Y"-6y'+9y=2 with y(0)=y'(0)=0

solve using Laplace transforms only.
we are not supposed to use partial fractions (if applicable)

Have a look at this example:
http://www.sosmath.com/diffeq/laplace/ap…

Let's go through the procedure for this problem:

First, apply the Laplace Transform:
L(y") - 6L(y') + 9L(y) = L(2)

Noting that L(y") = s²L(y) - sy(0) - y'(0), and since y(0) = y'(0) = 0, this reduces to
L(y'') = s²L(y)

Also L(y') = sL(y) - y(0), which is just equal to sLy) for the same reason,

and L(2) = 2/s.

So that gives s²L(y) - 6sL(y) + 9L(y) = 2/s, or

L(y)[s² - 6s + 9] = 2/s, or

L(y) = (2/s)(1/[s-3]²).

I know you're probably not meant to use http://www.eecircle.com/applets/007/ILap… ,

but it gives an inverse Laplace Transform of 2/9 + 2t/3 e^(3t) - 2/9 e^(3t).

You can easily solve as partial fractions and it appears you need to. Resolving

2/s(s-3)² as partial fractions gives A/s + B/(s-3) + C/(s-3)² with C = -2, A = 2/9 and B = -2/9.

So the A part is (2/9)s and the inverse LT of this part is 2/9.

The B part is -2/9(s-3) = -2/9(1/[s-3]) and looking up a table gives an inverse LT of -2/9 e^(3t)

The C part is -2/(s-3)² = -2(1/[s-3]²) and taking the inverse LT on a table gives -2t/3 e^(3t)