Math Is Fun Forum

For the first problem, I think the one I was missing was

[0,1:0,0:0,0]

Anyone have thoughts on the second problem?

~Derek

So far as having the RHS not equal to a constant....

There are general methods for solving the equations depending on the type of function on the right hand side.  For instance... if you've got

left hand side = cos(t)

your particular solution would look something like

A*cos(t) + B*sin(t)

I don't know if you have access to a Diff.Eq. book, but there should be a list of the different things that you can end up with depending on the right hand side.  Another common one would be something like

left hand side = A*t^n + ... +B*t + C

then the particular solution would be in the form

D*t^(n-1) + ... + E*t + F

fill in the '...' with the descending powers of t

as you can see, the more complicated the right hand side, the more constants you have to solve for in the end (using the initial conditions).

And college, well... I'm in my senior (4th) year at the University of Michigan majoring in Mechanical Engineering.  Anyone out there looking for an intern for next summer???

~Derek

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