Diff. Eq's Help?

[PhillDubya]

Problem:
f(t)=10D x(t) + 3 x(t)
with f(t) = 2sin8t, and
x(0) = 7

D≡d/dt

These are my equations. I understand how to solve each individually but I have never been given both at the same time. Should I solve my first f(t)=10D x(t) + 3 x(t) first, and then somehow substitute in the second f(t)= 2sin8t?

Then, when I have a general solution, get my particular solution by using the x(0) = 7?

The second part: f(t)= 2sin8t is confusing me.

Your time is greatly appreciated.

 

[MrAl]

Hi Phill,

You say you are a little confused by f(t)=2 sin (8t), but what would make you
think that shouldnt replace f(t) with 2 sin (8t) and then solve?
What gives you the idea that you *wouldnt* want to do this?
Do you need to approach the problem this way for some reason,
or is it that you are looking for a simpler way to solve the entire equation?

Also, what is stopping you from trying to use BOTH ways of doing it?

 

[PhillDubya]

Well, after looking at it more...

I was in a hurry and a little rusty and initially assumed that it was a typical first order homogeneous differential equation. However it is actually (and as you have stated):
inhomogeneous with both a y(h) + y(p) solution.

f(t)=10D x(t) + 3 x(t), with f(t) = 2sin8t, and x(0) = 7

2 sin 8t = 10 Dx(t) + 3x(t) = 10 d/dt x(t) + 3x(t) =
(10 d/dt + 3) x(t)
with the characteristic polynomial equal to: (10m + 3) = 0, and m=-3.
But there is where I am confused, do you go ahead and integrate both sides, now that you have your char. eq. solved?






This is where I went wrong originally..
Ok, this is long, so bare with me here please.

See, I had originally planed (this is where I got confused), to solve the initial equation: f(t)=10D x(t) + 3 x(t), by using the form: dy/dx + P(x)y = Q(x). Well there would be no y term, or so it seemed, however, I get to thinking and talk myself into pushing methods. So I was using 3 as my P(x), and got: e^∫3dx =IF, and used that.

Well then I ended up with:

e^3t 2sin8t = e^3t(10 d/dt x(t) + 3 x(t))

∫e^3t 2sin8t = ∫e^3t(10 d/dt x(t) + 3 x(t))

and then ended up WAAAAAY out in left field somewhere.

 

[speyman]

Hi Phil,
I have a tool for you: the Fourier transform.
First, the factoring out of x(t) away from d/dt cannot be done.

Set your forcing function, f(t), to the impulse delta function. F(jw)=1.
Next take the fourier transform of the left-hand side by consulting your fourier tables.
Solve for X(jw) and you have a transfer function.
At this point, you've done the hard part.
The output will be a sinusoid at the same frequency as the input. Multiply your transfer function by the amplitude of the forcing function and evaluate at w=8rad/s. This will give you a complex number. Convert the complex number to a phasor, and that is the amplitude and phase of the sinusoidal output at a frequency of 8 rad/s.

What makes this tool powerful is that it is very fast and we can use phasors instead of the fourier transform of the sinusoid based on the absolute fact that the system maintains sinusoidal fidelity. (output frequency=input frequency)

Now, to evaluate at the initial condition, use laplace. Now you have a causal system, whereas before, you had a sinusoid going forever and you just stepped into the middle to analyze it. Now, you are starting at t=0. Its basically the same drill, use laplace transform tables, probably even easier than the fourier. Here you will get one or more decaying exponentials.

Hope this helps.

 

[MrAl]

Hi again,


I dont want to start posting 'answers' just yet, but i did want to
put the equation into some different form and see if it helps you
with a direct solution...

The original:

f(t)=10D x(t) + 3 x(t) , with f(t)=2*sin(8t), and D stands for: d/dt

first simply replacing all the x(t) with y:
f(t)=10 dy/dt + 3 y , and with f(t)=2*sin(8t)

now with f(t) set equal to 2sin(8t):

2*sin(8t)=10 dy/dt + 3 y

or leave it alone and put into 'IF' form:

dy/dt + 0.3*y = 0.1*f(t)

Now the equation is in 'IF' form so i would bet you can solve it now
as it's just a matter of following the IF procedure.


Just curious if you have ever tried a numerical solution to these kinds
of problems? The only reason i ask is because a numerical solution
i think shows the basic workings of this kind of system better than
a purely procedural approach, even though both are informative.
Also, numerical techniques are usually the only recourse once
the system order gets higher and higher.

 

[PhillDubya]

Alright MrAl, maybe I got this completed. Sorry it took so long, I am certainly no mathematician.

You got me thinking with not sticking to one method...
This particular equation apparently requires Laplacian transforms.

I tried working it using the: dy/dx + P(x) = Q(x) method with an IF, and ended up with: 0.1e^0.3x + C. Which that may not even be the correct answer to "that" method, but either way, I didn't have NEARlY enough constants in the answer so I stopped there.

Ok, here is what I came up with using a Laplace transform:

2sin8t = 10Dx(t) + 3Dx(t), x(0) =7

divide all by 10, to isolate the highest order derivative operator and get: 1/5sin 8t = Dx(t) + 3/10x(t)

*L = Laplace operator

L{1/5 sin8t} = L{d/dt x(t)} + L{3/10 x(t)}

1/5 (8/s² + 8²) = sX(s) - x(0) + 3/10 X(s)

1/5 (8/s² + 8²) = sX(s) - 7 + 3/10 X(s)

1/5 (8/s² + 8²) + 7 = sX(s) + 3/10X(s) = (s+3/10)X(s)


X(s) = 1/5 (8/s² + 8²) + 7 / (s+3/10) = (8/5) / ((s^2 + 8^2)(s+3/10)) + 7/(s+3/10)

Now inverse Laplace:

L^-1 {X(s)} = L^-1{ (8/5) ÷ ((s² + 8²)(s+3/10)) + 7/(s+3/10) }

where: (s² + 8²) = (s+j8)(s-j8)

and the poles are: (s+j8)(s-j8)(s+3/10)

Part. Frac. Decomp.:

A1/(s+j8) + A2/(s-j8) + A3/(s+3/10) + B/(s+3/10) =
(8/5) ÷ (s+j8)(s-j8)(s+3/10)


Do a lot a canceling and setting s = N, where N equals a number to cancel your 2 other A's, you know the drill......

Do Laplace again to get back to the time domain for every A(n) and then finally for your B:

ANSWER ===>
x(t) = (8/5) [ A1 (e^-j8t x e^-3/10t) + A2(e^j8t x e^3/10t) + A3(e^-j8t x e^j8t) + 7e^3/10t)]

x(t) = (8/5) [ A1 (e^-j8t x e^-3/10t) + A2(e^j8t x e^3/10t) + (A3 + 7e^3/10t)]

Blue = constant sin f(x), and red = transient which eventually goes away.


FINALLY!


Thanks MrAl, and Speyman for the help!

 

[MrAl]

Hi again,

I didnt check your answer and it is good that you used Laplace too,
(and so i assume you got it right)
but you really should do this problem using the integrating factor
too because it works out pretty simple that way. If you didnt get
it right the first time it's probably just a slight oversight in one little
part of the procedure. If you were able to get so far with the other
method then you should be able to breeze through the IF method.
This is just a first order DE anyway.

 

[Sceadwian]

English only in the forums please =)