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Laplace Transform (Should be easy)

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[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

[tex]\Rightarrow s^2*Y(s)-s*y(0)-y'(0)-s*Y(s)+y(0)-6*Y(s)=0[/tex]

[tex]\Rightarrow s^2*Y(s)-2s+1-s*Y(s)+2-Y(s)=0[/tex]

[tex]\Rightarrow s^2*Y(s)-s*Y(s)-Y(s)=-3+2s[/tex]

[tex]\Rightarow Y(s)=\frac{-3+2s}{s^2-s-1}[/tex]

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
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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893
What happened to the "6" on 6Y(x)?

s2- s- 6 certainly does factor!
 
2,981
2
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 = [tex]\frac{1}{s}[/tex] - [tex]\frac{11}{3s+2}[/tex]

Does this transform to

I(t) = 1-11[tex]e^{-2t}[/tex]

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 ([tex]e^{-2/3t}[/tex])
 
Last edited:
2,981
2
I'm also having a problem with laplace transforms.

I have an equation in the form

I = [tex]\frac{1}{s}[/tex] - [tex]\frac{11}{3s+2}[/tex]

Does this transform to

I(t) = 1-11[tex]e^{-2t}[/tex]

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 ([tex]e^{-2/3t}[/tex])


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

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