#1

okay all you math buffs, help me out here...

i'm supposed to solve the initial value problem y'' + 2y' + y = x

the homogeneous solution is y_h = c_1*e^-x + c_2*x*e^-x

they get that from the characteristic equation (r^2 + 2r +1), so r_1=r_2= -1

that's all good, i get it, just some background.

now there is also a particular solution

y_p = A + Bx (why is it not Ax + B?)

so the A and B must satisfy

1. (r^2 + 2r +1)(A + Bx) = x or 2. 2B + A + Bx = x

how the hell do they get #2 from #1, i cannot see it, it's driving me nuts!! anyone wanna help?

i'm supposed to solve the initial value problem y'' + 2y' + y = x

the homogeneous solution is y_h = c_1*e^-x + c_2*x*e^-x

they get that from the characteristic equation (r^2 + 2r +1), so r_1=r_2= -1

that's all good, i get it, just some background.

now there is also a particular solution

y_p = A + Bx (why is it not Ax + B?)

so the A and B must satisfy

1. (r^2 + 2r +1)(A + Bx) = x or 2. 2B + A + Bx = x

how the hell do they get #2 from #1, i cannot see it, it's driving me nuts!! anyone wanna help?

#2

bumpitty bumpitty

#3

I took Linear Algebra and hated it,,, sorry

#4

I'm guessing there are very few math majors here. I'm about a year off from learning this stuff so I'm not of much help. If it were 2y' + y = x i'd definately be of help.

#5

Refresh my memory. What is a particular solution again?

EDIT: I'm getting a little flashback. The particular solution is when you are solving for a function/value and not the homogeneous one?

EDIT: I'm getting a little flashback. The particular solution is when you are solving for a function/value and not the homogeneous one?

*Last edited by darkstar2466 at Jun 13, 2008,*

#6

^just that, it's a specific solution of the equation in question with all the constants (c_1 etc.) all worked out and everything. i just don't know how the simplify 1 to get 2

EDIT: just saw your edit, and yes that's correct. worded it better than i could haha.

EDIT: just saw your edit, and yes that's correct. worded it better than i could haha.

#7

There's a few methods you can use to solve for the particular solution. "Variation of Parameters" is the first one I remember.

Look at the method from this site. I'll try to solve it meanwhile. It's been a while, so I might need a couple of minutes...

http://www.sosmath.com/tables/diffeq/diffeq.html

And what the hell are they doing teaching this to you in Linear Algebra? This is clearly for a Diff. Eq. course.

Look at the method from this site. I'll try to solve it meanwhile. It's been a while, so I might need a couple of minutes...

http://www.sosmath.com/tables/diffeq/diffeq.html

And what the hell are they doing teaching this to you in Linear Algebra? This is clearly for a Diff. Eq. course.

#8

There's a few methods you can use to solve for the particular solution. "Variation of Parameters" is the first one I remember.

yeah, there are few methods. but my question is how they get from

(r^2 + 2r + 1)(A+Bx) to 2B + A + Bx = x.

further more they get B to be 2. wtf? lol i've been studying all day, maybe i'm just not seeing clearly

EDIT: and the class is actually called linear analysis, we do both a linear algebra and differential equations.

*Last edited by psychokiller99 at Jun 13, 2008,*

#9

Your question is confusing the hell out of me.

Unfortunately, I did Linear Algebra/Diff Eq. more than a year ago and I don't have my books with me, or I could help you out better.

Unfortunately, I did Linear Algebra/Diff Eq. more than a year ago and I don't have my books with me, or I could help you out better.

#10

okay, well this is where i'm getting the problem from

http://www.calpoly.edu/~dhartig/Pages/244/Handouts/Midterm2BSolns.pdf

maybe you can understand the solution better than i can.

http://www.calpoly.edu/~dhartig/Pages/244/Handouts/Midterm2BSolns.pdf

maybe you can understand the solution better than i can.

#11

You forgot the most important part, the boundary values!!! Explaining this is easy now.

#12

yeah, i left out the intial values cuz i knew how to use them. but you see the part where the take the characteristic equation and multiply it by the particular solution and put it equal to x? then they simplify and get 2B + A + Bx = x.

that's what i don't get. how do they do that? and how do they get B = 1

that's what i don't get. how do they do that? and how do they get B = 1

#13

actually nevermind, i get it now. i just had a moment of clarity, thanks for the help man. btw, anytime there's a math thread, i notice you manage to make your way into it. you a math major or something?

#14

2B + A + Bx = x, so group it like this:

(2B+A) + Bx = 0 + x

The 2B+A corresponds to the constant on the other side, which is 0, and the Bx corresponds to the x, which has a coefficient of 1. So B HAS to have a value of 1, and 2B + A = 0, so A = -2.

EDIT: Whoops, sorry for the belated reply. I'm an Electrical Engineering/Computer Science major. I just haven't had to apply this stuff in a while, so I'm kinda stagnant...

(2B+A) + Bx = 0 + x

The 2B+A corresponds to the constant on the other side, which is 0, and the Bx corresponds to the x, which has a coefficient of 1. So B HAS to have a value of 1, and 2B + A = 0, so A = -2.

EDIT: Whoops, sorry for the belated reply. I'm an Electrical Engineering/Computer Science major. I just haven't had to apply this stuff in a while, so I'm kinda stagnant...

*Last edited by darkstar2466 at Jun 13, 2008,*

#15

2B + A + Bx = x, so group it like this:

(2B+A) + Bx = 0 + x

The 2B+A corresponds to the constant on the other side, which is 0, and the Bx corresponds to the x, which has a coefficient of 1. So B HAS to have a value of 1, and 2B + A = 0, so A = -2.

EDIT: Whoops, sorry for the belated reply. I'm an Electrical Engineering/Computer Science major. I just haven't had to apply this stuff in a while, so I'm kinda stagnant...

cool dude thanks a bunch, and you're double majoring? that's pretty crazy, where you going to school at?

*Last edited by psychokiller99 at Jun 13, 2008,*

#16

Oh shi.... you are from Turlock. My roommate from last year was from Turlock. Chances are, you know him.

<_<

>_>

I'm at UC Davis.

<_<

>_>

I'm at UC Davis.

#17

and what's your roommates name? maybe i do know him, depends on how old he is i guess haha.

#18

Same year as us. His name's Helio.

#19

no ****, helio costa?

#20

The one and only.

#21

that's pretty cool, yeah i went to high school with him. we had quite a few classes together my junior and senior year, pretty cool guy i think. it definitely is a small world haha.

#22

i had an exam on this yesterday! fun times...

#23

Definitely. What a coinkidinks!

I'm out. Good luck on the class man. I hear you guys have a lab for every class at Cal Poly. I would fu

I'm out. Good luck on the class man. I hear you guys have a lab for every class at Cal Poly. I would fu

**cking kill myself if that happened here.**
#24

Definitely. What a coinkidinks!

I'm out. Good luck on the class man. I hear you guys have a lab for every class at Cal Poly. I would fucking kill myself if that happened here.

well not for every class, but a lot of them do have labs, and yeah i do feel like killin myself sometimes. the gay part is that most instructors don't have a lab for linear analysis, but mine does (once a week), f

__uck__in aSShole. anyways, thanks man peace