# Laplace Transform (Should be easy)

[SOLVED] Laplace Transform (Should be easy)

1. Homework Statement
I know there are other ways to solve this, but I HAVE to use Laplace Transform

y''-y'-6y=0 where y(0)=2 and y'(0)=-1

3. The Attempt at a Solution

Written as Laplace:

y''-y'-6y=0

$$\Rightarrow s^2*Y(s)-s*y(0)-y'(0)-s*Y(s)+y(0)-6*Y(s)=0$$

$$\Rightarrow s^2*Y(s)-2s+1-s*Y(s)+2-Y(s)=0$$

$$\Rightarrow s^2*Y(s)-s*Y(s)-Y(s)=-3+2s$$

$$\Rightarow Y(s)=\frac{-3+2s}{s^2-s-1}$$

Am I missing something? The numerator and denominator don't factor. Is a separation by partial fractions my next step? Or did I miss something that would make it way easier?

Thanks

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HallsofIvy
Homework Helper
What happened to the "6" on 6Y(x)?

s2- s- 6 certainly does factor!

Splendid!! Now too bad the numerator wasn't -3+s then I would be in business.

I'm also having a problem with laplace transforms.

I have an equation in the form

I = $$\frac{1}{s}$$ - $$\frac{11}{3s+2}$$

Does this transform to

I(t) = 1-11$$e^{-2t}$$

or do i need to try and do something with the 3?

EDIT:
I think I just got it, sorry for hi-jacking you're thread.
I(t) = 1 - 11/3 ($$e^{-2/3t}$$)

Last edited:
I'm also having a problem with laplace transforms.

I have an equation in the form

I = $$\frac{1}{s}$$ - $$\frac{11}{3s+2}$$

Does this transform to

I(t) = 1-11$$e^{-2t}$$

or do i need to try and do something with the 3?

EDIT:
I think I just got it, sorry for hi-jacking you're thread.
I(t) = 1 - 11/3 ($$e^{-2/3t}$$)

It's cool. And you got it right it looks like!