Solving Laplace Transform y''-6y'+13y=0 | Help Needed

[andrewdavid]

I have this laplace transform that I need to solve: y''-6y'+13y=0 y'(0)=2 y(0)=-3


I figured out my Y(s)=(-3s+20)/(s^2-6s+12). All I need to do is take the inverse laplace of this but I can't figure it. I know I need to split it into two fractions, but after that I'm lost. I'd appreciate any help.

 

[Pyrrhus]

Yes Mr Beagss, it's right.

 

[Mr. Beags]

Try doing complete the square on the bottom... see what happens

 

[andrewdavid]

Ok I got -3s/((s-3)^2)+4 + 20/((s-3)^2)+4) after completing the square and splitting up the fraction.

 

[Pyrrhus]

Ok now look at your Laplace Table of inverse and convert them.

For example

e^{at} \sin (bt) = \frac{b}{(s-a)^{2} + b^{2}}

10 \frac{2}{(s-3)^{2} + (2)^{2}} = 10 e^{3t} \sin (2t)

 

[Pyrrhus]

For the other Laplace inverse is:

e^{at} \cos (bt) = \frac{s-a}{(s-a)^{2} + b^{2}}

\frac{-3s}{(s-3)^{2} + 4} = \frac{-3s + 9 - 9}{(s-3)^{2} + 4}

thus

\frac{-3s + 9 - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3) - 9}{(s-3)^{2} + 4}

\frac{-3(s - 3) - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3)}{(s-3)^{2} + 4} + \frac{-9}{(s-3)^{2} + 4}

and finally

\frac{-3(s - 3)}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t)

\frac{-9}{2} \frac{2}{(s-3)^{2} + (2)^{2}} = \frac{-9}{2} e^{3t} \sin (2t)

so for the end Laplace inverse of

\frac{-3s}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t) + \frac{-9}{2} e^{3t} \sin (2t)