Question

Use the Laplace transformation to solve the initial value problem:

Answer 1

\[\huge y''+6y'+18y=0\]

Answer 2

I was able to plug everything in and reduce it down to\[\large Y(s)=\frac{ 4s+13 }{ s^2+6s+18 }\]
(if someone could check this, that'd be awesome)
I don't know where to go from here since the denominator is irreducible.

Answer 3

Oops... initial values..\[y(0)=4, ~~~y'(0)=-11\]

Answer 4

Hmm I got -4s-13 in the numerator.
Sec checking my work...

Answer 5

Oh oh but I have to add them to the right side, derp.
Ok ya looks good so far :)

Answer 6

Everything looks good here

Answer 7

So we have something like...\[\large\rm Y(s)=\frac{4s+13}{(s+3)^2+3^2}\]ya?

Answer 8

I'll bet we can make an (s+3) in the numerator somehow,
and then split this up into (s+3)/(stuff) which turns into cosine, ya?
And another fraction that we have to figure out.

Answer 9

Oh you completed the square. Yeah, that makes sense X)

Answer 10

@zepdrix for the win lol

Answer 11

4s+13 = 4s+12 +1

= 4(s+3) +1

Answer 12

Gotcha, okay still following you!

Answer 13

\[\large\rm Y(s)=\frac{4(s+3)+1}{(s+3)^2+3^2}\]Giving us,\[\large\rm Y(s)=4\frac{(s+3)}{(s+3)^2+3^2}+\frac{1}{(s+3)^2+3^2}\]And that second term, hmmm...
Does that look like anything useful?
I can't remember these things sometimes.
Oh it's close to sine, isn't it?

Answer 14

I mean, it's close to sine, with an exponential shift or whatever (from the s+3)

Answer 15

Isn't it close to e^at?

Answer 16

\[\large\rm Y(s)=4\frac{(s+3)}{(s+3)^2+3^2}+\frac{1}{3^2}\frac{3^2}{(s+3)^2+3^2}\]

Answer 17

Oh wait, e^at doesn't have a^2

Answer 18

Or even an s squared right? :D

Answer 19

True :P Thank you! This helped a ton