Question

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Answer 1

\[\large \lambda^2+4\lambda+5=0\]
Hmm, looks like this one will factor nicely.

Answer 2

Woops I was mistaken :) lol

Answer 3

Time for the Quadratic Formula :3

Answer 4

You will probably end up with complex roots.
Telling us that we'll be dealing with sines and cosines..

I guess the initial conditions kind of hint towards that also though :)

Answer 5

\[\large \lambda = \alpha+\beta i\]Gives us solutions of the form,\[\large y=c_1e^{\alpha x}\sin\left(\beta x\right)+c_2 e^{\alpha x}\cos\left(\beta x\right)\]
Something like that, right?
So we plug all the goodies in, then given initial conditions, we can solve for the missing constants.

Answer 6

I'm coming up with \(\large \lambda=-2\pm i\)

hmm, lemme check again real quick. Maybe I made a mistake.

Answer 7

Ok good :) Yah I forget that sometimes also.

Answer 8

\[\large y=c_1e^{-2x}\sin\left(x\right)+c_2 e^{-2x}\cos\left(x\right)\]

So we have this so far :O Neato

Answer 9

Understand how to use the initial conditions to find the c's? :O

Answer 10

Hmm I didn't get a 2 on my cosine. Lemme try that again though.
Oh oh you factored a 1/2 out of each term. Nice.

Ok I came up with the same thing! :) Cool.
Err wait shouldn't the cosine have a negative? Hmmmm.

Answer 11

Cause I came up with \(\large c_2=-e^{\pi}\).

Answer 12

pshh i doubt that ^^ looks like yer on the right track.
That's a fairly minor mistake, depending on which one of us made it.

Answer 13

regardless of which one* blah i cant speaks :P

Answer 14

When you took the derivative of your general solution, did you absorb a negative into the c_2 maybe?

Answer 15

Ok maybe I messed it up c: lemme check.

\[\large y=c_1e^{-2t}\sin t+c_2 e^{-2t}\cos t\]Derivative gives us,\[\large y'=2c_2e^{-2t}\sin t-2c_1e^{-2t}\cos t\]

Does your derivative look the same?
Remember, derivative of cos t = -sin t !

Answer 16

whut? +_+

Answer 17

I did? It looks like I put -2's in the exponents...

Answer 18

This is our general solution, agreed or no? :D

\[\large y=c_1e^{-2t}\sin t+c_2 e^{-2t}\cos t\]

Answer 19

Because our roots gave us \(\large \alpha=-2\) and \(\large \beta=1\)

Answer 20

Minus the minus 2?? Did I get the formula mixed up maybe? :O

Answer 21

\[\large y=c_1e^{-\alpha x}\sin\left(\beta x\right)+c_2 e^{-\alpha x}\cos\left(\beta x\right)\]So the form should be like this? Hmm that doesn't look right :O I better google that, maybe I'm all mixed up.

Answer 22

Ya look for the formula with the big box around it or something XD hehe

Answer 23

kk c:

Answer 24

ah XD

Answer 25

Hmm weak sauce.. -_-
Ok ok ok I musta screwed up somewhere lol

anyway, good job :3 i shouldn't had drawn that out so long lol