Help on Solving this Eq.

[jrbigfish]

I hope my posting is not against the rules.
The equation is not that difficult but I had being out of school for 20 years.

y"+2y'+4y=0 Then I do
r^2+2r+4=0
then you can not do
(r+2)(r+2)=0 because do not work. can I have some help plz?

[tiny-tim]

Hi jrbigfish! Fishy welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
r^2+2r+4=0
then you can not do
(r+2)(r+2)=0 because do not work. can I have some help plz?

use good ol' (-b ± √(b2 - 4ac))/2a …

in this case -1 ± i√3. :wink:

we need more fish! :biggrin:

[HallsofIvy]

Most polynomial equations cannot be factored (with integer coefficients). Complete the square or use the quadratic formula.

Completing the square:
r^2+ 2r+ 4= r^2+ 2r+ 1- 1+ 4= r^2+ 2r+ 1+ 3= (r+1)^2+ 3= 0

Quadratic formula:
r= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}
with a= 1, b= 2, c= 4.

[jrbigfish]

Ok the quadratic formula! Yes! But what is this roots of the characteristic equation are
r1,2=-a+/-(a^2-b)^1/2 what is this and when I use it?

[tiny-tim]

Ok the quadratic formula! Yes! But what is this roots of the characteristic equation are
r1,2=-a+/-(a^2-b)^1/2 what is this and when I use it?

Do you mean r1,2 = -a ± √(a2 - b) ?

hmm :rolleyes: … that looks like the roots of x2 +2ax + b = 0.

I'd forget that formula if I were you.

[jrbigfish]

Ok thank you mag.

[jrbigfish]

Ok after I use the quadratic formula (-2+/-((12))^1/2)/2 which solves 0.732i and -2.73i. So the solution is of the form underdamped. So I use the equation y=e^(∝x) (C_1 cosβx+C_2 sinβx). then what?

[tiny-tim]

Ok after I use the quadratic formula (-2+/-((12))^1/2)/2 which solves 0.732i and -2.73i.

(copy the ± and √ symbols)

Nooo … that's i(-1 ± √3) … you should have got -1 ± i√3. :redface:

[jrbigfish]

Ok -1 ± i√3 then what I do?

[tiny-tim]

Ok -1 ± i√3 then what I do?

?? :confused:

you solve y' = (-1 ± i√3)y.

[jrbigfish]

so we end with y=Ce^(1±√3)t ?

[tiny-tim]

so we end with y=Ce^(1±√3)t ?

what happened to i? :confused:

[jrbigfish]

so we end with y=Ce^(1±i√3)t ?? Help me here it has being 12 years after the B.S. is this correct?

[tiny-tim]

(try using the X2 tag just above the Reply box :wink:)
so we end with y=Ce^(1±i√3)t ??

Sort of, but you won't get many marks if you write it like that.

e(1±i√3)t = ete±i√3t,

so the general solution is … ? :smile:

[tiny-tim]

Hi jrbigfish! Thanks for the PM. :smile:

The general solution will be Aetei√3t + Bete-i√3t …

though it would be more usual (and much easier if the solutions are going to be real anyway) to write it in the form Aetcos√3t + Betsin√3t :wink: