## johnny0929 Group Title Solve the initial value problem$u''+6u'+12u=0$with $$u(0)=1$$ and $$u'(0)=0$$ one year ago one year ago

1. modphysnoob Group Title

use characteristic equation a^2 + 6a+12 solve for a

2. abb0t Group Title

this is a second order homogeneous differential equation. do you know the general solution? $y_p = e^{rt}c_1+e^{rt}c_2$

3. abb0t Group Title

@modphysnoob is correct. except he used "a" so your solution would be: $y_p = e^{at}c_1+e^{-at}c_2$

4. abb0t Group Title

5. johnny0929 Group Title

I actually have the answer but I am just a little confused about how to do it... It's a spring mass problem and the answer is $u=e^{-3t}cos\sqrt{3}t+\sqrt{3}e^{-3t}sin\sqrt{3}t$

6. abb0t Group Title

then, plug in the initial conditions. take a few derivatives and violê

7. abb0t Group Title

solve for u(0)=1 take the derivative of your particular solution and plug in u(0)=0

8. johnny0929 Group Title

ok. when I factor $$a^2+6a+12=0$$ I get $$(-i a+\sqrt{3}-3 i) (i a+\sqrt{3}+3 i)$$

9. modphysnoob Group Title

that mean it is sinosoid

10. johnny0929 Group Title

which means it's $$(\sqrt{3}\pm(3+a)i)=0$$ how do i make it look like the general solution?

11. johnny0929 Group Title

can you help? because I'm so confused

12. modphysnoob Group Title

hold on, my computer is apparently running on steam engine

13. modphysnoob Group Title

a=-3 (+-) i* sqrt(3) which is $Ae^{(-3+ i \sqrt{3})t}+Be^{(-3- i \sqrt{3})t}$

14. modphysnoob Group Title

use euler identity and you will get A sin( ) + B cos ()

15. johnny0929 Group Title

now i can see it's $Ae^{-3}sin(\sqrt{3}t)+Be^{-3}cos(\sqrt{3}t)$then all i have to do is plug in the initial values, right?

16. modphysnoob Group Title

you are missing t e^(-3t)

17. johnny0929 Group Title

oh yeah. ok there's a t.

18. johnny0929 Group Title

$u=e^{−3t}cos\sqrt{3}t+\sqrt{3}e^{−3t}sin\sqrt{3}t$one more question. how do i determine if this equation is over damped, underdamped, or critically damped?

19. modphysnoob Group Title

so sin() and cos(x) just go up and down at t go on

20. modphysnoob Group Title

|dw:1362965561866:dw|

21. modphysnoob Group Title

while e(-3t) |dw:1362965602529:dw|

22. modphysnoob Group Title

just read it http://en.wikipedia.org/wiki/Damping#Critical_damping_.28.CE.B6_.3D_1.29 it will explain much better than I ever can

23. johnny0929 Group Title

oh ok i am reading from my book that all i need to do is compare the coefficient in front of $$u'$$ to $$\sqrt{4ac}$$ from the equation $$a^2+6a+12=0$$

24. modphysnoob Group Title

crticle damp goes to zero without oscillating under dampm oscillate

25. johnny0929 Group Title

thank you!