Question

what it the laplace transform of y'' + 6y' + 9y = 0; initial values of y(0) = -1 and y'(0) = 6

Answer 1

yay! come on goku!

Answer 2

ha ha, just a min

Answer 3

haha, no rush

Answer 4

We know that
\[L(y'')=s(sL(y)-y(0))-y'(0)\]
and
\[L(y')=sL(y)-y(0)\]
Let's use them here

Answer 5

I'm down to....

Answer 6

Y(s) = -s/(s+3)^2

Answer 7

\[ y'' + 6y' + 9y = 0\]
Let's take laplace transform both sides
\[L(y'')+6L(y')+9L(y)=0\]
\[(s^2L(y)-sy(0)-y'(0))+6sL(y)-6y(0)+9L(y)=0\]
\[L(y)(s^2+6s+9)+s-6+6=0\]

\[L(y)=\frac{-s}{(s+3)^2}\]
Yeah you're correct:D

Answer 8

so how does all of the equal -e^(-3t) + 3te^(-3t)

Answer 9

You have to take inverse Laplace transform to get that

Answer 10

i know, but how? I need to get a better grasp on the concepts

Answer 11

it's where i get stuck most of the time

Answer 12

Let me try

Answer 13

If we have
\[L(e^{at})=\frac{1}{s-a}\]
and
\[L(te^{at})=\frac{-d}{ds} (\frac{1}{s-a})\]
or
\[L(te^{at})=\frac{1}{(s-a)^2}\]

Answer 14

We have
\[L(y)=\frac{-s}{(s-3)^2}\]

it can be written as
\[L(y)=\frac{-s+3-3}{(s-3)^2}=\frac{-3}{(s-3)^2}-\frac{(s-3)}{(s-3)^2}\]
Do you get this?

Answer 15

damnit... i always forget to add to the numerator...

Answer 16

well, at least make it look alike. do you have time to walk me through a harder one?

Answer 17

Oops, I gotta go now. Sorry
Mention me there like this @IStutts , I'll try to come and help

Answer 18

thank you