solving a simple DE using the Laplace Transform - am i doing it right?

Hi all. I have been trying to relearn the laplace transform. I want to solve this second order ode

$\displaystyle y^{''}+y^{'}+y=1$

I know this can be easier done using the characteristic equation but I want to use the laplace tansform for learning purposes

here is my solution - is this right? have i done something stupid?

step 1 - $\displaystyle L(y^{''})+L(y^{'})+L(y)=L(1)$

$\displaystyle => S^{2}L(y) - Sy(0) - y^{'}(0) + SL(y)-y(0)+L(y)=L(1)$

*here i am confused. With no initial conditions does the equation become*

$\displaystyle => S^{2}L(y) + SL(y) +L(y)=1/S$

then $\displaystyle L(y) = \frac{1}{S\cdot(S^{2}+S+1)}$

step 2 - I factorise the quadratic and that becomes

$\displaystyle L(y) = \frac{1}{S\cdot(S+(\frac{1}{2}+\frac{\sqrt {3}}{2}))(S+(\frac{1}{2}-\frac{\sqrt{3}}{2}))}$

*i am sure what i did here with partial fractions is wrong*

$\displaystyle L(y) = \frac{1}{S}+\frac{A}{(S+(\frac{1}{2}+\frac{\sqrt {3}}{2}))}+\frac{B}{(S+(\frac{1}{2}-\frac{\sqrt{3}}{2}))}$

step 3 - take the inverse laplace transform

if i take the inverse laplace transform i get the result from the characteristic equation

$\displaystyle Ae^{\frac{1}{2}+\frac{\sqrt {3}}{2}} + Be^{\frac{1}{2}-\frac{\sqrt {3}}{2}} + 1$

are my steps correct?

thanks for any help