Question

find the general solution

\[(4 \text D^4 - 8\text D^3 - 7\text D^2 + 11 \text D + 6) y = 0\]

Answer 1

\[x=-1,x=-1/2,x=3/2,x=2?\]

Answer 2

wow fourth order DE...

Answer 3

uhmm could you explain @gracemere

Answer 4

Rational root theorem says \( \pm \frac{6,3,2,1}{4,2,1} \) are only possible roots for \( 4D^4-8D^3-7D^2+11D+6=0\)

Answer 5

rational root?

Answer 6

only possible rational roots***

Answer 7

\[4m^4-8m^3-7m^2+6m=0\]

Answer 8

they're using auxilaries to solve it, @mukushla

Answer 9

remind me what rational root theorem is @mukushla

Answer 10

@lgbasallote i mean*

Answer 11

@lgbasallote , the rational roots theorem is a theorem applying to polynomial factorization and root finding. You use it to tell if a given polynomial expression has rational roots, and which rational roots it might be.

Answer 12

Useful here because we are trying to factor this quartic polynomial into something more managable. Trust me, it's better than solving the quartic with a formula. That's just plain nasty.

Answer 13

isnt there any other way besides rational root theorem? i dont want to rely on a method i can barely remember

Answer 14

Well igbiw, there really isn't. You should know the rational roots theorem.

Answer 15

There is another way, which is solving that quartic polynomial. However, I can tell you that it will take you at least ten times as long to do that then t just learn the rational roots theorem and apply it.

Answer 16

so this needs to be factored out right? to get m

Answer 17

to get your m's yes

Answer 18

and there are 2 positive roots here right..

Answer 19

Rational roots theorem is the way to go for this job. Only when you have determined that the roots are not rational do you try to do something like a cubic or quartic formula. Even then, you will usually be able to take out one root and make the job a whole lot easier.

Answer 20

\[y=k _{1}e^{-x/2}+k_{2}e^{3x/2}+k_{3}e^{-x}+k_{4}e^{2x}\]
4(x-2)(x+1)(x-3/2)(x+1/2)

Answer 21

then 3 negative roots

Answer 22

@Herp_Derp what the heck just happened o.O

Answer 23

What do you mean? I solved it.

Answer 24

i dont see a solution anywhere

Answer 25

he solved it lgba

Answer 26

if you could tell me how you came up with that with no solution i would be glad to hear it. i believe it can aid me a lot

Answer 27

lgba, are you taking DE right now or trying to learn it?

Answer 28

taking it

Answer 29

by yourself

Answer 30

so do you know auxilary equations and how they work.

Answer 31

depends what does that mean?

Answer 32

im not good in remembering fancy words

Answer 33

when you switch to m and solve for the roots

Answer 34

ahh yes i am familiar

Answer 35

Well, observe that x=2 is a solution to 4x^4-8x^3-7x^2+11x+6 (For example, look at its graph, conjecture that 2 is a solution, then prove it). Then use polynomial long division to divide that out.

Answer 36

Then repeat, and you have a quadratic.

Answer 37

howd you know x = 2 would be a solution?

Answer 38

I told you, I looked at a graph. (Calculator). It's a perfectly legit method, as long as you plug it back in to prove that it's a zero.

Answer 39

well we are not allowed to use a calculator so i need another way to find the roots

Answer 40

Rational zeroes test is the only way

Answer 41

rational zero test? what's that?

Answer 42

rational roots theorem igbiw. just learn it.

Answer 43

it's a way of finding all possible roots

Answer 44

and then you use synthetic division to check each possible

Answer 45

it's required material for algebra two. and correction, @Outkast3r09 , it is not a way. It is a way of finding all the *rational* roots

Answer 46

All Possible, not every single way is a possible root

Answer 47

well i learned rational theorem long ago but i dont recall it anymore

Answer 48

Wrong. the root could be irrational. In that case, you would need incredible insight or the quartic/cubic/quadratic/linear equation depending on how many roots you already know.

Answer 49

can you demonstrate the rational theorem in this problem?

Answer 50

yes but i'm saying that the rational roots theorem will find all POSSIBLE rational roots of the equation

Answer 51

i'm not saying that there isn't a complex root within the equation

Answer 52

*ahem* someone asking for help here...

Answer 53

Yes, that's what I meant outkast. I thought you had implied that it would give you all the solutions (including the irrational ones). Nevermind then.

Answer 54

Igbiw, this is algebra two...

Answer 55

http://en.wikipedia.org/wiki/Rational_root_theorem

Answer 56

your point being?

Answer 57

Well, I forgot it too. xD

Answer 58

i dont need to see a wiki page -_- just show me a demonstration and i'll be able to get it

Answer 59

The wikipage SHOWS you a demonstration.

Answer 60

wiki loves overcomplicating things...

Answer 61

Besides, if you are going to be solving DE's like this one, you will need to know this theorem well. You aren't going to cut it just skimming the material, because you will need to use it a lot.

Answer 62

http://www.purplemath.com/modules/rtnlroot.htm

Answer 63

exactly why i need to see a demo :)

Answer 64

There. Two demos.

Answer 65

they say he who keeps posting links is he who knows not the answer lol

Answer 66

Btw, I don't even know how to solve DEs. But I do know how to solve polynomials. hahahaha

Answer 67

ill read the purplemath thing

Answer 68

Igbiw, you need to learn the material well. It's not a good idea getting someone to explain it real-time, then just consulting sources.

Answer 69

*than

Answer 70

just one question though...is it possible to find zeroes using derivatives or something?

Answer 71

Well, I'm pretty sure you can use calculus to aid yourself, but I'm more sure that it won't make it easier, seeing as I have yet to find calculus aided solutions of the quartic. You can find ones for cubics that have one real solution, as well as quadratics.

Answer 72

No, in general Newton's method would not converge. You would have to guess so well that you might as well have already known the solution.

Answer 73

Higher order polynomials have been proved unsolvable with a single formula involving a finite amount of radicals apparently. if you see a quintic or above, you are better off approximating or using the rational roots theorem.

Answer 74

maybe take the derivative until there is only a quadratic equation

Answer 75

Igbiw, how would that help you?

Answer 76

i do not know

Answer 77

And herp, you sure the newton-raphson method doesn't converge? Whatabout the picard?

Answer 78

uhmm okay i guess i'll just study rational roots theorem =_=

Answer 79

Yup. It's much faster. ^^ also note that approximation solutions are not really necessary usually if you have determined that it has a rational root.

Answer 80

@inkyvoyd the Newton method *could* converge, if x0 was close enough, but in general it would not.

I am not familiar with how Picard's method is used to approximate roots of polynomials. Perhaps you are talking about a different method than I am thinking:
http://www.math.tamu.edu/~mpilant/math308H/Projects/PicardMethod.pdf

Answer 81

Hm. I got the method from my textbook, which I would consult, but it is late and I have to go to bed. Uhh, you might have it, it's "calculus and analytic geometry" http://www.amazon.com/Calculus-Analytic-Geometry-9th-Edition/dp/0201531747 . I think I have like edition 6 on paper, and I have the 9th on ebook. Note that I'm actually still learning most of the material. But yeah.