## anonymous 3 years ago Solve the initial value problem$u''+6u'+12u=0$with $$u(0)=1$$ and $$u'(0)=0$$

1. anonymous

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

2. abb0t

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

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

4. abb0t

5. anonymous

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

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

7. abb0t

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

8. anonymous

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

9. anonymous

that mean it is sinosoid

10. anonymous

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

11. anonymous

can you help? because I'm so confused

12. anonymous

hold on, my computer is apparently running on steam engine

13. anonymous

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

14. anonymous

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

15. anonymous

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. anonymous

you are missing t e^(-3t)

17. anonymous

oh yeah. ok there's a t.

18. anonymous

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

19. anonymous

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

20. anonymous

|dw:1362965561866:dw|

21. anonymous

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

22. anonymous

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. anonymous

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. anonymous

crticle damp goes to zero without oscillating under dampm oscillate

25. anonymous

thank you!