X(s), you want to invert it to find x(t). Partial fractions will probably be useful to get X(s) into a form where you can invert it using a table.

 

The Laplace transform is a linear operator. That means if you have two transform pairs x(t), X(s) and y(t), Y(s), then L[ax(t)+by(t)] = aL[x(t)]+bL[y(t)] = aX(s) + bY(s). What you can't say is that x(t)y(t) corresponds to X(s)Y(s).

You want to use partial fractions to split your X(s) up into a sum of terms you can invert using the tables.
$$X(s) = \frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$

 

It looks wrong. The L.T. of a convolution is the product of the individual L.Ts, so the inverse of a product would be a convolution, not a product. You could, of course, actually carry out the convolution operation to get the correct answer.

Just use partial fractions; it is easier, and it avoids convolutions.

 

Multiply both sides by ##s^2(s^2+\omega^2)##. You can then set s=0 in the resulting equation. All terms except for the one with A will vanish, so you can solve for A. The rest of the coefficients will take a little more work. Note that with ##s=\omega##, you get ##s^2+\omega^2 = 2\omega^2##, not 0, so setting ##s=\omega## doesn't help you.

 

I meant multiply this equation
$$\frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$ by ##s^2(s^2+\omega^2)##, and then set s=0 to solve for A.

 

The other terms don't turn into 0. Why do you think they do?

 

Why are you making such heavy weather of this question? You just need the partial-fraction expansion of
[tex] \frac{1}{s^2(s^2+\omega^2)}.[/tex]
and surely you must have done lots of those in Calculus 101. If not, just Google 'partial fractions' to see lots of on-line sources showing you how to do it. It really is not that hard; it just takes a bit of practice.

 

Right. This is a lot different than saying that you get ##1=A(s^2+\omega^2)##. I hope you can see that. (If you're going to say you were just setting s=0 in the other terms, then I'd just ask why didn't you set s=0 in the first term then?)

To get the rest of the constants, you can collect the terms:
$$ 1 = (B+D)s^3 + (A+C)s^2 + (B\omega^2)s + A\omega^2$$ and then match coefficients on the two sides. You should eventually get B=D=0 and C=-1/ω².