does anyone have the notes for type 2 problems in hongs diff eq class. i don't have the notes for that section or if you have a good example problem that would be great. thanks

10/24/2005 6:56:46 PM

L(cos(Bt)) = s/(s^2 + B^2)

L(sin(Bt)) = B/(s^2 + B^2)

L(e^(at)*cos(Bt)) = (s - a)/((s-a)^2 + B^2)

L(e^(at)*sin(Bt)) = B/((s-a)^2 + B^2)

when it comes to factoring, if it has a non-real root, don't factor. instead carry it out and use

Ae^... + Be^... + C(s/(s^2 + B^2)) + D(B/(s^2 + B^2))

which as you can see are the Laplace transformations.

if you need any more help, just ask

[Edited on October 24, 2005 at 7:57 PM. Reason : use the same concept of not factoring for later types as well]

10/24/2005 7:57:03 PM

ok, i started problem 1 of type 2 and got stuck, honestly i'm sure if i'm on the right track or not so here is where i'm at.....

1/((s+3)(s-2)) + 2s/(s^2+16)

I don't know if this is right or not but if i am on the right track can you help me out with where to go next. thanks.

10/24/2005 8:46:47 PM

where did that plus come from? you're trying to get Y by itself, so you factor out the s terms on the left side as you did, but you're supposed to divide them so they end up on the right side of the equation... much like you did in Type 1 problems.....

edit: actually, I think I know where you're confused. You solve the equation the same way you would Type 1, but instead of factoring everything down before decomposing the fraction, don't factor the ones with the non-real roots.

[Edited on October 24, 2005 at 10:16 PM. Reason : .]

[Edited on October 24, 2005 at 10:19 PM. Reason : .]

10/24/2005 10:16:18 PM

ok so i've got it to this point.....

Y(s)= 2/(s^2+s-6)(s^2+16) so what is getting me know is breaking down the s^2 +16 part so i can solve for my coefficients. from that point i know i go back and do and inverse laplace on it to get y=........

10/24/2005 11:26:20 PM

... you don't break it down. You decompose the fraction as:

A(1/s..) + B(1/s..) + C(s/(s^2 + B^2)) + D(B/(s^2 + B^2))

solve for the coefficients (you'll have to use your answers to solve for the others in some cases,) then invert the transformation using

L(cos(Bt)) = s/(s^2 + B^2)

L(sin(Bt)) = B/(s^2 + B^2)

Look at the answers and see how they compare.


(sorry, I just realized I messed up the A and B part in my first post, but even so)

10/24/2005 11:54:31 PM

I'll help, I am pulling an all nighter for this one.

(OK, I made a bunch of mistakes by not doing my HW ahead of time, now, with no memory of how to do 3/7 of those methods, I need to review them really damned well, plus I have laundry and some stuff to clean in the kitchen, so it's excusable... *cough* back to me helping you)

You should be able to break that down to ( A/(S+3) + B/(S-2) + C*S/(S^2+4^2) + D*4/(S^2+4^2) ) = 2/(S^2+S-6)(S^2+16). I think, double check.

REMEMBER THOUGH, as this is what was holding me up, that the S in C*S/(S^2+B^2) + D*B/(S^2+B^2) can equal "(S-3)" or something. Kinda important when you get to those ones where you've got y'' + 4y' + 4y = e^(-3t) * cos(4t) problems... I'll help you if I can. I'm trying to memorize this stuff too.

[Edited on October 24, 2005 at 11:57 PM. Reason : slow... d'oh :p]

10/24/2005 11:57:19 PM

sounds good. i'll be up a while myslef. I can do some of these but type 2 is holding me up, specifically this prob because i don't have the notes for this part. it is slowly coming togather but i think i will need some more assitance. let me mess with this some more for a few and i'll ghet back to you. which parts were you having trouble with, i may be able to help you with those.

10/25/2005 12:07:40 AM

It's cool, I broke through it. It was pissing me off but I figured it out, heh.

I REALLY wish I had a textbook for this class (that followed with the class).

It would've made this a bit easier (requiring less attendance and more out of class work).

10/25/2005 12:13:44 AM

you're going to stay up? awesome. I want to sleep. I do plan on studying more in the morning since I work better then. 'night guys. Good luck tomorrow.

10/25/2005 12:25:09 AM

Yeah, I'll stay up.

Sure as hell hope I don't regret this tomorrow... I will...

It's cool, as long as I cover a B on the test.

10/25/2005 12:27:04 AM

ok, what you have from your first post, ( A/(S+3) + B/(S-2) + C*S/(S^2+4^2) + D*4/(S^2+4^2) ) = 2/(S^2+S-6)(S^2+16), is what i have now where do i take it from there as far as the algebra goes to get my coefficients right?

10/25/2005 12:45:21 AM

Gotta multiply both sides by (S+3)(S-2)(S^2+16), that'll make it so you can plug in different values of S to allow you to solve for A,B,C,D.

10/25/2005 12:50:40 AM

I have been all over it all night, yet i was missing something small. ok say for my B coefficient, i have:
s=-3: B(s-2)(s^2+16) ; i get the 125 on the bottom but am i supposed to multiply the 2 by 's' as well since it with the 'c' coefficient?

10/25/2005 12:57:24 AM

while im at it this is what i have to solve my coefficients:

2=A(s-2)(s^2+16) s=-3
2=B(s+3)(s^2+16) s=2
2=Cs(s+3)(s-2) s=???
2=D*4(s+3)(s-2) s=???

What am i missing in these, and what values should i use fo rs in the last two since s is squared?

10/25/2005 1:03:45 AM

Uh, I think something's badly wrong in your calculations.

After you multiply in (S+3)(S-2)(S^2+16) it should look like this:

2 = A*(S-2)*(S^2+16) + B*(S+3)*(S^2+16) + C*S*(S+3)*(S-2) + D*4*(S+3)*(S-2)

You'd solve from there.

If you plug in s=-3, you'll have:

2 = A*(-5)*25 + 0 + 0 + 0

A = -2/125

However, I'm starting to think there's another error in the calculations somewhere... yeah! yeah OK that "2" is supposed to be "2*s"... How'd that slip out? Anyway, yeah...

2*S = A*(S-2)*(S^2+16) + B*(S+3)*(S^2+16) + C*S*(S+3)*(S-2) + D*4*(S+3)*(S-2)

S=-3

-6 = A * (-5) * 25

A = 6/125

Boom, just had an error somewhere such that the S went missing.

10/25/2005 1:04:30 AM

For D, S=0.

For C, plug in some arbitrary value for S, say, S=1. Then plug in the values you have so far for A,B,D. Then figure out what C is.

[Edited on October 25, 2005 at 1:05 AM. Reason : I'm starting to wonder how the fuck we're going to do all these calculations in 7 minutes per prob.]

10/25/2005 1:05:13 AM

well at 7 mins per prob we will be done in 50 mins, but the last test was easy and it took me all of class plus some to get 7 of those done. ok i was on the right track there, i'm gonna play with these for a min and i'll get back atcha when i move on up to the next one. thanks.

10/25/2005 1:09:37 AM

for "d" when i plug in 0 for s im not gettin the 2/125, or -11/125, which ever case it should be....

2*s+d*4(s-2)(s+3) that's my equation.

10/25/2005 1:26:43 AM

actually the equationm says 2*s=d.....

10/25/2005 1:27:16 AM

You forget! The only the crap following C will be zero on this one. You'll still have A * -2 * 16 + B * 3 * 16 added onto what you have there.

Just plug in what you have for A and B and you'll be able to find D.

(I made that same mistake, it was holding me back too. You're on the right track.)

[Edited on October 25, 2005 at 1:36 AM. Reason : (The only the crap? I lose...)]

10/25/2005 1:33:48 AM

ok now on to #2 of type 2

10/25/2005 2:09:41 AM

Still making it? Cool.

10/25/2005 2:10:32 AM

im hangin in there, ill get it before dawn.http://brentroad.com/images/rolleyes.gif

10/25/2005 2:11:47 AM

so here i am again.

2/((s-3)^2+4)(s-4)(s-1).....when i break it down into A/... + B/....; where does this 4 come into play.

10/25/2005 2:20:40 AM

You can't factor ((S-3)^2+4), you'll have to do that C(s/(s^2 + B^2)) + D(B/(s^2 + B^2)) stuff, where s = (S-3) and B = 2.

10/25/2005 2:24:59 AM

ok, now where did you get that from?

10/25/2005 2:27:52 AM

From the last problem you did.

Your "s" in that problem has to be the (s-3) part of that 1/((S-3)^2+2^2) mess, and the "B" in that problem has to be the "2" part of that mess. Use that C(s/(s^2 + B^2)) + D(B/(s^2 + B^2)) stuff for whenever you do the laplacian transform of sin or cos times e ^ something. (Yeah, that's a bit vague, but that's when you're supposed to use it. )

10/25/2005 2:31:59 AM

so i've got this here:

Ae^4t+Be^t+C((s-3)/(s-3)^2+4)+D(2/(s-3)^2+4) =2

this is what the girl who was helping me earlier told me, i'm probably way off...

but if i am right how do i go about solving for a and b. then i just solve for d and c the same way i did last prob right.

10/25/2005 2:39:02 AM

Quote :

"... you don't break it down. You decompose the fraction as: A(1/s..) + B(1/s..) + C(s/(s^2 + B^2)) + D(B/(s^2 + B^2)) solve for the coefficients (you'll have to use your answers to solve for the others in some cases,) then invert the transformation using L(cos(Bt)) = s/(s^2 + B^2) L(sin(Bt)) = B/(s^2 + B^2) Look at the answers and see how they compare. (sorry, I just realized I messed up the A and B part in my first post, but even so)"

The "Ae^4t+Be^t" is wrong. It should be something like A/(s-4) + B/(s-1).

She said she'd made a mistake in her first post.

10/25/2005 2:46:52 AM

ok i've got the same values for a and b as he has but they are oppisite sign, and i still can't get d, even though i set s to 0 and played with the signs.

2=a(s-1)((s-3)^2+4)+b(s-4)((s-3)^2+4)=c((s-3)/(s-3)^2+4)+d(2/(s-3)^2+4) that's what i have.....


gettin tired.....

10/25/2005 3:15:24 AM

Yeah... I hink you missed another sign in there. It should be equal to negative two, not positive two... so...

Also I don't think you multiplied through correctly. It should look like this.

-2=a(s-1)((s-3)^2+4)+b(s-4)((s-3)^2+4)+c(s-3)(s-1)(s-4)+d*2*(s-1)(s-4)

10/25/2005 3:21:13 AM

for the third problem of type 2 my equation should look like:2

2s-4=a(s+2)((s-1)^2+16) + b(s+3)((s-1)^2+16) + c*2s(s3)(s+2) - d*4(s+2)(s+3) ??

10/25/2005 3:42:37 AM

More like 2s-6 = A*(S+2)*((S-1)^2 + 16) + B*(S+3)((S-1)^2 + 16) + C*(S-1)*(S+3)*(S+2) + D*4*(s+3)*(S+2)

And yeah, your variables A,B,C,D might be in different spots (I might've just labelled mine differently, just make sure yours have got those numbers in em).

10/25/2005 3:52:02 AM

why is it -6 and not -4 ??

10/25/2005 3:55:18 AM

Because you calculated something wrong down the line.

10/25/2005 3:59:13 AM

found the mistake on that one.....my brain is doing everything it can to quit on me

btw, before you take off thanks for all the help, i really appreciate it.

10/25/2005 4:05:24 AM

np, I think I'm about to take a nap...

Getting a bit fuzzy in the head, making more mistakes than I like...

10/25/2005 4:06:30 AM

yea me 2

10/25/2005 4:07:22 AM

this stuff aint that bad after all

10/25/2005 8:46:08 AM

no suprises. but graphing that stuff takes way too much time.

10/25/2005 11:47:35 AM

Oh yeah, that graphing was some half-assery by me.

I'm so glad he didn't do the type 7 problem with the requirement that we solve with the QDH method instead of the laplace method. I would've been fucked.

I think I did good on that test overall, though.

I feel like shit having stayed up cramming for it though.

10/25/2005 8:09:26 PM

my graphs weren't pretty. I didn't practice enough. I also didn't have time to check over anything, but overall I don't feel too bad about the test. Of course whenever that happens I end up doing poorly, so I'm not getting my hopes up. If he wanted us to do type 7 the QDH way I would've been so fucked as well. Not only would I not know what to do but I wouldn't have enough time to do it.

10/25/2005 8:49:23 PM

I barely was able to graph the last one. Its so messy it probably wont even be counted.

10/25/2005 9:57:11 PM

okay, question on the homework.. figured I'd better do it after class today.

For type 1 problems, how are we supposed to find the A, B values when decomposing the fractions? The y values in it are confusing me and I don't have how in my notes.

Thanks.

11/10/2005 11:20:28 PM

Well, if you have something like -1 = A*(y-1) + B*(y+1) and you want to solve for A and B, just set y equal to a value that cancels out the other variables. So y=1 gets you B, y=-1 gets you A. Just figure out what variable you have to use for certain problems. On some problems, you might just have to solve for which variables you can at first, then plug in an arbitrary value for y (and plug in the values for A, B, C, whatever you have solved for so far) then solve for your other unknown variable.

You know, we had to do that for the problems with the what what and the cosine and sine crap mixed up in the partial fraction decompos.... OK I'm just going to stop talking now, but you get the idea from my first paragraph.

And now for something completely different...

11/11/2005 12:22:12 AM

So do we have to solve for the C in the = t + C part. I know how to get to here but coulda sworn we had to solve for the C, but he only gives up to here. Does anyone know?

11/11/2005 2:05:45 PM

I don't think so since he didn't give any initial conditions.

11/11/2005 11:06:59 PM