Alright... I can show how I would solve this w/o using Laplace transforms. There might be a more streamlined way to do it, but if there is, someone else will have to show you

Here goes...

First we'll create what I think is called the "characteristic polynomial" for the differential equation. Basically what that means is that we're going to neglect the right hand side (300) for now, and work on the left hand side. By neglecting the right hand side, we're dealing with something called a homogenous differential equation.

Now, to start, we'll replace all d^n(q)/dt^n with m^n... so the equation becomes

m^2+8m+25 = 0

Now, solve for m, and you get -4 ± 3j (use quadratic formula, or calculator, or whatever...)

What this means is that we've got complex roots of the homogenous equation. The solution to a differential equation is usually in the form

C*exp(r1*t) + D*exp(r2*t)

where C and D are constants, and r1 and r2 are the roots of the homogenous equation. Because we have complex roots, we have to apply Euler's formula, which states something like

exp(j*K) = cos(K) + j*sin(K)

You can go through the algebra, but what this ends up boiling down to for our purposes is something that looks like this

C*exp(r*t)*cos(k*t) + D*exp(r*t)*sin(k*t)

where "r" is the real part of the complex root, and "k" is the imaginary part, so right now we've got

q(t)_homogenous = exp(-4*t)*[D*cos(3*t)+E*sin(3*t)]

Now let's deal with the right hand side of the equation that we neglected earlier. Because it is just a number, there's a bit of a trick that you can use. Essentially the right hand side is describing the "input" to the system... which means that after the transient behavior of the system dies off (the differential terms) you're left with the "input" and the linear terms of q(t). What this means in terms of our equation is that

25*q(t) = 300

and if you solve for q(t) you get q(t) = 12, which we will call our "particular" solution... because it is particular to the "input" or something like that. To get q(t) we'll add the homogenous and particular solutions

Ok, now we're basically done, we've got

q(t) = exp(-4*t)*[D*cos(3*t)+E*sin(3*t)] + 12

Now it's just a matter of plugging in initial conditions (you need (order of differential equation - 1) initial conditions) and solving for the constants, D and E.

If the initial conditions were

q'(0) = 0 and q(0) = 0

You should end up with

q(t) = exp(-4*t)*[-12*cos(3*t) - 16*sin(3*t)] + 12

I'll spare you the algebra, I'm sure you can do it (assuming you can take the derivative of q(t).

So I think that's it, if something's unclear, which I'm sure there's at least a few things, post back and I'll see if I can try to clear it up. Enjoy!

~Derek