Question

linear differential equation question:
y"'-6y"+12y' -8y = 0

Answer 1

i need to find the general equation

Answer 2

find the roots of its associated quartic equation

Answer 3

y"'-6y"+12y' -8y

or is that a triple? which would correspond to a cubic

Answer 4

r^3 -6r^2+12r -8 = 0

or ... may tryout a power solution since i think those are pretty awesome :)

Answer 5

its r^3-^r2+12r-8 = 0 is correct but what do i do from here?

Answer 6

determine the values of r that produce zeros for the function. after quadritic forms things can get complicated

Answer 7

we are in luck tho, this is really a perfect cube

Answer 8

r = 2,2,2

with a repetition of roots, the solutions i believe are modified with an extra x

Answer 9

so using quadratic formula we can find it?

Answer 10

to make the solutions linearly independant, we modify each :\[y=c_1e^{2x}+c_2~xe^{2x}+c_3~x^2e^{2x}\]

Answer 11

no, qubics cant be found by using the quadratic formula

Answer 12

or cubics even lol

Answer 13

r^3 -6r^2+12r -8 = 0

if it has rational roots, which would be nice, the rational roots thrm says that they must be in the form of the factors of the last term divided by factors of the first term

since the first term is a 1; that leaves us with potential roots of: 1,2,4,8

Answer 14

to see if one of these options works out, synthetic division is a nice thing to know

r^3 -6r^2+12r -8
0 2 -8 8
----------------
2 ) 1 -4 4 0 ; since we have a remainder of zero, 2 is a root
and the equation reduces to a quadratic

r^2 -4r + 4 = 0 ; which is a perfect square to begin with of the form
(r-2)^2

so as it turns out, the whole thing was a perfect cube of the form: (r-2)^3 whose roots are r=2

Answer 15

so bascially since it is not factorable we try using the synthetic division to try out? and get the final general solution?

Answer 16

basically, yes :) which why an arsenal of algebraic techniques is useful to have at your disposal. For the most part, Its not the calculus that gets you in these things - its working thru the accursed algebra of it all ;)

Answer 17

oh geez but they are many options how do i know which one to choose =(

Answer 18

its not really about choosing one, it about choosing many of them and whittling them down to the one the actually works. Dont be afraid if your first attempt steers down the wrong way; you can always turn around and try another method

Answer 19

okAY THANK you so much~

Answer 20

can you name few that are mostly used when finding this kind of question i know that quadratic formula is one of them but not really sure about the rest

Answer 21

sum and difference of squares and cubes; perfect squares and cubes can be helpful as well as this one demonstrates

Answer 22

if you cardanos method for cubics ... which hardly anyone ive seen actually does since its a pain

Answer 23

okay thank you i will try few more and ask again thanks =)

Answer 24

okay i get it but how did we get for the answer (cl+clx+c3x^2)e^2x??

Answer 25

we require 3 linearly independant solutions since this is a third level derivative. The problem is we only obtain a single root for the e^part. to rectify this issue we adjust each solution by a factor of x to keep it linearly independant from one another

Answer 26

i see so x is more like trying to balance everything out in order to make it linearly independent?

Answer 27

yes.

if we just say that the 3 solutions are c e^2x that only accounts for a "single" family set. the adjustment by a factor of x ensures us that the basic structure still works but that we cant create a similar solution from the 3 that we need

Answer 28

theres prolly a more stringent reason as to why, but i dont recall that details of it :)

Answer 29

i see wow thanks but how do i know exactly when to apply x? do i just apply x on each questions ?

Answer 30

if you see that you can only create one family set for a solution; and you require more; then you adjust the others with an additional product of x each time

Answer 31

\[y=c_1e^{2x}\]is one family of solution that satisfies this setup; but since its a third derivative setup, its has 3 linearly independant solutions to be found

Answer 32

\[y=c_1e^{2x}; y=c_2~xe^{2x}; y=c_3~x^2e^{2x}\]

Answer 33

http://college.cengage.com/mathematics/larson/calculus_early/3e/shared/chapter15/clc7eap1503.pdf

that seems to be an easy read that should better explain it

Answer 34

no i perfectly understood what you said THANK you so much for helping me understand =)

Answer 35

youre welcome :) and good luck

Answer 36

one last question i did the same similar question but then i got r = \[\pm1\]
would the answer be \[y = c1e^1x + c2e^-1x + c3xe^1x + c4x^2e^-1x \]

Answer 37

its the same thing i did above the orginal equation is y"'-y"-3y'+2y = 0

Answer 38

since e^x and e^-x are different functions, they they are already linearly independant; they stand apart from each other and can be rightfully called separate function in their own right;

so as long as you only need 2 family solutions and not any more, then there would be no need to produce any extra adjustments

Answer 39

\[y=c_1e^x+c_2e^{-x}\]

Answer 40

oh i see thank you so much after reading through the link you provided it really cleared everything out now do you know stuff about undetermined coefficients?

Answer 41

the method of undetermined coeffs i believe deals with finding general and specific solutions to nonhomogenous diffyQs; there is the long hand version (works it all out by hand) and the short hand version (Wronskian).

You determine the homogenous solutions and use those to define functions that are then worked out to find solutions to the nonhomogenous stuff