ODE - constant coefficients

Printable View

December 27th 2008, 12:49 PM
zokomoko

1 Attachment(s)

ODE - constant coefficients

Hello :-)
We've learned how to solve second order ODE's with constant coefficients by guessing the exponential solution and plugging it into the equation.
I've tried doing the same here, but it seems impossible to guess the roots of the characteristic polynomial. I've used some website to get the roots and they were pretty ugly (if it's correct), so there must be some other way.
much help is appreciated with this (see attachment below).
December 27th 2008, 01:23 PM
Soroban

Hello, zokomoko!

I can get you started . . .

Quote:

$\frac{d\;\!^6y}{dx^t} - 4\frac{d\;\!^5y}{dx^5} + 8\frac{d\;\!^4y}{dx^4} - 8\frac{d\;\!^3y}{dx^3} + 4\frac{d\;\!^2y}{dx^2} \:=\:0$

The characteristic equation is: . $m^6 - 4m^5 + 8m^4 - 8m^3 + 4m \:=\:0$

. . which factors: . $m^2(m^4 - 4m^3 + 8m^2 - 8m + 4) \:=\:0$

. . and factors further: . $m^2(m^2-2m + 2)^2 \:=\:0$

The roots are: . $0,\;0,\;1\pm i,\;1\pm i$

Can you write the solution now?
December 27th 2008, 01:35 PM
Constatine11

Quote:

Originally Posted by zokomoko

Hello :-)
We've learned how to solve second order ODE's with constant coefficients by guessing the exponential solution and plugging it into the equation.
I've tried doing the same here, but it seems impossible to guess the roots of the characteristic polynomial. I've used some website to get the roots and they were pretty ugly (if it's correct), so there must be some other way.
much help is appreciated with this (see attachment below).

The charateristic polynomial is:

$p(\lambda)=\lambda^6-4 \lambda^5+8 \lambda^4-8\lambda^3+4\lambda^2$

which factors:

$p(\lambda)=\lambda^2(\lambda-1+i)^2(\lambda-1-i)^2$

from which you should be able to proceed.

.
December 28th 2008, 12:30 AM
zokomoko

:-)

thank you :-)
I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?
December 28th 2008, 02:50 AM
mr fantastic

Quote:

Originally Posted by zokomoko

thank you :-)
I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?

$= \lambda^2 (\lambda^4 - 4 \lambda^3 + 8 \lambda^2 - 8 \lambda + 4)$

Now note that

$\lambda^4 - 4 \lambda^3 + 8 \lambda^2 - 8 \lambda + 4$

$= (\lambda^4 + 8 \lambda^2 + 4) - 4 \lambda^3 - 8 \lambda$

$= (\lambda^2 + 2)^2 + 4 \lambda^2 - 4 \lambda (\lambda^2 + 2)$

$= (\lambda^2 + 2)^2 + 4 \lambda^2 - 2 \lambda (\lambda^2 + 2) - 2 \lambda (\lambda^2 + 2)$

$= (\lambda^2 + 2)(\lambda^2 + 2 - 2 \lambda) + 4 \lambda^2 - 2 \lambda (\lambda^2 + 2)$

$= (\lambda^2 + 2)(\lambda^2 + 2 - 2 \lambda) - 2 \lambda (-2 \lambda + \lambda^2 + 2)$

$= (\lambda^2 + 2 - 2 \lambda)(- 2 \lambda + \lambda^2 + 2)$

$= (\lambda^2 - 2 \lambda + 2)(\lambda^2 - 2 \lambda + 2)$

$= (\lambda^2 - 2 \lambda + 2)^2$ .

It's times like this that having access to a CAS is ....... helpful.
December 28th 2008, 04:03 AM
Constatine11

Quote:

Originally Posted by zokomoko

thank you :-)
I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?

The quartic obviously has no real roots, and so we know it factors into a pair of monic quadratics neither of which has real roots. Writing out the equations that the coefficients of the quadratics must satisfy quickly allows one to find a solution for the coefficients which gives the factorisation given in the earlier posts.

.
December 28th 2008, 04:25 AM
zokomoko

1 Attachment(s)

thanks

thank you very much

now you have brought it on yourself, I have another question ..
consider the attached ODE, it has non-constant coefficients. We were asked to solve it using variation of parameters, but I couldn't find the solutions to the associated homogenous equation =/
December 28th 2008, 07:18 AM
Danny

Quote:

Originally Posted by zokomoko

thank you very much

now you have brought it on yourself, I have another question ..
consider the attached ODE, it has non-constant coefficients. We were asked to solve it using variation of parameters, but I couldn't find the solutions to the associated homogenous equation =/

The homegenous is

$t^3 x''' + t^2 x'' - 2tx' + 2x = 0.$
December 28th 2008, 07:29 AM
Danny

Quote:

Originally Posted by danny arrigo

The homegenous is

$t^3 x''' + t^2 x'' - 2tx' + 2x = 0.$

Substitute the form $x = t^m$ gives the characteristic equation

$m(m-1)(m-2) + m(m-1) - 2m +2 = 0$ which factors giving

$m = -1, 1, 2$ The complimentary solution is therefore

$x = \frac{c_1}{t} + c_2 t + c_3 t^2$
December 28th 2008, 07:46 AM
zokomoko

thank you :)