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johnny0929
Solve the initial value problem\[u''+6u'+12u=0\]with \(u(0)=1\) and \(u'(0)=0\)
use characteristic equation a^2 + 6a+12 solve for a
this is a second order homogeneous differential equation. do you know the general solution? \[y_p = e^{rt}c_1+e^{rt}c_2\]
@modphysnoob is correct. except he used "a" so your solution would be: \[y_p = e^{at}c_1+e^{-at}c_2\]
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 \[u=e^{-3t}cos\sqrt{3}t+\sqrt{3}e^{-3t}sin\sqrt{3}t\]
then, plug in the initial conditions. take a few derivatives and violê
solve for u(0)=1 take the derivative of your particular solution and plug in u(0)=0
ok. when I factor \(a^2+6a+12=0\) I get \((-i a+\sqrt{3}-3 i) (i a+\sqrt{3}+3 i)\)
that mean it is sinosoid
which means it's \((\sqrt{3}\pm(3+a)i)=0\) how do i make it look like the general solution?
can you help? because I'm so confused
hold on, my computer is apparently running on steam engine
a=-3 (+-) i* sqrt(3) which is \[Ae^{(-3+ i \sqrt{3})t}+Be^{(-3- i \sqrt{3})t}\]
use euler identity and you will get A sin( ) + B cos ()
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?
you are missing t e^(-3t)
oh yeah. ok there's a t.
\[u=e^{−3t}cos\sqrt{3}t+\sqrt{3}e^{−3t}sin\sqrt{3}t\]one more question. how do i determine if this equation is over damped, underdamped, or critically damped?
so sin() and cos(x) just go up and down at t go on
|dw:1362965561866:dw|
while e(-3t) |dw:1362965602529:dw|
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
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 \(a^2+6a+12=0\)
crticle damp goes to zero without oscillating under dampm oscillate