Question

Compute the Laplace Transform.
y(0)=1 ; y'(0)=-1

y'' + 5y' + 3y = 0

Answer 1

I evaluated it to
L[y] = -(s+4) / (s^2 + 5s + 3)

I am pretty certain my algebra is correct, I am just not sure how I would evaluate this further.

Answer 2

maybe I did some algebra wrong but I got
\[Y(s)=\frac{s+4}{s^2+5s+3}\]
anyways try completing the square on bottom
\[Y(s)=\frac{s+4}{s^2+5s+(\frac{5}{2})^2+3-(\frac{5}{2})^2} \\ Y(s)=\frac{s+4}{(s+\frac{5}{2})^2-\frac{13}{4}}\]

Answer 3

there is a couple of other tricks
you could use 5/2+3/2=4
and then you will need have to do a fancy multiplication of 1
not incredibly fancy, just a smidgen fancy

Answer 4

ah got it, thank you very much!