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Mathematics 10 Online
OpenStudy (fibonaccichick666):

ODE question again. So I need to determine for what r values the ode has a solution of the form \(y=t^r\) for t>0. What method yields only \(t^r\)? I know ce^rt, but not one for t and cannot locate it in the book.

OpenStudy (fibonaccichick666):

@ganeshie8 @Marki

OpenStudy (fibonaccichick666):

the actual problem, I can do, but I would use an integrating factor for this problem not whatever method this is

OpenStudy (anonymous):

so for matrices or equations ?

OpenStudy (fibonaccichick666):

it's an equation, specifically:\[y''+4t^{-1}y'+2t^{-2}y=0\]

OpenStudy (fibonaccichick666):

I am assuming there is some method similar to the characteristic eq. then solve for r and have a generic answer

OpenStudy (anonymous):

this is special equation is it of Legendre or Bessel i dont memorize these stuff ;-;

OpenStudy (fibonaccichick666):

no clue, We haven't learned it. This is something for the first week's homework...

OpenStudy (anonymous):

ok i wanna refresh m mind what topics u have for the HW ?

OpenStudy (fibonaccichick666):

uhm, well a lot there is like 50 problems, so first was, convert to canonical form, second verify something is the solution of the de, third determine r values which give an answer of ce^rt, fourth, this stuff with the t^r, fifth all of the numerical methods, sixth translate the dif eq tot he origin, seventh method of successive approximations, eighth sequence convergence(idk i haven't tried those yet)

OpenStudy (anonymous):

ok i'll try something with Bessel

OpenStudy (fibonaccichick666):

ok, and can you link to something, I've never heard of him

OpenStudy (anonymous):

check equation 2 http://mathworld.wolfram.com/BesselDifferentialEquation.html

OpenStudy (anonymous):

interesting @wio

OpenStudy (anonymous):

sowe should convert ur equation to this form \(y''+t^{-1}y'+(1-n^2y^{-2})y=0\)

OpenStudy (anonymous):

I like to guess, because I've forgotten a lot\[ y=t^{r}\\ y' = rt^{r-1}\\y'' = r(r-1)t^{r-2} \]And then: \[ r(r-1)t^{r-2} +(4t^{-1})rt^{r-1} + (2t^{-2})t^r=0 \]

OpenStudy (fibonaccichick666):

@Marki that seems a little complicated, this is the first week... I would need some explanations in english. I'm a little confused by the terminology. @wio cool it looks like a derivation

OpenStudy (anonymous):

ok then forget it :P this is the only way i remember tbh for this equation idk maybe @ganeshie8 could also know

OpenStudy (anonymous):

Hmm....\[ \bigg(r(r-1)+4r+2\bigg)t^{r-2} = 0t^{r-2}\implies r^2+3r+2=0 \]

OpenStudy (fibonaccichick666):

lol it's cool,

OpenStudy (fibonaccichick666):

so I guess to use that we would need an n, and to +1 and -1?

OpenStudy (anonymous):

Now we can verify whatever solution we get.

OpenStudy (fibonaccichick666):

@wio, I like your work, but I am confused a little let me reread it

OpenStudy (fibonaccichick666):

ok, so that is awesome wio! I get it! and I think it will work perfectly!

OpenStudy (anonymous):

What is the method you are supposed to be using?

OpenStudy (fibonaccichick666):

no clue

OpenStudy (fibonaccichick666):

lol that's the issue

OpenStudy (fibonaccichick666):

the only thing similar is the method where you solve the char eq then have a solution of the form ce^rt, but the book doesn't really cover that either

OpenStudy (fibonaccichick666):

this teacher is sort of a discombobulated, barely speaks english mess

OpenStudy (anonymous):

Are you supposed to put it in canonical form?

OpenStudy (fibonaccichick666):

lol I wish, but no not on this one. That would have been much simpler

OpenStudy (anonymous):

Okay, I think that doing what I did is the answer then?

OpenStudy (fibonaccichick666):

if it helps the original eq is this, \[t^2y''+4ty'+2y=0\]

OpenStudy (fibonaccichick666):

I think it's possible

OpenStudy (fibonaccichick666):

I have convinced myself that you are 100% correct wio, thanks, really really great reasoning skills!!!

OpenStudy (anonymous):

ok here is wolfram solution

OpenStudy (anonymous):

but term of x instead of t xD save it u might use it some where xd

OpenStudy (fibonaccichick666):

that's awesome, it is wio's method!

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