anonymous
  • anonymous
Solve the initial value problem\[u''+6u'+12u=0\]with \(u(0)=1\) and \(u'(0)=0\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
use characteristic equation a^2 + 6a+12 solve for a
abb0t
  • abb0t
this is a second order homogeneous differential equation. do you know the general solution? \[y_p = e^{rt}c_1+e^{rt}c_2\]
abb0t
  • abb0t
@modphysnoob is correct. except he used "a" so your solution would be: \[y_p = e^{at}c_1+e^{-at}c_2\]

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abb0t
  • abb0t
solve the quadratic
anonymous
  • 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 \[u=e^{-3t}cos\sqrt{3}t+\sqrt{3}e^{-3t}sin\sqrt{3}t\]
abb0t
  • abb0t
then, plug in the initial conditions. take a few derivatives and violê
abb0t
  • abb0t
solve for u(0)=1 take the derivative of your particular solution and plug in u(0)=0
anonymous
  • anonymous
ok. when I factor \(a^2+6a+12=0\) I get \((-i a+\sqrt{3}-3 i) (i a+\sqrt{3}+3 i)\)
anonymous
  • anonymous
that mean it is sinosoid
anonymous
  • anonymous
which means it's \((\sqrt{3}\pm(3+a)i)=0\) how do i make it look like the general solution?
anonymous
  • anonymous
can you help? because I'm so confused
anonymous
  • anonymous
hold on, my computer is apparently running on steam engine
anonymous
  • anonymous
a=-3 (+-) i* sqrt(3) which is \[Ae^{(-3+ i \sqrt{3})t}+Be^{(-3- i \sqrt{3})t}\]
anonymous
  • anonymous
use euler identity and you will get A sin( ) + B cos ()
anonymous
  • 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?
anonymous
  • anonymous
you are missing t e^(-3t)
anonymous
  • anonymous
oh yeah. ok there's a t.
anonymous
  • 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?
anonymous
  • anonymous
so sin() and cos(x) just go up and down at t go on
anonymous
  • anonymous
|dw:1362965561866:dw|
anonymous
  • anonymous
while e(-3t) |dw:1362965602529:dw|
anonymous
  • 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
anonymous
  • 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 \(a^2+6a+12=0\)
anonymous
  • anonymous
crticle damp goes to zero without oscillating under dampm oscillate
anonymous
  • anonymous
thank you!

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