1. ## Laplace Transform Question

Given g(t)= 4t for 0<t<1 and 4 for t>1 ;

Also given that L(g(t))= 4s^-2(1-e^-s)

Using this result solve

y''-4y=g(t) for y(0)=y'(0)=0

I first started it off like this

Let F(s)= y(t)

then,

s^2F(s)-4F(s)=4s^-2(1-e^-s)

F(s)=4s^-2(1-e^-s) / (s^2 -4)

I'm just having trouble separting the right hand side before finding it's inverse.

Thanks

2. Result (with which I agree):

$\displaystyle F(s)=\dfrac{4(1-e^{-s})}{s^{2}(s^{2}-4)}.$

Well, you need to break apart things. For example, you need to do partial fractions on the fraction

$\displaystyle \dfrac{4}{s^{2}(s-2)(s+2)}.$

Then you can write the LT as

$\displaystyle F(s)=\dfrac{4}{s^{2}(s-2)(s+2)}-\dfrac{4e^{-s}}{s^{2}(s-2)(s+2)}.$

The exponential you can handle using a shifting theorem.

Does this make sense?

3. Yeah, that makes perfect sense. I did that and then I separted that too.. It would of been alot easier if you didn't expand the 1-e^-s..

I made 1/s^2(s^-4) = a/s + b/s^2 + c/s-2 + d/s+2 ...

Is this correct ??

4. It would of been alot easier if you didn't expand the 1-e^-s...
Not that much easier. The work you do in separating out the fraction using partial fractions can apply to both terms in the LT. You'll just multiply one of the functions by an appropriate factor to take into account the exponential.

Your expansion of the fraction will work fine. What do you get for a, b, c, d?