Question

I'm trying to solve an IVP using laplace transforms and must specifically use s-axis and t-axis shifting properties, but am running into issues putting the inverse laplace in the right form. Posted below in a moment.

Answer 1

\[y''-6y'+13y=0, \ \ \ y(0)=0, \ y'(0)=-3\]

Answer 2

\[S^2Y(s)-sy(0)-y'(0)-6[sY(s)-y(0)]+13 Y(s)=0\]

Answer 3

CORRECT SO FAR

Answer 4

\[Y(s)[s^2-6s+13]=-3\]

Answer 5

\[Y(s)=\frac {-3}{s^2-6s+13}\]

Answer 6

\[L^{-1} \left\{\vphantom{} \frac{-3}{(s-\sqrt{6})^2+7} \right\} = L^{-1} \left\{\vphantom{} \frac{\sqrt{7}}{(s-\sqrt{6})^2+7} \right\}-L^{-1} \left\{\vphantom{} \frac{?}{(s-\sqrt{6})^2+7} \right\}\]

Answer 7

I guess I'm having trouble figuring out how to non-messily do that in the denominator, putting the inverse laplace in the form necessary to use exponential shifting in combination with the sin(kt) form, to do that.

Answer 8

\[\frac{ 1 }{ s^2-6s+13 }=\frac{ 1 }{ s^2-2.s.3+3^2+4 }=\frac{ 1 }{ (s-3)^2+4 }\]
try this

Answer 9

-3 is just a constant term

Answer 10

getting this?
i will become\[f(t)=-3.L^{-1}\frac{ 1 }{ (s-3)^2+4 }=\frac{ -3 }{ 2 }.L^{-1}\frac{ 2 }{ (s-3)^2+2^2 }\\=\frac{ -3 }{ 2 }.e^{3t}\sin(2t)\]
looks good?

Answer 11

*it will become :P

Answer 12

@Mendicant_Bias

Answer 13

I have no idea what you did with all those periods in the original post, heh.

Answer 14

ok
first step: i made (s^2-6s+13)=(s-3)^2+4
ok?

Answer 15

Oh my god, what the heck was I even thinking!

Answer 16

Yeah, no, I don't even know where I was going with the root six thing, heh, that makes no sense.

Answer 17

yup it was that easy LOL

Answer 18

Alright, gonna try to finish it now. Man. Need to look out for those errors.

Answer 19

yeah :)

Answer 20

Oh, nevermind, you finished it in your original post, but yeah, I got it. Thank you.

Answer 21

at your service :)