# Help on Solving this Eq.

1. Aug 3, 2009

### 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?

2. Aug 3, 2009

### tiny-tim

Welcome to PF!

Hi jrbigfish! Fishy welcome to PF!

(try using the X2 tag just above the Reply box )
use good ol' (-b ± √(b2 - 4ac))/2a …

in this case -1 ± i√3.
we need more fish!

3. Aug 4, 2009

### HallsofIvy

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

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

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

4. Aug 5, 2009

### 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?

5. Aug 5, 2009

### tiny-tim

Do you mean r1,2 = -a ± √(a2 - b) ?

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

I'd forget that formula if I were you.

6. Aug 6, 2009

### jrbigfish

Ok thank you mag.

7. Aug 6, 2009

### 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?

8. Aug 6, 2009

### tiny-tim

(copy the ± and √ symbols)

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

9. Aug 7, 2009

### jrbigfish

Ok -1 ± i√3 then what I do?

10. Aug 7, 2009

### tiny-tim

??

you solve y' = (-1 ± i√3)y.

11. Aug 7, 2009

### jrbigfish

so we end with y=Ce^(1±√3)t ?

12. Aug 7, 2009

### tiny-tim

what happened to i?

13. Aug 10, 2009

### 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?

14. Aug 10, 2009

### tiny-tim

(try using the X2 tag just above the Reply box )
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 … ?

15. Sep 5, 2009

### tiny-tim

Hi jrbigfish! Thanks for the PM.

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