Question

Solve the initial value problem\[u''+6u'+12u=0\]with \(u(0)=1\) and \(u'(0)=0\)

Answer 1

use characteristic equation

a^2 + 6a+12

solve for a

Answer 2

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

Answer 3

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

Answer 4

solve the quadratic

Answer 5

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 \[u=e^{-3t}cos\sqrt{3}t+\sqrt{3}e^{-3t}sin\sqrt{3}t\]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

that mean it is sinosoid

Answer 10

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

Answer 11

can you help? because I'm so confused

Answer 12

hold on, my computer is apparently running on steam engine

Answer 13

a=-3 (+-) i* sqrt(3)


which is

\[Ae^{(-3+ i \sqrt{3})t}+Be^{(-3- i \sqrt{3})t}\]

Answer 14

use euler identity and you will get

A sin( ) + B cos ()

Answer 15

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?

Answer 16

you are missing t

e^(-3t)

Answer 17

oh yeah. ok there's a t.

Answer 18

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

Answer 19

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